problem_id stringlengths 3 7 | contestId stringclasses 660
values | problem_index stringclasses 27
values | programmingLanguage stringclasses 3
values | testset stringclasses 5
values | incorrect_passedTestCount float64 0 146 | incorrect_timeConsumedMillis float64 15 4.26k | incorrect_memoryConsumedBytes float64 0 271M | incorrect_submission_id stringlengths 7 9 | incorrect_source stringlengths 10 27.7k | correct_passedTestCount float64 2 360 | correct_timeConsumedMillis int64 30 8.06k | correct_memoryConsumedBytes int64 0 475M | correct_submission_id stringlengths 7 9 | correct_source stringlengths 28 21.2k | contest_name stringclasses 664
values | contest_type stringclasses 3
values | contest_start_year int64 2.01k 2.02k | time_limit float64 0.5 15 | memory_limit float64 64 1.02k | title stringlengths 2 54 | description stringlengths 35 3.16k | input_format stringlengths 67 1.76k | output_format stringlengths 18 1.06k ⌀ | interaction_format null | note stringclasses 840
values | examples stringlengths 34 1.16k | rating int64 800 3.4k ⌀ | tags stringclasses 533
values | testset_size int64 2 360 | official_tests stringlengths 44 19.7M | official_tests_complete bool 1
class | input_mode stringclasses 1
value | generated_checker stringclasses 231
values | executable bool 1
class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
113/A | 113 | A | Python 3 | TESTS | 136 | 92 | 102,400 | 212181856 | a=input().split()
i=0
n=len(a)
while i<n and (len(a[i])>=4 and a[i][len(a[i])-4:]=="lios"):
i+=1
i1=i
while i<n and (len(a[i])>=3 and a[i][len(a[i])-3:]=="etr"):
i+=1
if i-i1!=1 and i!=0 and n>1:
print("NO")
exit()
while i<n and (len(a[i])>=6 and a[i][len(a[i])-6:]=="initis"):
i+=1
#print(i)
if i==0... | 360 | 78 | 307,200 | 4075975 | s=input().split()
if(len(s)==1):
if(s[0].endswith("lios") or s[0].endswith("etr") or s[0].endswith("liala") or s[0].endswith("etra") or s[0].endswith("inites") or s[0].endswith("initis")):
print("YES")
else:
print("NO")
elif(s[0].endswith("lios") or s[0].endswith("etr")):
n=len(s)
i=... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | Python 3 | TESTS | 65 | 92 | 102,400 | 151378536 | def isPetyaLanguage():
read = input().split()
if len(read) == 1:
if read[0].endswith('lios') or read[0].endswith('lialia') or read[0].endswith('etr') or read[0].endswith('etra') or read[0].endswith('initis') or read[0].endswith('inites'):
print('YES')
return
else:
... | 360 | 92 | 102,400 | 151322401 | def isPetyaLanguage():
words = input().split()
if len(words) == 1:
if words[0].endswith("etr") or words[0].endswith("etra") or words[0].endswith("lios") or words[0].endswith("liala") or words[0].endswith("initis") or words[0].endswith("inites"):
print("YES")
return
else:
... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | PyPy 3-64 | TESTS | 146 | 156 | 716,800 | 171980063 | from collections import defaultdict as dc
words=input().split()
ends=['lios','liala','etr','etra','initis','inites']
adjm,adjf,verbm,verbf,nounm,nounf,=set(),set(),set(),set(),set(),set()
for s in words:
n=len(s)
for i in ends:
idx=len(s)-len(i)
if idx<0:continue
j,flag=0,0
while... | 360 | 92 | 102,400 | 214687216 | from sys import stdin, stdout
x = stdin.readline().replace("\n","")
lst = x.split(' ')
genderLst = []
typeLst = []
valid = 1
for x in lst:
if x.endswith("lios") or x.endswith("etr") or x.endswith("initis"):
genderLst.append(1)
elif x.endswith("liala") or x.endswith("etra") or x.endswith("inites"... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | PyPy 3-64 | TESTS | 101 | 124 | 204,800 | 212179196 | def get_type(slovo):
slovo = slovo[::-1]
slovo += '#######'
if slovo[:4] == 'soil':
return [0, 0]
if slovo[:5] == 'alail':
return [0, 1]
if slovo[:3] == 'rte':
return [1, 0]
if slovo[:4] == 'arte':
return [1, 1]
if slovo[:6] == 'sitini':
return [2, 0]
... | 360 | 92 | 102,400 | 227284427 | # LUOGU_RID: 128399015
a = input().split()
c = 5 #设置一个变量，记录当前的状态
for b in a:
if c == 5:
if len(a) == 1:
if b[-4:] == 'lios' or b[-5:] == 'liala' or b[-3:] == 'etr' or b[-4:] == 'etra' or b[-6:] == 'initis' or b[-6:] == 'inites': #输入只有一个任意合法单词时也特判为合法
c = 0
elif b[-4:] == ... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | PyPy 3-64 | TESTS | 7 | 124 | 614,400 | 171980252 | from collections import defaultdict as dc
words=input().split()
ends=['lios','liala','etr','etra','initis','inites']
adjm,adjf,verbm,verbf,nounm,nounf,=set(),set(),set(),set(),set(),set()
frq=dc(lambda:0)
for s in words:
n=len(s)
for i in ends:
idx=len(s)-len(i)
if idx<0:continue
j,flag=... | 360 | 122 | 4,710,400 | 157359820 | n = input().split()
w = 0
x=y=z=0
r=s=t=0
while w<len(n):
if n[w][-4:]=="lios" or n[w][-5:]=="liala":
if n[w][-4:]=="lios":
x+=1
w+=1
else:
r+=1
w+=1
else:
break
while w<len(n):
if n[w][-3:]=="etr" or n[w][-4:]=="etra":
if n[w][... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | PyPy 3 | TESTS | 7 | 184 | 0 | 11937161 | def pos(w):
if len(w) < 3:
return False
if w == 'etr' or w[-3:] == 'etr':
return 'MN'
if len(w) == 3:
return False
if w == 'lios' or w[-4:] == 'lios':
return 'MA'
if w == 'etra' or w[-4:] == 'etra':
return 'FN'
if len(w) == 4:
return False
if w... | 360 | 124 | 102,400 | 151380136 | '''
def isPetyaLanguage():
read = input().split(' ')
if len(read) == 1:
if read[0].endswith('lios') or read[0].endswith('lialia') or read[0].endswith('etr') or read[0].endswith('etra') or read[0].endswith('initis') or read[0].endswith('inites'):
print('YES')
return
else:
... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | Python 3 | TESTS | 7 | 186 | 307,200 | 91305797 | s=list(map(str,input().split(' ')))
g=[['lios','liala'],['etr','etra'],['initis','inites']]
if len(s)==1:
n=len(s[0])
if s[0][n-len(g[0][0]):]==g[0][0] or s[0][n-len(g[0][1]):]==g[0][1] or s[0][n-len(g[1][0]):]==g[1][0] or s[0][n-len(g[1][1]):]==g[1][1] or s[0][n-len(g[2][0]):]==g[2][0] or s[0][n-len(g[2][1]):]... | 360 | 124 | 204,800 | 208405599 | class Context:
def __init__(self, words):
self.words = words
self.index = 0
def get_token(self):
return self.words[self.index]
def next(self):
self.index += 1
def end(self):
return self.index == len(self.words)
def dump(self):
return self.index
... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | Python 3 | TESTS | 66 | 156 | 204,800 | 22654372 | ms = input().split()
al = [['lios','etr','initis'],['liala','etra','inites']]
s = ms[0]
i=0;
ans = True;
if s.endswith(al[0][0]) or s.endswith(al[0][1]):
fs = 0;
elif s.endswith(al[1][0]) or s.endswith(al[1][1]):
fs = 1;
else:
ans=False;
if ans==True:
for s in ms:
if i==0:
if s... | 360 | 124 | 614,400 | 162396833 | import re;t=input();p=[r'([^ ]*lios )*([^ ]*etr)( [^ ]*initis)*',r'([^ ]*liala )*([^ ]*etra)( [^ ]*inites)*',r'[^ ]*(li(os|ala)|etra?|init[ie]s)'];print(['NO','YES'][any(re.fullmatch(q,t)for q in p)]) | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | Python 3 | TESTS | 66 | 216 | 8,396,800 | 127696390 | import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect, ... | 360 | 154 | 204,800 | 15182995 | s, n, m, f = input().split(), False, False, False
def cc(w, me, fe):
global m, f
if w.endswith(me):
m = True
return True
elif w.endswith(fe):
f = True
return True
else:
return False
def ad(w): return cc(w, 'lios', 'liala')
def nn(w): return cc(w, 'etr', 'etra')
de... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
113/A | 113 | A | Python 3 | TESTS | 66 | 218 | 614,400 | 99499790 | lista = input().split()
passadj = False
passsub = False
gender = -1
can = True
# verificando os generos das palavras
for i in range(0,len(lista)):
if (len(lista[i]) >= 3 and lista[i][len(lista[i])-3::] == "etr") or (len(lista[i]) >= 4 and lista[i][len(lista[i])-4::] == "lios") or (len(lista[i]) >= 6 and lista[i][le... | 360 | 216 | 716,800 | 15383496 | #!/usr/bin/python3
import re
al = re.compile(r'^1*23*$')
def getType(word):
if word.endswith("lios"): return 1
elif word.endswith("liala"): return -1
elif word.endswith("etr"): return 2
elif word.endswith("etra"): return -2
elif word.endswith("initis"):return 3
elif word.endswith("inites"): return -3
else: retur... | Codeforces Beta Round 86 (Div. 1 Only) | CF | 2,011 | 5 | 256 | Grammar Lessons | Petya got interested in grammar on his third year in school. He invented his own language called Petya's. Petya wanted to create a maximally simple language that would be enough to chat with friends, that's why all the language's grammar can be described with the following set of rules:
- There are three parts of spee... | The first line contains one or more words consisting of lowercase Latin letters. The overall number of characters (including letters and spaces) does not exceed 105.
It is guaranteed that any two consecutive words are separated by exactly one space and the input data do not contain any other spaces. It is possible tha... | If some word of the given text does not belong to the Petya's language or if the text contains more that one sentence, print "NO" (without the quotes). Otherwise, print "YES" (without the quotes). | null | null | [{"input": "petr", "output": "YES"}, {"input": "etis atis animatis etis atis amatis", "output": "NO"}, {"input": "nataliala kataliala vetra feinites", "output": "YES"}] | 1,600 | ["implementation", "strings"] | 360 | [{"input": "petr\r\n", "output": "YES\r\n"}, {"input": "etis atis animatis etis atis amatis\r\n", "output": "NO\r\n"}, {"input": "nataliala kataliala vetra feinites\r\n", "output": "YES\r\n"}, {"input": "qweasbvflios\r\n", "output": "YES\r\n"}, {"input": "lios lios petr initis qwe\r\n", "output": "NO\r\n"}, {"input": "... | false | stdio | null | true |
818/C | 818 | C | Python 3 | TESTS | 16 | 763 | 26,828,800 | 28155256 | from sys import stdin, stdout
k = int(stdin.readline())
n, m = map(int, stdin.readline().split())
left, right, down, up = [], [], [], []
challengers = []
for i in range(k):
x1, y1, x2, y2 = map(int, stdin.readline().split())
if x1 == x2:
if y1 < y2:
challengers.append((x1, y1, x2, y2,... | 212 | 482 | 40,140,800 | 170725525 | import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
d = int(input())
n, m = map(int, input().split())
L = []
R = []
T = []
B = []
XY = []
for i in range(d):
x1, y1, x2, y2 = map(int, input().split())
L.append((min(x1, x2)))
R.append(-max(x1, x2))
T.append(min(y1, y2))... | Educational Codeforces Round 24 | ICPC | 2,017 | 1 | 256 | Sofa Thief | Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs ... | The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa... | Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1. | null | Let's consider the second example.
- The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
- The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
- The third sofa has cntl = 2, cntr = 1, c... | [{"input": "2\n3 2\n3 1 3 2\n1 2 2 2\n1 0 0 1", "output": "1"}, {"input": "3\n10 10\n1 2 1 1\n5 5 6 5\n6 4 5 4\n2 1 2 0", "output": "2"}, {"input": "2\n2 2\n2 1 1 1\n1 2 2 2\n1 0 0 0", "output": "-1"}] | 2,000 | ["brute force", "implementation"] | 212 | [{"input": "2\r\n3 2\r\n3 1 3 2\r\n1 2 2 2\r\n1 0 0 1\r\n", "output": "1\r\n"}, {"input": "3\r\n10 10\r\n1 2 1 1\r\n5 5 6 5\r\n6 4 5 4\r\n2 1 2 0\r\n", "output": "2\r\n"}, {"input": "2\r\n2 2\r\n2 1 1 1\r\n1 2 2 2\r\n1 0 0 0\r\n", "output": "-1\r\n"}, {"input": "1\r\n1 2\r\n1 1 1 2\r\n0 0 0 0\r\n", "output": "1\r\n"}, ... | false | stdio | null | true |
818/C | 818 | C | PyPy 3 | TESTS | 8 | 841 | 33,689,600 | 142736287 | import sys
input = sys.stdin.readline
d = int(input())
n, m = map(int, input().split())
xl = [[] for _ in range(n + 2)]
xr = [[] for _ in range(n + 2)]
yt = [[] for _ in range(m + 2)]
yb = [[] for _ in range(m + 2)]
for i in range(1, d + 1):
x1, y1, x2, y2 = map(int, input().split())
if x1 > x2:
x1, x2... | 212 | 498 | 17,305,600 | 93759517 | import sys
try:
fin=open('in')
except:
fin=sys.stdin
input=fin.readline
d = int(input())
n, m = map(int, input().split())
x1, y1, x2, y2 = [], [], [], []
T=[]
for _ in range(d):
u, v, w, x = map(int, input().split())
if u>w:u,w=w,u
if v>x:v,x=x,v
x1.append(u)
y1.append(v)
x2.append(-w)#... | Educational Codeforces Round 24 | ICPC | 2,017 | 1 | 256 | Sofa Thief | Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs ... | The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa... | Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1. | null | Let's consider the second example.
- The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
- The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
- The third sofa has cntl = 2, cntr = 1, c... | [{"input": "2\n3 2\n3 1 3 2\n1 2 2 2\n1 0 0 1", "output": "1"}, {"input": "3\n10 10\n1 2 1 1\n5 5 6 5\n6 4 5 4\n2 1 2 0", "output": "2"}, {"input": "2\n2 2\n2 1 1 1\n1 2 2 2\n1 0 0 0", "output": "-1"}] | 2,000 | ["brute force", "implementation"] | 212 | [{"input": "2\r\n3 2\r\n3 1 3 2\r\n1 2 2 2\r\n1 0 0 1\r\n", "output": "1\r\n"}, {"input": "3\r\n10 10\r\n1 2 1 1\r\n5 5 6 5\r\n6 4 5 4\r\n2 1 2 0\r\n", "output": "2\r\n"}, {"input": "2\r\n2 2\r\n2 1 1 1\r\n1 2 2 2\r\n1 0 0 0\r\n", "output": "-1\r\n"}, {"input": "1\r\n1 2\r\n1 1 1 2\r\n0 0 0 0\r\n", "output": "1\r\n"}, ... | false | stdio | null | true |
797/C | 797 | C | Python 3 | TESTS | 32 | 467 | 1,331,200 | 111796189 | from collections import deque
s = input()
t = deque([])
u = ""
ind = [-1]*26
for i, ss in enumerate(s):
ind[ord(ss)-97] = i
for i,ss in enumerate(s):
now = ord(ss)-97
for j in range(0,now):
if ind[j] > i:
t.append(ss)
break
else:
while t:
now =... | 189 | 93 | 14,336,000 | 196924702 | from collections import Counter, deque
s = input()
n = len(s)
mn = ['z'] * (n + 1)
for i in range(n - 1, -1, -1):
mn[i] = min(mn[i + 1], s[i])
t, u = [], []
for i in range(n):
while t and t[-1] <= mn[i]:
u.append(t.pop())
if s[i] == mn[i]:
u.append(s[i])
else:
... | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Minimal string | Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:
- Extract the first character of s and append t with this character.
- Extract the last character of t and append u with this chara... | First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters. | Print resulting string u. | null | null | [{"input": "cab", "output": "abc"}, {"input": "acdb", "output": "abdc"}] | 1,700 | ["data structures", "greedy", "strings"] | 189 | [{"input": "cab\r\n", "output": "abc\r\n"}, {"input": "acdb\r\n", "output": "abdc\r\n"}, {"input": "a\r\n", "output": "a\r\n"}, {"input": "ab\r\n", "output": "ab\r\n"}, {"input": "ba\r\n", "output": "ab\r\n"}, {"input": "dijee\r\n", "output": "deeji\r\n"}, {"input": "bhrmc\r\n", "output": "bcmrh\r\n"}, {"input": "aaaaa... | false | stdio | null | true |
691/B | 691 | B | Python 3 | TESTS | 11 | 46 | 0 | 172134941 | s = input()
mu = ['A','H','I','M','O','o','T','U','V','v','W','w','X','x','Y']
slen = len(s)
flag = 1
for i in range(slen):
if s[i]==s[-i-1] and s[i] in mu:
pass
elif (s[i]=='b' and s[-i-1]=='d') or (s[i]=='d' and s[-i-1]=='b'):
pass
else:
flag = 0
if flag == 1:
print("TAK")
else:
print("NIE") | 168 | 46 | 0 | 158963335 | t = input()
k = len(t) // 2 + 1
p = zip(t[:k], t[-k:][::-1])
u, v = 'AHIMOoTUVvWwXxY', 'bd db pq qp'
s = all(a in u and a == b or a + b in v for a, b in p)
print(['NIE', 'TAK'][s]) | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
691/B | 691 | B | Python 3 | TESTS | 11 | 62 | 307,200 | 109822485 | chars = {
'A': {'A'},
'b': {'d'},
'd': {'b'},
'H': {'H'},
'M':{'M'},
'O':{'O'},
'o':{'o'},
'T':{'T'},
'U':{'U'},
'V':{'V'},
'v':{'v'},
'W':{'W'},
'w':{'w'},
'X':{'X'},
'x': {'x'},
'Y':{'Y'},
}
s = input()
ans = True
for i in range(len(s)//2):
if s[... | 168 | 46 | 0 | 185154340 | s=input()
p='AbdHIMOopqTUVvWwXxY'
p1='AdbHIMOoqpTUVvWwXxY'
j=1
n=len(s)
for i in range(n):
if s[i] not in p:
j=0
break
elif p1[p.index(s[i])]!=s[n-i-1]:
j=0
break
if j==1:
print('TAK')
else:
print('NIE') | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
691/B | 691 | B | PyPy 3 | TESTS | 11 | 62 | 19,968,000 | 36211090 | symmetric_letters = ['A', 'H', 'I', 'M', 'O', 'o', 'U', 'V',
'v', 'W', 'w', 'X', 'x', 'Y']
letter_trans = {
'b': 'd',
'd': 'b',
'p': 'q',
'q': 'p',
}
for letter in symmetric_letters:
letter_trans[letter] = letter
word = input()
if all([x in letter_trans for x in word]) and \... | 168 | 62 | 0 | 19079997 | d = "AHIMOoTUVvWwXxY"
s = input()
for i in range(len(s) // 2 + 1):
if not ((s[i] in d) and (s[i] == s[-i-1])) and not ((s[i] + s[-i-1]) in ["bd", "db", "pq", "qp"]):
print("NIE")
break
else:
print("TAK") | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
691/B | 691 | B | Python 3 | TESTS | 11 | 46 | 0 | 171744136 | a='AAHHMMTTUUVVWWXXYYOOMMbddboovvxxww'
a1=[]
for i in range(0,len(a),2):
a1.append(a[i]+a[i+1])
s=input()
t='AHMTUVWMXYOMoxvw'
x=True
if len(s)%2==1:
if s[len(s)//2] in t:
for i in range(len(s)//2):
if s[i]+s[len(s)-i-1] in a1:
x=True
else:
x=False... | 168 | 62 | 0 | 19080023 | s = input()
p = 'aBCcDEeFfGghiJjKkLlmNnPQRrSstuyZz'
for c in s:
if c in p:
print('NIE')
exit()
def trans(c):
d = {
'b': 'd',
'd': 'b',
'p': 'q',
'q': 'p',
}
if c in d:
return d[c]
return c
t = ''.join(map(trans, s[::-1]))
if t == s:
pri... | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
797/C | 797 | C | PyPy 3 | TESTS | 20 | 140 | 29,900,800 | 123806207 | import sys
import heapq
def get_string(): return sys.stdin.readline().strip()
# Output for list
# sys.stdout.write(" ".join(map(str, final)) + "\n")
# Output for int or str
# sys.stdout.write(str(best) + "\n")
def solve(s):
length = len(s)
current = 97
final = []
start = 0
end = length - 1
extra = []
... | 189 | 93 | 14,540,800 | 175279224 | from collections import defaultdict
from collections import Counter
def parse():
return(int(input()))
def parse2():
return(list(map(int,input().split())))
def parsestr():
return input()
def solve():
s = parsestr()
n = len(s)
best = [0]*(n+1)
best[-1] = 'z'
for i in range... | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Minimal string | Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:
- Extract the first character of s and append t with this character.
- Extract the last character of t and append u with this chara... | First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters. | Print resulting string u. | null | null | [{"input": "cab", "output": "abc"}, {"input": "acdb", "output": "abdc"}] | 1,700 | ["data structures", "greedy", "strings"] | 189 | [{"input": "cab\r\n", "output": "abc\r\n"}, {"input": "acdb\r\n", "output": "abdc\r\n"}, {"input": "a\r\n", "output": "a\r\n"}, {"input": "ab\r\n", "output": "ab\r\n"}, {"input": "ba\r\n", "output": "ab\r\n"}, {"input": "dijee\r\n", "output": "deeji\r\n"}, {"input": "bhrmc\r\n", "output": "bcmrh\r\n"}, {"input": "aaaaa... | false | stdio | null | true |
797/C | 797 | C | Python 3 | TESTS | 20 | 639 | 921,600 | 136697802 | s = list(input())
#s = ['a']*100000
u = []
index = 0
while(len(s)!= 0):
for j in range(index, len(s)):
if(len(u) != 0 and s[j] == u[len(u)-1]):
index = j
break
if(s[j] < s[index]):
index = j
u.append(s.pop(index))
index = max(0, index-1)
... | 189 | 108 | 14,131,200 | 220321279 | stack = input()
n = len(stack)
aux = [0] * n
aux[-1] = stack[-1]
for i in range(n - 2, -1, -1):
aux[i] = min(aux[i + 1], stack[i])
second_stack = []
result_stack = []
for i in range(n):
if not second_stack:
second_stack.append(stack[i])
else:
while second_stack:
if second_sta... | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Minimal string | Petya recieved a gift of a string s with length up to 105 characters for his birthday. He took two more empty strings t and u and decided to play a game. This game has two possible moves:
- Extract the first character of s and append t with this character.
- Extract the last character of t and append u with this chara... | First line contains non-empty string s (1 ≤ |s| ≤ 105), consisting of lowercase English letters. | Print resulting string u. | null | null | [{"input": "cab", "output": "abc"}, {"input": "acdb", "output": "abdc"}] | 1,700 | ["data structures", "greedy", "strings"] | 189 | [{"input": "cab\r\n", "output": "abc\r\n"}, {"input": "acdb\r\n", "output": "abdc\r\n"}, {"input": "a\r\n", "output": "a\r\n"}, {"input": "ab\r\n", "output": "ab\r\n"}, {"input": "ba\r\n", "output": "ab\r\n"}, {"input": "dijee\r\n", "output": "deeji\r\n"}, {"input": "bhrmc\r\n", "output": "bcmrh\r\n"}, {"input": "aaaaa... | false | stdio | null | true |
691/B | 691 | B | Python 3 | TESTS | 24 | 62 | 4,505,600 | 134395076 | s = input()
f = 1
if len(s) % 2 == 1:
k = len(s) // 2
if s[k] != 'H' and s[k] != 'A' and s[k] != 'I' and s[k] != 'M' and s[k] != 'O' and s[k] != 'o' \
and s[k] != 'T' and s[k] != 'V' and s[k] != 'v' and s[k] != 'W' and s[k] != 'w' \
and s[k] != 'X' and s[k] != 'x' and s[k] != 'Y' and s[k... | 168 | 62 | 0 | 19080106 | midsym = 'AHIMOoTUVvWwXxY'
sym = 'pqbd'
s = input()
if len(s) % 2 == 1:
t = len(s) // 2
if s[t] not in midsym:
print('NIE')
exit()
s = s[:t] + s[t+1:]
#print(s)
f = s[:int(len(s)/2)]
l = s[int(len(s)/2):]
l = l[::-1]
for k in range(len(f)):
if f[k] not in midsym and f[k] not in sym:
print('NIE')
exit()... | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
691/B | 691 | B | PyPy 3 | TESTS | 24 | 124 | 23,244,800 | 20033974 | #!/usr/bin/env python3
def check(s):
chars = 'AHIMOoTUVvWwXxY'
sr = list(reversed(s))
for i in range(len(s)):
if s[i] not in chars + 'dbpq':
return False
for a, b in [('d', 'b'), ('b', 'd'),
('q', 'p'), ('p', 'q')]:
if s[i] == a and sr[i] != b:
... | 168 | 62 | 0 | 19084773 | s = str(input())
n = len(s)
ss = ['A', 'H', 'I', 'M', 'O','o','T','U','V','v','W','w','X','x','Y']
ss2 = [('d','b'),('q','p'),('p','q'),('b','d')]
for i in range(0, n//2):
if s[i]==s[n-1-i]:
if not (s[i] in ss):
print("NIE")
exit()
else:
if not ((s[i],s[n-1-i]) in ss2)... | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
858/E | 858 | E | Python 3 | TESTS | 1 | 46 | 204,800 | 30971427 | n = int(input())
top = set()
bottom = set()
for i in range(n):
name, type = input().split()
if type == '1':
top.add(name)
else:
bottom.add(name)
a = len(top)
b = len(bottom)
sa = set(str(i) for i in range(1, a + 1))
q = sa & top
sa -= q
top -= q
sb = set(str(i) for i in range(a + 1, b + a +... | 165 | 295 | 27,136,000 | 230751905 | n = int(input())
t = [1] + [0] * n
b, a = d = [], []
h, s = [], []
for i in range(n):
f, k = input().split()
d[int(k)].append(f)
m = len(a)
for i in a:
if i.isdigit() and i[0] != '0':
j = int(i)
if 0 < j <= m:
t[j] = 1
elif m < j <= n:
t[j] = -1
else... | Технокубок 2018 - Отборочный Раунд 1 | CF | 2,017 | 2 | 256 | Tests Renumeration | The All-Berland National Olympiad in Informatics has just ended! Now Vladimir wants to upload the contest from the Olympiad as a gym to a popular Codehorses website.
Unfortunately, the archive with Olympiad's data is a mess. For example, the files with tests are named arbitrary without any logic.
Vladimir wants to re... | The first line contains single integer n (1 ≤ n ≤ 105) — the number of files with tests.
n lines follow, each describing a file with test. Each line has a form of "name_i type_i", where "name_i" is the filename, and "type_i" equals "1", if the i-th file contains an example test, and "0" if it contains a regular test. ... | In the first line print the minimum number of lines in Vladimir's script file.
After that print the script file, each line should be "move file_1 file_2", where "file_1" is an existing at the moment of this line being run filename, and "file_2" — is a string of digits and small English letters with length from 1 to 6. | null | null | [{"input": "5\n01 0\n2 1\n2extra 0\n3 1\n99 0", "output": "4\nmove 3 1\nmove 01 5\nmove 2extra 4\nmove 99 3"}, {"input": "2\n1 0\n2 1", "output": "3\nmove 1 3\nmove 2 1\nmove 3 2"}, {"input": "5\n1 0\n11 1\n111 0\n1111 1\n11111 0", "output": "5\nmove 1 5\nmove 11 1\nmove 1111 2\nmove 111 4\nmove 11111 3"}] | 2,200 | ["greedy", "implementation"] | 165 | [{"input": "5\r\n01 0\r\n2 1\r\n2extra 0\r\n3 1\r\n99 0\r\n", "output": "4\r\nmove 3 1\r\nmove 01 5\r\nmove 2extra 4\r\nmove 99 3\r\n"}, {"input": "2\r\n1 0\r\n2 1\r\n", "output": "3\r\nmove 1 dytuig\r\nmove 2 1\r\nmove dytuig 2\r\n"}, {"input": "5\r\n1 0\r\n11 1\r\n111 0\r\n1111 1\r\n11111 0\r\n", "output": "5\r\nmove... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline())
files = []
type1_count = 0
for _ in range(n):
line = f.readline().strip()
name, typ = line.s... | true |
691/B | 691 | B | PyPy 3 | TESTS | 38 | 93 | 0 | 110982063 | s = input()
d = {"b" : "d","d":"b","l":"l","o":"o","p":"q","q":"p","v":"v","w":"w","x":"x","A":"A","H":"H","I":"I","M":"M","O":"O","T":"T","U":"U","V":"V","W":"W","X":"X","Y":"Y"}
n = len(s)//2
flag = 0
l = []
if len(s)%2 == 0:
i = n-1
j = n
else:
i = n
j = n
while i > -1:
if s[i] not in d:
... | 168 | 62 | 0 | 19099266 | '''
Created on Jul 14, 2016
@author: Md. Rezwanul Haque
'''
s = input()
D = {'A': 'A', 'b': 'd', 'd': 'b', 'H': 'H', 'I': 'I', 'M': 'M', 'O': 'O', 'o': 'o', 'p': 'q', 'q': 'p', 'T': 'T', 'U': 'U', 'V': 'V', 'v': 'v', 'W': 'W', 'w': 'w', 'X': 'X', 'x': 'x', 'Y': 'Y'}
for(c1,c2) in zip(s, s[::-1]):
if(D.get(... | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 37 | 140 | 7,987,200 | 25858647 | import sys
s = list(input())
digsum = 0
rem = [0 for i in range(len(s))]
for i in range(0, len(s)):
dig = ord(s[i]) - ord('0');
digsum += dig
rem[i] = dig % 3
currem = digsum % 3
if currem == 0:
print(''.join(s))
sys.exit()
if len(s) == 1:
print(-1)
sys.exit()
inds1 = []
inds2 = []
for i... | 162 | 77 | 6,860,800 | 26445505 | import re
s = input()
t = s.translate(str.maketrans("0123456789", "0120120120"))
x = (t.count('1') + t.count('2') * 2) % 3
if x:
res = ['']
l = t.rsplit("12"[x == 2], 1)
if len(l) == 2:
a = len(l[0])
res.append((s[:a], s[a + 1:]))
l = t.rsplit("12"[x == 1], 2)
if len(l) == 3:
... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
818/C | 818 | C | Python 3 | TESTS | 6 | 61 | 5,632,000 | 34566012 | def main():
# import sys
d = int(input())
n, m = [int(x) for x in input().split()]
point = []
for x in range(d):
point.append([int(x) for x in input().split()])
point[-1].append(x+1)
q = [int(x) for x in input().split()]
def cntl(n):
# print(max(n[0],n[2]))
retu... | 212 | 560 | 17,920,000 | 93166980 | import sys
from bisect import bisect_left, bisect_right
d = int(sys.stdin.buffer.readline().decode('utf-8'))
n, m = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
a = [list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
for _ in range(d)]
cnt = list(map(int, sys.stdin.buffer.readlin... | Educational Codeforces Round 24 | ICPC | 2,017 | 1 | 256 | Sofa Thief | Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs ... | The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa... | Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1. | null | Let's consider the second example.
- The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
- The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
- The third sofa has cntl = 2, cntr = 1, c... | [{"input": "2\n3 2\n3 1 3 2\n1 2 2 2\n1 0 0 1", "output": "1"}, {"input": "3\n10 10\n1 2 1 1\n5 5 6 5\n6 4 5 4\n2 1 2 0", "output": "2"}, {"input": "2\n2 2\n2 1 1 1\n1 2 2 2\n1 0 0 0", "output": "-1"}] | 2,000 | ["brute force", "implementation"] | 212 | [{"input": "2\r\n3 2\r\n3 1 3 2\r\n1 2 2 2\r\n1 0 0 1\r\n", "output": "1\r\n"}, {"input": "3\r\n10 10\r\n1 2 1 1\r\n5 5 6 5\r\n6 4 5 4\r\n2 1 2 0\r\n", "output": "2\r\n"}, {"input": "2\r\n2 2\r\n2 1 1 1\r\n1 2 2 2\r\n1 0 0 0\r\n", "output": "-1\r\n"}, {"input": "1\r\n1 2\r\n1 1 1 2\r\n0 0 0 0\r\n", "output": "1\r\n"}, ... | false | stdio | null | true |
691/B | 691 | B | PyPy 3 | TESTS | 28 | 109 | 20,480,000 | 127190638 | from collections import defaultdict
import sys
input = sys.stdin.readline
d = defaultdict(lambda : "?")
for i in list("AHIMmOoTUVvWwXxY"):
d[i] = i
d["b"] = "d"
d["d"] = "b"
d["p"] = "q"
d["q"] = "p"
s = list(input().rstrip())
n = len(s)
ans = "TAK"
for i in range(n):
if not d[s[i]] == s[n - i - 1]:
an... | 168 | 62 | 0 | 19110383 | import sys
pairs = [
'AA', 'HH', 'II', 'MM', 'OO', 'TT', 'UU', 'VV', 'WW', 'XX', 'YY',
'bd', 'db', 'oo', 'pq', 'qp', 'vv', 'ww', 'xx'
]
s = input()
l, r = 0, len(s) - 1
while l <= r:
if s[l] + s[r] not in pairs:
print("NIE")
sys.exit(0)
l += 1
r -= 1
print("TAK") | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
691/B | 691 | B | PyPy 3 | TESTS | 28 | 140 | 20,172,800 | 80008474 | def f(s):
i=0
n=len(s)
j = int(n)-1
S = {'A','H','I','M','m','O','o','T','U','V','v','W','w','X','x','Y'}
while(i<=j):
if((s[i]=='b' and s[j]=='d') or (s[i]=='d' and s[j]=='b')):
i+=1
j-=1
continue
if((s[i]=='p' and s[j]=='q') or (s[i]=='q' and s[j... | 168 | 62 | 0 | 148521856 | s = input()
print(all([
s[i] + s[-i-1] in "AA.HH.II.MM.OO.oo.TT.UU.VV.vv.WW.ww.XX.xx.YY.bd.db.pq.qp"
for i in range(len(s))
]) and "TAK" or "NIE") | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
792/C | 792 | C | PyPy 3 | TESTS | 36 | 139 | 30,208,000 | 26737037 | def f(n, r, k):
tmp = n
for t in range(k):
for i in range(len(tmp) - 1, -1, -1):
if int(tmp[i]) % 3 == r:
tmp = tmp[:i] + tmp[i + 1:]
break
for i in range(len(tmp) - 1):
if int(tmp[i]) != 0:
return ''.join(tmp[i:])
if len(tmp) == 0 ... | 162 | 109 | 6,246,400 | 28276110 | n = list(input())
leng = 0
mp=[0,0,0]
for x in n:
leng += 1
mp[int(x)%3]+=1
tot = (mp[1]+2*mp[2])%3
if tot == 0:
print("".join(n))
exit()
if mp[tot] == 0:
do = tot ^ 3
cnt = 2
else:
if mp[tot] == 1 and int(n[0])%3==tot and n[1:3] == ['0','0']:
do =tot^3
cnt = 2
if mp[do] == 0:
do = tot
cnt = 1
else:... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
691/B | 691 | B | Python 3 | TESTS | 23 | 109 | 307,200 | 55610602 | d = {'o': 'o', 'I': 'I', 'H': 'H', 'i': 'i', 'b': 'd', 'x': 'x',
'X': 'X', 'V': 'V', 'v': 'v', 'U': 'U', 'W': 'W', 'w': 'w',
'M': 'M', 'A': 'A', 'Y': 'Y', 'T': 'T', 'p': 'q', 'q': 'p',
'd': 'b', 'O': 'O'}
c = ['U', 'I', 'O', 'T', 'H', 'i', 'X', 'x', 'o', 'M', 'A', 'W', 'w',
'V', 'v', 'Y', ]
s = in... | 168 | 62 | 4,608,000 | 19725920 | def check(s):
p='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
q='A HI M O TUVWXY d b oqp vwx '
for i in range(len(s)):
if q[ p.find(s[i]) ]!=s[-i-1]:
return False
return True
print('TAK' if check(input()) else 'NIE') | Educational Codeforces Round 14 | ICPC | 2,016 | 1 | 256 | s-palindrome | Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.
English alphabet
You are given a string ... | The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters. | Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise. | null | null | [{"input": "oXoxoXo", "output": "TAK"}, {"input": "bod", "output": "TAK"}, {"input": "ER", "output": "NIE"}] | 1,600 | ["implementation", "strings"] | 168 | [{"input": "oXoxoXo\r\n", "output": "TAK\r\n"}, {"input": "bod\r\n", "output": "TAK\r\n"}, {"input": "ER\r\n", "output": "NIE\r\n"}, {"input": "o\r\n", "output": "TAK\r\n"}, {"input": "a\r\n", "output": "NIE\r\n"}, {"input": "opo\r\n", "output": "NIE\r\n"}, {"input": "HCMoxkgbNb\r\n", "output": "NIE\r\n"}, {"input": "v... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 59 | 140 | 9,523,200 | 37535803 | #!/usr/bin/env python3
# solution after hint:(
sn = input().strip()
n = len(sn)
def rem3(sn):
return sum(map(int, sn)) % 3
def buildn(sn, com, start=0):
return ''.join(sn[start + i] for i in com)
def subminbeau2(sn, start=0):
n = len(sn)
r3 = rem3(sn[start:])
nums = set(sn[1 + start:])
nums3 = set((int(i) % ... | 162 | 109 | 6,553,600 | 25896991 | def sumstr(s):
sums = 0
for i in range(len(s)):
sums += int(s[i])
return sums
def mod3(s):
sums = sumstr(s)
return sums % 3
def onlyzeros(s):
i = 0
while i < len(s):
if s[i] != '0':
return False
i += 1
return True
def lastin(s, s1):
maxf = -1
#d = '-'
for d in s1:
f = s.rfind(d)
#print(d, f,... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
837/B | 837 | B | Python 3 | TESTS | 107 | 77 | 4,915,200 | 29204823 | N, M = map(int, input().split())
flag = [input() for n in range(N)]
def get_rect_of_color(color):
rect_points = []
for i, row in enumerate(flag):
first = True
tmp = [None] * 4
for j, char in enumerate(row):
if char == color:
if first:
fir... | 153 | 46 | 0 | 188298470 | # First of all, we should validate the input by checking the eligibilities in a function called 'validate_flag'
def validate_flag(field: list) -> bool:
# First, we validate the number of colors in the field
total_colors = set()
for row in field:
total_colors.update(row)
# If it is not exac... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
837/B | 837 | B | Python 3 | TESTS | 127 | 93 | 5,120,000 | 29167646 | import re
n,m =map(int,input().split())
data = [ input() for _ in range(n)]
temp = ''.join([data[i][0] for i in range(n)])
f1=False
count={}
i=0
mark=temp[0]
while i<n and temp[i]==mark:
i+=1
count[mark]=i
j=i
if j<n:
mark=temp[j]
while j<n and temp[j]==mark :
j+=1
count[mark]=j-i
if j<n... | 153 | 61 | 4,608,000 | 29264658 | def c(f, n, m):
if n % 3 != 0:return 0
s=n//3;s=f[0],f[s],f[2*s]
if set(s)!=set(map(lambda x:x*m,'RGB')):return 0
for i in range(n):
if f[i] != s[i*3 // n]:return 0
return 1
n,m=map(int, input().split())
f=[input() for _ in range(n)]
print('Yes'if c(f,n,m) or c([''.join(f[j][i] for j in rang... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
837/B | 837 | B | PyPy 3 | TESTS | 127 | 234 | 3,174,400 | 81700781 | import math as mt
import sys,string
input=sys.stdin.readline
from collections import defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
n,m=M()
l=[]
flag=0
for i in range(n):
s=input().strip()
x=set([])
for j i... | 153 | 62 | 0 | 30220071 | def solve_h(lines):
'''
>>> solve_h(['R', 'G', 'B'])
True
>>> solve_h(['RRR', 'GGG', 'BBB'])
True
>>> solve_h(['RRR', 'RRR', 'GGG', 'GGG', 'BBB', 'BBB'])
True
>>> solve_h(['RRR', 'RRR', 'RRR', 'GGG', 'BBB', 'BBB'])
False
>>> solve_h(['R', 'G', 'A'])
False
'''
if len(l... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
837/B | 837 | B | Python 3 | TESTS | 113 | 124 | 0 | 40912880 | n,m=list(map(int,input().split()))
l=[0]*n
for i in range(n):
l[i]=input()
types=['B','G','R']
first=[]
second=[]
third=[]
counter=0
container=0
ans=[]
ans1=[]
if l[0].count('B')==l[0].count('R') and l[0].count('B')==l[0].count('G') :
first.append(l[0][0])
types.remove(first[0])
for i in range(len(l[... | 153 | 62 | 307,200 | 31198872 | import sys
#sys.stdin = open("input2","r")
n,m = map(int,input().split())
s = []
for i in range(n):
string = input()
s.append(string)
no = 0
# row mix check
r_mix = 0
for i in range(n):
cnt_r = s[i].count('R')
cnt_b = s[i].count('B')
cnt_g = s[i].count('G')
if ((cnt_r and cnt_b) or (cnt_b and ... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 34 | 155 | 9,318,400 | 26250901 | def checkBeauty (digi):
total = 0
for i in range(0,len(digi)):
total += int(digi[i])
return total % 3
def modThree (digi):
zero = []
one = []
two = []
for i in range(0,len(digi)):
if int(digi[i]) % 3 == 0:
zero.append(i)
elif int(digi[i]) % 3 == 1:
one.append(i)
else:
two.append(i)
modResult =... | 162 | 109 | 7,168,000 | 32921983 | import re
f = lambda t: re.sub(r'^0+(\d)', r'\1', t)
g = lambda k: n - p.index(s, k)
t = input()
n = len(t) - 1
p = [int(q) % 3 for q in t][::-1]
s = sum(p) % 3
if s == 0:
print(t)
exit()
u = v = ''
if s in p:
i = g(0)
u = f(t[:i] + t[i + 1:])
s = 3 - s
if p.count(s) > 1:
i = g(0)
j = g(n - ... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
807/A | 807 | A | Python 3 | TESTS | 43 | 109 | 102,400 | 47254713 | n=int(input())
x,y=[],[]
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
z=x[:]
z.sort()
z.reverse()
if z==y:
print("maybe")
exit()
if x==y:
print("unrated")
exit()
if x!=y:
print("rated") | 150 | 46 | 0 | 131984875 | t=int(input())
c=0
l=[]
for i in range(t):
a,b=map(int,input().split())
l.append(b)
if a==b:
c=c+1
if c!=t:
print("rated")
else:
r=l[:]
r.sort(reverse=True)
if l==r:
print("maybe")
else:
print("unrated") | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
598/C | 598 | C | Python 3 | TESTS | 113 | 841 | 22,016,000 | 116184762 | import sys
from math import atan2, degrees
from decimal import Decimal
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7
def resolve():
n = int(input())
XY = []
for i in range(n):
x, y = map(int, input().split())
theta = Decimal(degrees(atan2... | 163 | 779 | 41,369,600 | 175306822 | from sys import stdin
input=lambda :stdin.readline()[:-1]
def compare(a,b):
# 比較関数 (a<=b)
# 偏角ソート
ax, ay = a[:2]
bx, by = b[:2]
if ay < 0:
return by >= 0 or ax * by - ay * bx > 0
if ay == 0:
return ax >= 0 and (by > 0 or (by == 0 and bx < 0))
return by >= 0 and (ax * by - ay * bx) > 0
def me... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
837/B | 837 | B | PyPy 3-64 | TESTS | 17 | 77 | 2,969,600 | 210295461 | import sys
input = lambda: sys.stdin.readline().rstrip()
N,M = map(int, input().split())
S = []
for _ in range(N):
S.append(input())
INF = float('inf')
A = [[INF,INF,-1,-1],[INF,INF,-1,-1],[INF,INF,-1,-1]]
for i in range(N):
for j in range(M):
if S[i][j]=='R':
A[0][0] = min(A[0][0],i)
... | 153 | 62 | 307,200 | 110106847 | def rotate_90_degree_anticlckwise(matrix):
new_matrix = []
for i in range(len(matrix[0]), 0, -1):
new_matrix.append(list(map(lambda x: x[i-1], matrix)))
return new_matrix
n,m = map(int,input().split())
mat = []
for i in range(n):
temp = input()
mat.append([i for i in temp])
horiz, verti = Fa... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
807/A | 807 | A | PyPy 3 | TESTS | 69 | 233 | 2,150,400 | 97199173 | from sys import maxsize, stdout, stdin, stderr
# mod = int(1e9 + 7)
import re # can use multiple splits
tup = lambda: map(int, stdin.readline().split())
I = lambda: int(stdin.readline())
lint = lambda: [int(x) for x in stdin.readline().split()]
S = lambda: stdin.readline().replace('\n', '').strip()
def grid(r, c): ret... | 150 | 46 | 0 | 132422536 | a=int(input())
d=[]
e=[]
f=[]
for i in range(a):
b,c=map(int,input().split())
if b==c:
d.append(b)
e.append(b)
e.sort()
d=d[::-1]
if len(d)!=a:
print("rated")
else:
for j in range(len(d)):
if e[j]==d[j]:
f.append(e[j])
if len(f)==a:
print("maybe")
else... | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
598/C | 598 | C | PyPy 3-64 | TESTS | 118 | 483 | 17,715,200 | 147382716 | import sys,math
input = lambda: sys.stdin.readline()[:-1]
get_int = lambda: int(input())
get_int_iter = lambda: map(int,input().split())
get_int_list = lambda: list(map(int,input().split()))
n = get_int()
vectors = []
mod = math.pi * 2
for i in range(1,n+1):
x,y = get_int_iter()
if x != 0:
vectors.appe... | 163 | 1,060 | 35,328,000 | 154855646 | from math import atan2
s = lambda a, b: a[0] * b[0] + a[1] * b[1]
v = lambda a, b: a[0] * b[1] - a[1] * b[0]
p = []
for i in range(int(input())):
x, y = map(int, input().split())
p.append((atan2(x, y), (x, y), i + 1))
p.sort()
d = [(s(a, b), abs(v(a, b)), i, j) for (x, a, i), (y, b, j) in zip(p, p[1:] + p... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | Python 3 | TESTS | 113 | 530 | 8,704,000 | 173893303 | import math
def nearest_vec(L):
L.sort()
L.append(L[0])
min_val = float('inf')
values = [-1, -1]
for r in range(1, len(L)):
x = L[r][0] - L[r - 1][0]
if x < 0:
x += 360
if x < min_val:
min_val = x
values[0] = L[r][1]
values[1]... | 163 | 1,153 | 20,992,000 | 112977689 | from math import atan2
inner = lambda a, b: a[0] * b[0] + a[1] * b[1]
outer = lambda a, b: a[0] * b[1] - a[1] * b[0]
N = int(input())
vectors = []
for i in range(N):
x, y = map(int, input().split())
vectors.append((atan2(y, x), (x, y), i + 1))
vectors.sort()
diff = []
for i in range(N):
diff.append([inne... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | Python 3 | TESTS | 117 | 483 | 9,932,800 | 111295171 | """
You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always betwee... | 163 | 1,185 | 29,696,000 | 197816661 | from functools import cmp_to_key
def is_top(vector):
return vector[1] > 0 or vector[1] == 0 and vector[0] > 0
def cross(first, second):
return first[0] * second[1] - second[0] * first[1]
def dot(first, second):
return first[0]*second[0] + first[1]*second[1]
def polar_cmp(first, second):
if is_to... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | Python 3 | TESTS | 117 | 764 | 14,540,800 | 118716539 | import math
class vector:
def __init__(self, X, Y, P) -> None:
self.v = math.atan2(X, Y)
self.p = P
def work(v) -> float:
while v < 0:
v += math.pi * 2
while v > math.pi * 2:
v -= math.pi * 2
return min(v, math.pi * 2 - v)
n = int(input())
a = list()
for i in range(... | 163 | 1,201 | 23,347,200 | 112976028 | from math import atan2
inner = lambda a, b: a[0] * b[0] + a[1] * b[1]
outer = lambda a, b: a[0] * b[1] - a[1] * b[0]
vectors = []
for i in range(int(input())):
x, y = map(int, input().split())
vectors.append((atan2(y, x), (x, y), i + 1))
vectors.sort()
diff = [(inner(vec1, vec2), abs(outer(vec1, vec2)), i, ... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
858/E | 858 | E | Python 3 | PRETESTS | 0 | 46 | 0 | 30438694 | n = int(input())
tasks = []
e = 0
for i in range(n):
s = input().split()
if s[0].isdigit():
s[0] = int(s[0])
a = [s[0], int(s[1])]
tasks.append(a)
if a[1]:
e += 1
tasks.sort(key=lambda x: -x[1])
newSet = set(range(1, e + 1))
goodSet = set()
for i in range(e):
if tasks[i][0] in ne... | 165 | 295 | 27,136,000 | 230751905 | n = int(input())
t = [1] + [0] * n
b, a = d = [], []
h, s = [], []
for i in range(n):
f, k = input().split()
d[int(k)].append(f)
m = len(a)
for i in a:
if i.isdigit() and i[0] != '0':
j = int(i)
if 0 < j <= m:
t[j] = 1
elif m < j <= n:
t[j] = -1
else... | Технокубок 2018 - Отборочный Раунд 1 | CF | 2,017 | 2 | 256 | Tests Renumeration | The All-Berland National Olympiad in Informatics has just ended! Now Vladimir wants to upload the contest from the Olympiad as a gym to a popular Codehorses website.
Unfortunately, the archive with Olympiad's data is a mess. For example, the files with tests are named arbitrary without any logic.
Vladimir wants to re... | The first line contains single integer n (1 ≤ n ≤ 105) — the number of files with tests.
n lines follow, each describing a file with test. Each line has a form of "name_i type_i", where "name_i" is the filename, and "type_i" equals "1", if the i-th file contains an example test, and "0" if it contains a regular test. ... | In the first line print the minimum number of lines in Vladimir's script file.
After that print the script file, each line should be "move file_1 file_2", where "file_1" is an existing at the moment of this line being run filename, and "file_2" — is a string of digits and small English letters with length from 1 to 6. | null | null | [{"input": "5\n01 0\n2 1\n2extra 0\n3 1\n99 0", "output": "4\nmove 3 1\nmove 01 5\nmove 2extra 4\nmove 99 3"}, {"input": "2\n1 0\n2 1", "output": "3\nmove 1 3\nmove 2 1\nmove 3 2"}, {"input": "5\n1 0\n11 1\n111 0\n1111 1\n11111 0", "output": "5\nmove 1 5\nmove 11 1\nmove 1111 2\nmove 111 4\nmove 11111 3"}] | 2,200 | ["greedy", "implementation"] | 165 | [{"input": "5\r\n01 0\r\n2 1\r\n2extra 0\r\n3 1\r\n99 0\r\n", "output": "4\r\nmove 3 1\r\nmove 01 5\r\nmove 2extra 4\r\nmove 99 3\r\n"}, {"input": "2\r\n1 0\r\n2 1\r\n", "output": "3\r\nmove 1 dytuig\r\nmove 2 1\r\nmove dytuig 2\r\n"}, {"input": "5\r\n1 0\r\n11 1\r\n111 0\r\n1111 1\r\n11111 0\r\n", "output": "5\r\nmove... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline())
files = []
type1_count = 0
for _ in range(n):
line = f.readline().strip()
name, typ = line.s... | true |
598/C | 598 | C | PyPy 3-64 | TESTS | 117 | 1,045 | 16,588,800 | 142613751 | from math import atan2
arr = []
n = int(input())
for j in range(n):
x, y = [int(i) for i in input().split()]
arr += [(atan2(x, y), j)]
arr = sorted(arr)
gap = 10
for x, y in zip(arr, arr[1:] + [(arr[0][0] + 2*3.1415926, arr[0][1])]):
if abs(x[0] - y[0]) < gap:
gap = abs(x[0] - y[0])
ans =... | 163 | 1,699 | 26,009,600 | 154838623 | import math
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def quad(self):
return self.y > 0 or (self.y == 0 and self.x > 0)
def len(self):
return self.x * self.x + self.y * self.y
def __lt__(self, other):
if self.quad() != other.quad():
... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | PyPy 3 | TESTS | 113 | 483 | 10,547,200 | 168322502 | import sys
readline=sys.stdin.readline
import math
class Vector2():
def __init__(self,x,y):
self.x=x
self.y=y
def __lt__(self,other):
if (other.y,other.x)<(0,0)<(self.y,self.x):
return True
if (self.y,self.x)<(0,0)<(other.y,other.x):
return False
... | 163 | 779 | 41,369,600 | 175306822 | from sys import stdin
input=lambda :stdin.readline()[:-1]
def compare(a,b):
# 比較関数 (a<=b)
# 偏角ソート
ax, ay = a[:2]
bx, by = b[:2]
if ay < 0:
return by >= 0 or ax * by - ay * bx > 0
if ay == 0:
return ax >= 0 and (by > 0 or (by == 0 and bx < 0))
return by >= 0 and (ax * by - ay * bx) > 0
def me... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | PyPy 3-64 | TESTS | 118 | 1,419 | 14,950,400 | 202054936 | # https://codeforces.com/contest/598/problem/C
from math import atan2, pi
def diff(angle1, angle2):
delta = abs(angle1 - angle2)
return delta if delta <= pi else 2 * pi - delta
n = int(input())
vecs = []
for i in range(1, n + 1):
x, y = map(int, input().split())
angle = atan2(y, x)
if angle < 0:
... | 163 | 1,060 | 35,328,000 | 154855646 | from math import atan2
s = lambda a, b: a[0] * b[0] + a[1] * b[1]
v = lambda a, b: a[0] * b[1] - a[1] * b[0]
p = []
for i in range(int(input())):
x, y = map(int, input().split())
p.append((atan2(x, y), (x, y), i + 1))
p.sort()
d = [(s(a, b), abs(v(a, b)), i, j) for (x, a, i), (y, b, j) in zip(p, p[1:] + p... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | PyPy 3-64 | TESTS | 144 | 904 | 24,678,400 | 136345958 | import math
def dot_product(ax, ay, bx, by):
return ax * bx + ay * by
def cross_product(ax, ay, bx, by):
return ax * by - ay * bx
n = int(input())
A = []
for i in range(n):
a, b = map(int, input().split())
A.append((a, b, i))
A.sort(key=lambda x: math.atan2(cross_product(x[0], x[1], 1, 0), dot_prod... | 163 | 1,153 | 20,992,000 | 112977689 | from math import atan2
inner = lambda a, b: a[0] * b[0] + a[1] * b[1]
outer = lambda a, b: a[0] * b[1] - a[1] * b[0]
N = int(input())
vectors = []
for i in range(N):
x, y = map(int, input().split())
vectors.append((atan2(y, x), (x, y), i + 1))
vectors.sort()
diff = []
for i in range(N):
diff.append([inne... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
598/C | 598 | C | PyPy 3 | TESTS | 113 | 389 | 10,035,200 | 205154019 | import math
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
z = []
for i in range(n):
x, y = map(int, input().split())
u = math.degrees(math.atan2(y, x))
z.append((u, i + 1))
z.sort()
mi = 360
for i in range(n):
x1, u = z[i]
x2, v = z[(i + 1) % n]
... | 163 | 1,185 | 29,696,000 | 197816661 | from functools import cmp_to_key
def is_top(vector):
return vector[1] > 0 or vector[1] == 0 and vector[0] > 0
def cross(first, second):
return first[0] * second[1] - second[0] * first[1]
def dot(first, second):
return first[0]*second[0] + first[1]*second[1]
def polar_cmp(first, second):
if is_to... | Educational Codeforces Round 1 | ICPC | 2,015 | 2 | 256 | Nearest vectors | You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.
Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 ... | First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.
The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed t... | Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any. | null | null | [{"input": "4\n-1 0\n0 -1\n1 0\n1 1", "output": "3 4"}, {"input": "6\n-1 0\n0 -1\n1 0\n1 1\n-4 -5\n-4 -6", "output": "6 5"}] | 2,300 | ["geometry", "sortings"] | 163 | [{"input": "4\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n", "output": "3 4\r\n"}, {"input": "6\r\n-1 0\r\n0 -1\r\n1 0\r\n1 1\r\n-4 -5\r\n-4 -6\r\n", "output": "5 6\r\n"}, {"input": "10\r\n8 6\r\n-7 -3\r\n9 8\r\n7 10\r\n-3 -8\r\n3 7\r\n6 -8\r\n-9 8\r\n9 2\r\n6 7\r\n", "output": "1 3\r\n"}, {"input": "20\r\n-9 8\r\n-7 3\r\n0 10\r\... | false | stdio | import sys
import math
def read_vectors(input_path):
with open(input_path) as f:
n = int(f.readline())
vectors = []
for i in range(n):
x, y = map(int, f.readline().split())
vectors.append( (x, y, i+1) )
return vectors
def compute_angle(x, y):
return math.at... | true |
837/B | 837 | B | Python 3 | TESTS | 79 | 124 | 307,200 | 56131685 | altura, largura = map(int, input().split())
band = [input() for i in range(altura)]
new_bandeira = [''.join(coluna) for coluna in zip(*band)]
def confere(bandeira, altura):
flag = True
for linha in bandeira:
cor = linha[0]
if cor * len(linha) != linha:
flag = False
brea... | 153 | 62 | 4,608,000 | 29234268 | #! /usr/bin/python3
import sys
def verifyFlag(flagList, row, column):
if row%3 != 0:
return False
q = row // 3 #since there are 3 equal parts
u = [flagList[(0 * q)], flagList[(1 * q)], flagList[(2 * q)]]
us = set(u) #removes duplicates
patternToMatch = set()
for color in ['R', 'G', 'B... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 28 | 46 | 0 | 188425791 | def is_it_rated():
n = int(input())
before, after = list(), list()
for i in range(n):
b, a = map(int, input().split())
before.append(b)
after.append(a)
maybe = False
for i in range(n):
if before[i] != after[i]:
return "rated"
max_till = before[0]
for i in range(1,n):
if before[i] <= max_till:
... | 150 | 46 | 0 | 141227803 | l=[];l1=[]
t=int(input())
for i in range(t):
a,b=map(int,input().split())
l.append(a);l1.append(b)
p=l
x=[i-j for (i,j) in zip(l,l1)]
y=sorted(p,reverse=True)
if x.count(0)!=t:
print("rated")
elif x.count(0)==t and y!=l:
print("unrated")
else:
print("maybe") | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 28 | 46 | 0 | 145144892 | n=int(input())
a,b=0,0
counter=0
List=[]
for i in range(n):
a,b=map(int,input().split())
if a!=b:
counter+=1
List.append([a,b])
if counter!=0:
print("rated")
else:
minValue=min(List)
if minValue !=List[len(List)-1]:
print("unrated")
else:
print("maybe") | 150 | 46 | 0 | 144055840 | w=int(input())
l=[]
flag=0
for i in range(w):
t=list(map(int,input().split()))
if t[0]!=t[1]:
flag=1
else:
l.append(t[0])
if flag!=1:
y=list(l)
y.sort(reverse=True)
if y==l:
print('maybe')
else:
print('unrated')
else:
print('rated') | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
837/B | 837 | B | PyPy 3 | TESTS | 24 | 186 | 4,403,200 | 118945367 | import sys
#import random
from bisect import bisect_left as lb
from collections import deque
#sys.setrecursionlimit(10**8)
from queue import PriorityQueue as pq
from math import *
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda ... | 153 | 62 | 4,915,200 | 29161993 | import sys
def main():
n, m = map(int, sys.stdin.readline().split())
if n % 3 != 0 and m % 3 != 0:
print("NO")
return
f = []
for i in range(n):
f.append(sys.stdin.readline())
ok = True
if f[0][0] == f[n - 1][0]: # vertical
if m % 3 != 0:
ok = False... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
837/B | 837 | B | Python 3 | TESTS | 24 | 61 | 4,915,200 | 29167831 | n, m = map(int, input().split())
f = []
for i in range(n):
f.append(input()) #flag
rf1 = [] #real flag
rez = True
if (n % 3 == 0) or (m % 3 == 0):
if n % 3 == 0:
for i in range(n // 3):
rf1.append('R' * m)
for i in range(n // 3):
rf1.append('B' * m)
for i in range... | 153 | 62 | 4,915,200 | 29162004 | n, m = map(int, input().split())
flag = []
def letterwidth(i):
res = flag[i][0]
for item in flag[i]:
if item != res:
return None
return res
def letterheight(i):
res = flag[0][i]
for j in range(n):
if flag[j][i] != res:
return None
return res
for i in ra... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 21 | 46 | 0 | 225697559 | n=int(input())
before_round=[]
after_round=[]
sum_before=0
sum_after=0
for i in range(n):
before,after=map(int,input().split())
sum_before+=before
sum_after+=after
before_round.append(before)
after_round.append(after)
if(sum_before!=sum_after):
print("rated")
else:
if(before_round!=s... | 150 | 46 | 0 | 144294915 | a = []
b = []
for i in range(int(input())):
input2 = list(map(int,input().split()))
if input2[0] < input2[1]:
a.append(input2[0])
b.append(input2[1])
break
else:
a.append(input2[0])
b.append(input2[1])
if a != b:
print("rated")
else:
a.sort()
a = ... | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
837/B | 837 | B | Python 3 | TESTS | 62 | 62 | 4,915,200 | 29204655 | N, M = map(int, input().split())
flag = [input() for n in range(N)]
def get_rect_of_color(color):
rect_points = []
for i, row in enumerate(flag):
first = True
tmp = [None] * 4
for j, char in enumerate(row):
if char == color:
if first:
fir... | 153 | 62 | 4,915,200 | 29164140 | a, b = map(int, input().split())
rows = [list(input()) for x in range(a)]
columns = [[x[y] for x in rows] for y in range(b)]
def check(l):
line = []
for x in l:
p = x[0]
for y in x:
if y != p:
break
else:
line.append(p)
continue
... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
837/B | 837 | B | Python 3 | TESTS | 24 | 62 | 4,915,200 | 29273806 | '''input
4 4
RRRR
RRRR
BBBB
GGGG
'''
n, m = map(int, input().split())
if (n*m) % 3 != 0:
print("NO")
else:
f = [input() for _ in range(n)]
if n % 3 == 0:
for x in f:
if len(set(x)) != 1:
print("NO")
break
else:
if any(len(set(f[(n*(i-1))//3:(n*i)//3])) != 1 for i in range(1, 3)):
print("NO")
... | 153 | 62 | 4,915,200 | 29165939 | def check(n, m, fl, count):
global flag, tr_flag
if count == 3:
return 'NO'
num = n // 3
is_ok = set()
for k in range(0, n, num):
new_check = set()
for i in range(k, k + num):
new_check = new_check | set(fl[i])
if len(new_check) != 1:
flag, tr_flag = tr_flag, flag
if m % 3 ==... | Educational Codeforces Round 26 | ICPC | 2,017 | 1 | 256 | Flag of Berland | The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each c... | The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field. | Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes). | null | The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1. | [{"input": "6 5\nRRRRR\nRRRRR\nBBBBB\nBBBBB\nGGGGG\nGGGGG", "output": "YES"}, {"input": "4 3\nBRG\nBRG\nBRG\nBRG", "output": "YES"}, {"input": "6 7\nRRRGGGG\nRRRGGGG\nRRRGGGG\nRRRBBBB\nRRRBBBB\nRRRBBBB", "output": "NO"}, {"input": "4 4\nRRRR\nRRRR\nBBBB\nGGGG", "output": "NO"}] | 1,600 | ["brute force", "implementation"] | 153 | [{"input": "6 5\r\nRRRRR\r\nRRRRR\r\nBBBBB\r\nBBBBB\r\nGGGGG\r\nGGGGG\r\n", "output": "YES\r\n"}, {"input": "4 3\r\nBRG\r\nBRG\r\nBRG\r\nBRG\r\n", "output": "YES\r\n"}, {"input": "6 7\r\nRRRGGGG\r\nRRRGGGG\r\nRRRGGGG\r\nRRRBBBB\r\nRRRBBBB\r\nRRRBBBB\r\n", "output": "NO\r\n"}, {"input": "4 4\r\nRRRR\r\nRRRR\r\nBBBB\r\nG... | false | stdio | null | true |
99/B | 99 | B | PyPy 3 | TESTS | 136 | 139 | 1,433,600 | 110947011 | n = int(input())
d = {}
for i in range(n):
s = int(input())
if s in d:
d[s] += [i + 1]
else:
d[s] = [i + 1]
if len(d) == 1:
print("Exemplary pages.")
exit()
mx = max(d)
mn = min(d)
if len(d) > 3 or len(d[mx]) != 1 or len(d[mn]) != 1:
print("Unrecoverable configuration.")
... | 200 | 46 | 0 | 136745255 | n=int(input())
a=[0]*n
for i in range(n):
a[i]=int(input())
avg=sum(a)
if avg%n!=0:
print('Unrecoverable configuration.')
exit(0)
avg=avg//n
x=[]
for i in range(n):
if a[i]!=avg:
x.append(i)
if len(x)==0:
print('Exemplary pages.')
elif len(x)==2:
if (a[x[0]]+a[x[1]])//2==avg:
... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
99/B | 99 | B | Python 3 | TESTS | 136 | 77 | 512,000 | 111230120 | from operator import itemgetter
import functools as ft
if __name__ == '__main__':
n = int(input())
s = lambda v, a, b: "%s ml. from cup #%s to cup #%s." % (v, a, b)
a, d, ans = list(), dict(), ["Exemplary pages.", "Unrecoverable configuration."]
for i in range(n):
k = int(input())
a.ap... | 200 | 46 | 0 | 144160247 | n = int(input())
a = []
for i in range(n):
a.append(int(input()))
if(sum(a)%n != 0):
print("Unrecoverable configuration.")
else:
k = sum(a)//n
res = []
flag = 0
for i in range(n):
if(a[i] != k):
flag = 1
res.append([a[i],i])
if(flag == 0):
print("Exemplary pages.")
else:
if(len(res) == 2):
res.... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 34 | 62 | 307,200 | 106524131 | n=int(input())
a=[0]*n
flag=-1
for i in range(n):
if flag==1:continue
p, q=[int(y) for y in input().split()]
a[i]=p-q
if i==0:
temp1 = p
temp2 = a[i]
else:
if temp2!=a[i]:
flag=1
elif temp1<p:
flag=0
else:
temp1=p
if flag==... | 150 | 46 | 0 | 145145439 | n=int(input())
a,b=0,0
counter=0
List=[]
for i in range(n):
a,b=map(int,input().split())
if a!=b:
counter+=1
List.append([a,b])
if counter!=0:
List.clear()
print("rated")
else :
check=False
for i in range(1,len(List)):
if List[i][0]>List[i-1][0]:
print("unrated... | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 32 | 202 | 1,740,800 | 89772717 | # i dont understand this either
def countremove(n, mod):
res = 0
for i in range(len(n)):
if mod[len(n)-1-i] == 3:
continue
if n[i] == '0':
res += 1
else:
break
return res
n = input()
mod = [int(x)%3 for x in n][::-1]
excess = sum(mod) % 3
ones = mod.count(1)
twos = mod.count(2)
if excess == 1:
remo... | 162 | 124 | 6,860,800 | 105961030 | import sys
n = list(input(""))
l = [int(c)%3 for c in n][::-1]
d = sum(l)%3
if (len(n) == 1):
if (n == ['3']):
print(3)
else:
print(-1)
sys.exit(0)
def li(x):
global n
del n[len(n)-1-l.index(x)]
del l[l.index(x)]
def cond(x):
x = x.lstrip('0')
if (len(x) == 0): return "0"
return x
if (d == 0):
pass
e... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
99/B | 99 | B | PyPy 3 | TESTS | 8 | 77 | 307,200 | 149628016 | from collections import defaultdict
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
x = [int(input()) for _ in range(n)]
cnt = defaultdict(lambda : 0)
for i in x:
cnt[i] += 1
c = list(cnt.keys())
c.sort()
if len(c) == 1:
ans = "Exemplary pages."
elif len(c) == 3... | 200 | 77 | 307,200 | 106310001 | n=int(input())
ar=[int(input()) for i in range(n)]
if(len(set(ar))==1):
print("Exemplary pages.")
quit()
mn=min(ar)
mx=max(ar)
mni=ar.index(mn)
mxi=ar.index(mx)
vl=(mx-mn)//2
ar[mni]+=vl
ar[mxi]-=vl
if(len(set(ar))==1):
print(str(vl)+" ml. from cup #"+str(mni+1)+" to cup #"+str(mxi+1)+".")
quit()
else:
... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
99/B | 99 | B | Python 3 | TESTS | 8 | 46 | 0 | 146841297 | l=[]
n=int(input())
for z in range(n):
l.append(int(input()))
ma,mi=max(l),min(l)
um=-1
for i in range(min(n,3)):
if l.count(l[i])==n-2 and l[i] not in (ma,mi):
um=l[i]
break
if l.count(l[0])==n:
print("Exemplary pages.")
elif um>-1 and (ma-mi)%2==0:
print((ma-mi)//2,"ml. from cup #"+str... | 200 | 92 | 6,963,200 | 128839381 | def help_chef():
count = int(input())
volumes = []
total = 0
for _ in range(count):
volume = int(input())
volumes.append(volume)
total += volume
if total % count != 0:
print("Unrecoverable configuration.")
elif len(set(volumes)) == 1:
print("Exemplary pages.")
else:
diffs = []
vol = None
sourc... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
99/B | 99 | B | Python 3 | TESTS | 8 | 46 | 0 | 151407236 | n=int(input())
s=[]
for i in range(n):
s.append(int(input()))
x=list(set(s))
if len(x)==1:
print("Exemplary pages.")
elif len(x)==3:
x.sort()
if s.count(x[0])==1 and s.count(x[2])==1 and (x[0]+x[2])/2==sum(s)/len(s):
print(f"{x[1]-x[0]} ml. from cup #{s.index(min(x))+1} to cup #{s.index(max(x))+1}.")
else:
pri... | 200 | 109 | 307,200 | 93070393 | n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
a=set(l)
if(len(a)==1):
print("Exemplary pages.")
else:
m=min(l)
n=max(l)
r=(n-m)//2
q=l.index(n)
w=l.index(m)
l[q]-=r
l[w]+=r
b=set(l)
if(len(b)==1):
print(str(r)+" ml. from cup #"+str(w+1)+" to cup #"+str... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
99/B | 99 | B | Python 3 | TESTS | 8 | 31 | 0 | 151408789 | n = int(input())
list = [0 for j in range(n)]
list2 = [0 for J in range(n)]
list3 = [0 for K in range(4)]
for i in range(n):
list[i] = int(input())
list1 = sorted(list)
result = 0
x = S = 0
num = int(list1[n // 2])
for j in range(n):
if int(list[j]) != num:
result += 1
list2[j] = num - int(list[... | 200 | 109 | 1,433,600 | 106359561 | n=int(input())
s=[]
for i in range(n):
s.append(int(input()))
x=list(set(s))
x.sort()
if len(x)==1:
print('Exemplary pages.')
elif len(s)==2 and sum(x)%2==0:
print('{} ml. from cup #{} to cup #{}.'.format((x[1]-x[0])//2,s.index(x[0])+1,s.index(x[1])+1))
elif len(x)==3 and s.count(x[0])==1 and s.count(x[2])==1 and... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 35 | 46 | 0 | 173044894 | l1=[]
l2=[]
for _ in range(int(input())):
a,b=map(int,input().split())
l1.append(a)
l2.append(b)
cnt=0
c=0
c2=0
for i in range(len(l1)-1):
if l1[i]!=l2[i]: cnt+=1
elif l1[i]<l1[i+1]: c+=1
else: c2+=1
if cnt!=0: print('rated')
elif c!=0: print('unrated')
else: print('maybe') | 150 | 46 | 0 | 145487418 | n = int(input())
p = {1:"rated", 2:"unrated", 3:"maybe"}
d = []
t = 2
noninc = True
for i in range(n):
a,b = map(int, input().split())
if a == b and t != 1:
if len(d) > 0 and noninc:
if d[-1] >= a:
t = 3
else:
noninc = False
t = 2
... | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 35 | 61 | 0 | 173533502 | a=int(input())
arr=[]
for i in range(a):
sub=list(map(int,input().split()))
arr.append(sub)
c=0
j=0
for i in range(a-1):
if arr[i][0]>arr[i][1] or arr[i][0]<arr[i][1]:
c=c+1
if arr[i][0]<arr[i+1][0]:
j=j+1
if c>0:
print('rated')
elif j>0:
print('unrated')
else:
print('ma... | 150 | 46 | 0 | 145885952 | n = int(input())
f,l=0,[]
for i in range(n):
h,k = map(int,input().split())
l.append(h)
if h!=k:
f=1
l1 = list(l)
l.sort(reverse=True)
if f==0 :
print(['unrated','maybe'][l1==l])
else:
print("rated") | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 27 | 93 | 7,372,800 | 26228548 | s = list(input())
def db3(n,sm):
new = n.copy()
sm %= 3
if sm==0:
return ''.join(new)
else:
if sm==2:
for j in range(1,len(new)):
if int(new[j])%3==2:
new[j] = ''
return ''.join(new)
for j in range(1,len(new)):
if int(new[j])%3==1:
new[j]=''
sm-=1
if sm==0:
return ''.join(n... | 162 | 140 | 5,529,600 | 25861795 | s = input()
def rm(s,m):
if s == None or len(s) == 1:
return None
i = len(s) - 1
while i >= 0:
if int(s[i]) % 3 == m:
break
i -= 1
if i == -1:
return None
else:
if i == 0:
k = i+1
while k < len(s) and s[k] == "0":
... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
807/A | 807 | A | Python 3 | TESTS | 35 | 109 | 6,963,200 | 84974838 | n = int(input())
ls = []
res = ''
for i in range(n):
ls.append(list(map(int,input().split())))
for i in range(n-1):
if ls[i][0]!=ls[i][1]:
res = "rated"
break
if res != 'rated':
for i in range(n-1):
if ls[i][1]<ls[i+1][1]:
res = "unrated"
break
else:
res = "maybe"
print(res) | 150 | 46 | 0 | 146642324 | n = int(input())
f_r = 0
f_u = 0
before_list = []
after_list = []
for i in range(n):
before, after = map(int, input().split())
if before != after:
f_r = 1
elif len(before_list) and before > before_list[-1]:
f_u = 1
before_list.append(before)
after_list.append(after)
if f_r:
prin... | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
807/A | 807 | A | Python 3 | TESTS | 35 | 109 | 6,963,200 | 86453139 | a=int(input())
n1=[]
m1=[]
for i in range(a):
n,m=map(int, input().split())
n1.append(n)
m1.append(m)
ans=0
cnt=1
i=0
while i<(a-1):
if n1[i]!=m1[i]:
ans=1
print("rated")
break
elif n1[i]>=n1[i+1]:
cnt+=1
i=i+1
if cnt!=a and ans==0:
print('unrated')
elif cnt==... | 150 | 46 | 0 | 146977522 | n =int(input())
l=[]
res='maybe'
for x in range(n):
b,a=map(int,input().split())
l.append([b,a])
if a!=b:
res='rated'
if res!='rated' :
c=0
while c<len(l)-1 :
if l[c][0]<l[c+1][0] :
res='unrated'
break
c+=1
print(res) | Codeforces Round 412 (rated, Div. 2, base on VK Cup 2017 Round 3) | CF | 2,017 | 2 | 256 | Is it rated? | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before ... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of th... | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | null | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, som... | [{"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884", "output": "rated"}, {"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400", "output": "unrated"}, {"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699", "output": "maybe"}] | 900 | ["implementation", "sortings"] | 150 | [{"input": "6\r\n3060 3060\r\n2194 2194\r\n2876 2903\r\n2624 2624\r\n3007 2991\r\n2884 2884\r\n", "output": "rated\r\n"}, {"input": "4\r\n1500 1500\r\n1300 1300\r\n1200 1200\r\n1400 1400\r\n", "output": "unrated\r\n"}, {"input": "5\r\n3123 3123\r\n2777 2777\r\n2246 2246\r\n2246 2246\r\n1699 1699\r\n", "output": "maybe\... | false | stdio | null | true |
99/B | 99 | B | PyPy 3-64 | TESTS | 62 | 62 | 0 | 228951081 | n = int(input())
volumes = [int(input()) for _ in range(n)]
total_volume = sum(volumes)
if total_volume % n == 0:
target_volume = total_volume // n
pour_from = None
pour_to = None
for i, volume in enumerate(volumes):
diff = target_volume - volume
if diff > 0:
if pour_from i... | 200 | 124 | 307,200 | 73945356 | n = int(input())
a = []
sum = 0
for i in range(n):
x = int(input())
sum += x
a.append(x)
ave = sum / n
if int(ave) == ave:
ave = int(ave)
b = set()
cnt = 0
for i in range(n):
if abs(ave - a[i])!=0:
b.add(abs(ave - a[i]))
t = ave - a[i]
if t > 0:
... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
858/E | 858 | E | PyPy 3 | TESTS | 0 | 61 | 0 | 155070236 | import sys
input = sys.stdin.buffer.readline
def process(A):
"""
6 options:
1. test is already type 1
and from 1 <= x <= m1
no moves needed.
2. test is type 1
and from m1+1 <= x <= m1+m2
move needed but NOTE
3. test is type 1
and neither.
move needed.
4. t... | 165 | 295 | 27,136,000 | 230751905 | n = int(input())
t = [1] + [0] * n
b, a = d = [], []
h, s = [], []
for i in range(n):
f, k = input().split()
d[int(k)].append(f)
m = len(a)
for i in a:
if i.isdigit() and i[0] != '0':
j = int(i)
if 0 < j <= m:
t[j] = 1
elif m < j <= n:
t[j] = -1
else... | Технокубок 2018 - Отборочный Раунд 1 | CF | 2,017 | 2 | 256 | Tests Renumeration | The All-Berland National Olympiad in Informatics has just ended! Now Vladimir wants to upload the contest from the Olympiad as a gym to a popular Codehorses website.
Unfortunately, the archive with Olympiad's data is a mess. For example, the files with tests are named arbitrary without any logic.
Vladimir wants to re... | The first line contains single integer n (1 ≤ n ≤ 105) — the number of files with tests.
n lines follow, each describing a file with test. Each line has a form of "name_i type_i", where "name_i" is the filename, and "type_i" equals "1", if the i-th file contains an example test, and "0" if it contains a regular test. ... | In the first line print the minimum number of lines in Vladimir's script file.
After that print the script file, each line should be "move file_1 file_2", where "file_1" is an existing at the moment of this line being run filename, and "file_2" — is a string of digits and small English letters with length from 1 to 6. | null | null | [{"input": "5\n01 0\n2 1\n2extra 0\n3 1\n99 0", "output": "4\nmove 3 1\nmove 01 5\nmove 2extra 4\nmove 99 3"}, {"input": "2\n1 0\n2 1", "output": "3\nmove 1 3\nmove 2 1\nmove 3 2"}, {"input": "5\n1 0\n11 1\n111 0\n1111 1\n11111 0", "output": "5\nmove 1 5\nmove 11 1\nmove 1111 2\nmove 111 4\nmove 11111 3"}] | 2,200 | ["greedy", "implementation"] | 165 | [{"input": "5\r\n01 0\r\n2 1\r\n2extra 0\r\n3 1\r\n99 0\r\n", "output": "4\r\nmove 3 1\r\nmove 01 5\r\nmove 2extra 4\r\nmove 99 3\r\n"}, {"input": "2\r\n1 0\r\n2 1\r\n", "output": "3\r\nmove 1 dytuig\r\nmove 2 1\r\nmove dytuig 2\r\n"}, {"input": "5\r\n1 0\r\n11 1\r\n111 0\r\n1111 1\r\n11111 0\r\n", "output": "5\r\nmove... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline())
files = []
type1_count = 0
for _ in range(n):
line = f.readline().strip()
name, typ = line.s... | true |
792/C | 792 | C | Python 3 | TESTS | 28 | 93 | 5,734,400 | 25896525 | def sumstr(s):
sums = 0
for i in range(len(s)):
sums += int(s[i])
return sums
def mod3(s):
sums = sumstr(s)
return sums % 3
def onlyzeros(s):
i = 0
while i < len(s):
if s[i] != '0':
return False
i += 1
return True
def lastin(s, s1):
maxf = -1
#d = '-'
for d in s1:
f = s.find(d)
if f > maxf:
... | 162 | 140 | 28,262,400 | 32921730 | def f(t):
i, n = 0, len(t) - 1
while i < n and t[i] == '0': i += 1
return t[i:]
t = input()
n = len(t) - 1
p = [int(q) % 3 for q in t][::-1]
s = sum(p) % 3
if s == 0:
print(t)
exit()
u = v = ''
if s in p:
i = n - p.index(s)
u = f(t[:i] + t[i + 1:])
s = 3 - s
if p.count(s) > 1:
i = n - p.... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
750/B | 750 | B | PyPy 3-64 | TESTS | 28 | 62 | 0 | 197427400 | n = int(input())
dis=[]
direc = []
for j in range(n):
a,s = map(str,input().split())
dis.append(int(a))
direc.append(s)
if(direc[0]!='South'):
print('NO')
else:
lenght=0
flag=0
for k in range(n):
if(lenght<=20000):
if(direc[k]=='South'):
lenght =... | 140 | 46 | 0 | 213825326 | n = int(input())
k1 = 0
k2 = 0
for i in range(n):
s = input().split()
a = int(s[0])
b = s[1]
if k1 == 0 and b != 'South':
k2 = 1
if k1 == 20000 and b != 'North':
k2 = 1
if b =='South':
k1 += a
if b == 'North':
k1 -= a
if k1 > 20000 or k1 < 0:
k2 = ... | Good Bye 2016 | CF | 2,016 | 2 | 256 | New Year and North Pole | In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the N... | The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, $${ dir } _ { i } \in \{ \mathrm { N o r t h, ~ S o u t h, ~ W e s t, ~ E a s t } \}$$) — the length and the direction of the i-th part of the journey, according to the... | Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes. | null | Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole. | [{"input": "5\n7500 South\n10000 East\n3500 North\n4444 West\n4000 North", "output": "YES"}, {"input": "2\n15000 South\n4000 East", "output": "NO"}, {"input": "5\n20000 South\n1000 North\n1000000 West\n9000 North\n10000 North", "output": "YES"}, {"input": "3\n20000 South\n10 East\n20000 North", "output": "NO"}, {"input... | 1,300 | ["geometry", "implementation"] | 140 | [{"input": "5\r\n7500 South\r\n10000 East\r\n3500 North\r\n4444 West\r\n4000 North\r\n", "output": "YES\r\n"}, {"input": "2\r\n15000 South\r\n4000 East\r\n", "output": "NO\r\n"}, {"input": "5\r\n20000 South\r\n1000 North\r\n1000000 West\r\n9000 North\r\n10000 North\r\n", "output": "YES\r\n"}, {"input": "3\r\n20000 Sout... | false | stdio | null | true |
99/B | 99 | B | PyPy 3 | TESTS | 50 | 140 | 0 | 60445506 | __author__ = 'Esfandiar'
n = int(input())
a = list()
for i in range(n):
a.append([int(input()),i+1])
a.sort()
j = a[0][0]
f = 1
s = j
for i in range(1,n):
if a[i][0] != j:
f = 0
s+=a[i][0]
if f == 1:
print('Exemplary pages.')
exit()
yy = s/n
if int(yy) != yy:
print('Unrecoverable config... | 200 | 124 | 307,200 | 82599994 | n=int(input())
x=[]
for i in range(n):
x.append(int(input()))
y=sorted(x)
z=sorted(x,reverse=True)
if y==z:print('Exemplary pages.')
else:
z=y.copy();t=[]
while(y[n-1]>0):
y[0]+=1;y[n-1]-=1
if y[0]==y[n-1]:break
y.sort();t=sorted(y,reverse=True)
if y==t:print('%d ml. from cup #%d to ... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
792/C | 792 | C | Python 3 | TESTS | 28 | 139 | 11,673,600 | 25889964 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().spl... | 162 | 171 | 6,041,600 | 57314819 | def find(s):
if len(s)==1:
if int(s)%3==0:
return s
return -1
digit=[[] for i in range(3)]
zeros=[]
for i in range(len(s)):
digit[int(s[i])%3]+=[i]
if s[i]=="0":
zeros+=[i]
total=(len(digit[1])*1+len(digit[2])*2)%3
if total==0:
... | Educational Codeforces Round 18 | ICPC | 2,017 | 1 | 256 | Divide by Three | A positive integer number n is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zer... | The first line of input contains n — a positive integer number without leading zeroes (1 ≤ n < 10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print - 1. | null | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | [{"input": "1033", "output": "33"}, {"input": "10", "output": "0"}, {"input": "11", "output": "-1"}] | 2,000 | ["dp", "greedy", "math", "number theory"] | 162 | [{"input": "1033\r\n", "output": "33\r\n"}, {"input": "10\r\n", "output": "0\r\n"}, {"input": "11\r\n", "output": "-1\r\n"}, {"input": "3\r\n", "output": "3\r\n"}, {"input": "1\r\n", "output": "-1\r\n"}, {"input": "117\r\n", "output": "117\r\n"}, {"input": "518\r\n", "output": "18\r\n"}, {"input": "327\r\n", "output": ... | false | stdio | import sys
def is_subsequence(s, t):
t_iter = iter(t)
try:
for c in s:
while next(t_iter) != c:
pass
return True
except StopIteration:
return False
def is_beautiful(s):
if not s:
return False
if s[0] == '0' and len(s) > 1:
return ... | true |
99/B | 99 | B | Python 3 | TESTS | 8 | 31 | 0 | 167498988 | n = int(input())
s = list()
s1 = set()
for _ in range(n):
enter = int(input())
s.append(enter)
s1.add(enter)
if len(s1) == 1:
print("Exemplary pages.")
elif len(s1) == 3:
s1 = list(s1)
for i in range(3):
if s.count(s1[i]) > 1:
x = s1[(i + 1) % 3]
y = s1[(i - 1) % ... | 200 | 124 | 307,200 | 90197114 | n=int(input())
a=[]
for i in range(1,n+1): a.append(int(input()))
mina,maxa=min(a),max(a)
minai,maxai=a.index(mina),a.index(maxa)
if (maxa-mina)%2!=0:
print("Unrecoverable configuration.")
elif mina==maxa:
print("Exemplary pages.")
else:
mid=(maxa-mina)//2
a[maxai]-=mid
a[minai]+=mid
if len(set(... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
818/E | 818 | E | PyPy 3 | TESTS | 13 | 93 | 1,843,200 | 106262915 | n,k=map(int,input().split())
l=list(map(int,input().split()))
ans=0
last=0
now=1
for i in range(len(l)):
now=now*l[i]
while now%k==0 and last<=i:
ans+=n-i
now/=l[last]
last+=1
print(ans) | 135 | 77 | 13,414,400 | 170114604 | R,G=lambda:map(int,input().split()),range
n,k=R();a=[0]+[*R()];z,l,p=0,1,1
for r in G(1,n+1):
p=p*a[r]%k
if p==0:
p=1;i=r
while p*a[i]%k:p=p*a[i]%k;i-=1
z+=(n-r+1)*(i-l+1);l=i+1
print(z) | Educational Codeforces Round 24 | ICPC | 2,017 | 2 | 256 | Card Game Again | Vova again tries to play some computer card game.
The rules of deck creation in this game are simple. Vova is given an existing deck of n cards and a magic number k. The order of the cards in the deck is fixed. Each card has a number written on it; number ai is written on the i-th card in the deck.
After receiving th... | The first line contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the numbers written on the cards. | Print the number of ways to choose x and y so the resulting deck is valid. | null | In the first example the possible values of x and y are:
1. x = 0, y = 0;
2. x = 1, y = 0;
3. x = 2, y = 0;
4. x = 0, y = 1. | [{"input": "3 4\n6 2 8", "output": "4"}, {"input": "3 6\n9 1 14", "output": "1"}] | 1,900 | ["binary search", "data structures", "number theory", "two pointers"] | 135 | [{"input": "3 4\r\n6 2 8\r\n", "output": "4\r\n"}, {"input": "3 6\r\n9 1 14\r\n", "output": "1\r\n"}, {"input": "5 1\r\n1 3 1 3 1\r\n", "output": "15\r\n"}, {"input": "5 1\r\n5 5 5 5 5\r\n", "output": "15\r\n"}, {"input": "5 1\r\n5 4 4 4 4\r\n", "output": "15\r\n"}, {"input": "100 1\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1... | false | stdio | null | true |
99/B | 99 | B | PyPy 3-64 | TESTS | 39 | 77 | 614,400 | 149230725 | import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
from collections import defaultdict
arr = []
sm = 0
n = I()
for _ in range(n):
t = I()
sm+=t
arr.append(t)
each = sm//n
over_under = [0]*(n)
... | 200 | 124 | 307,200 | 103664743 | n = int(input())
a = []
sum = 0
for i in range(n):
a.append(int(input()))
sum+=a[i]
c = sum//n
fro = 0
to = 0
cou = 0
for i in range(n):
if c<a[i]:
to = i+1
cou+=1
elif c>a[i]:
fro = i+1
cou+=1
amt = (a[to-1]-a[fro-1])//2
a[to-1]-=amt
a[fro-1]+=amt
if fro == to:
print... | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 0.5 | 256 | Help Chef Gerasim | In a far away kingdom young pages help to set the table for the King. As they are terribly mischievous, one needs to keep an eye on the control whether they have set everything correctly. This time the royal chef Gerasim had the impression that the pages have played a prank again: they had poured the juice from one cup... | The first line contains integer n — the number of cups on the royal table (1 ≤ n ≤ 1000). Next n lines contain volumes of juice in each cup — non-negative integers, not exceeding 104. | If the pages didn't pour the juice, print "Exemplary pages." (without the quotes). If you can determine the volume of juice poured during exactly one juice pouring, print "v ml. from cup #a to cup #b." (without the quotes), where v represents the volume of poured juice, a represents the number of the cup from which the... | null | null | [{"input": "5\n270\n250\n250\n230\n250", "output": "20 ml. from cup #4 to cup #1."}, {"input": "5\n250\n250\n250\n250\n250", "output": "Exemplary pages."}, {"input": "5\n270\n250\n249\n230\n250", "output": "Unrecoverable configuration."}] | 1,300 | ["implementation", "sortings"] | 200 | [{"input": "5\r\n270\r\n250\r\n250\r\n230\r\n250\r\n", "output": "20 ml. from cup #4 to cup #1.\r\n"}, {"input": "5\r\n250\r\n250\r\n250\r\n250\r\n250\r\n", "output": "Exemplary pages.\r\n"}, {"input": "5\r\n270\r\n250\r\n249\r\n230\r\n250\r\n", "output": "Unrecoverable configuration.\r\n"}, {"input": "4\r\n200\r\n190\... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 154 | 6,963,200 | 124043707 | n = input()
num = n.split(".")
the_num = int(num[0])
dec_part = int(num[1])
p = (len(num[1]))
the_dec_part = int(num[1])/pow(10, p)
if(the_num%10 == 9):
print('GOTO Vasilisa.')
elif(the_num%10 != 9 and the_dec_part<0.5):
print(the_num)
elif(the_num%10 != 9 and the_dec_part>=0.5):
print(the_num+1) | 150 | 92 | 0 | 4701450 | t = input()
k = t.find('.')
if k < 0: print(int(t))
elif t[k - 1] == '9': print('GOTO Vasilisa.')
else: print(int(t[:k]) + int(t[k + 1] > '4')) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 124 | 0 | 116240399 | def f(s):
i=s.index(".")
if s[i-1]=="9":
return "GOTO Vasilisa."
if float(s[i:])<0.5:
return s[:i]
else:
return int(s[:i])+1
s=input()
print(f(s)) | 150 | 92 | 0 | 136128059 | a = input()
for i in range(len(a)):
if a[i] == "." and a[i-1] == "9":
print("GOTO Vasilisa.")
break
if a[i] == "." and a[i+1] >= "5":
print(int(a[:i])+1)
break
if a[i] == "." and a[i+1] <= "5":
print(a[:i])
break | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | PyPy 3-64 | TESTS | 134 | 124 | 0 | 195444322 | s = input()
pos = s.find('.')
i = int(s[0:pos])
dec = float(s[pos::])
if(i%10 != 9) and(dec<0.5):
print(i)
elif(i%10 != 9) and(dec>=0.5):
print(i+1)
else:
print("GOTO Vasilisa.") | 150 | 92 | 0 | 139120426 | m,n= input().split(".")
if m[-1]=="9":
print("GOTO Vasilisa.")
else:
l=n[0]>"4"
print(int(m)+l) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 122 | 0 | 186991953 | n = input()
if '.' not in n:
print(n)
exit()
else:
m = n.split('.')
if m[0][-1] == "9":
print("GOTO Vasilisa.")
exit()
elif float("0." + m[1]) < 0.5:
print(int(m[0]))
elif float("0." + m[1]) >= 0.5:
print(int(m[0]) + 1) | 150 | 92 | 0 | 142060203 | i, j=map(str,input().split("."))
print(["GOTO Vasilisa.", int(i)+(j[0]>'4')][i[-1]<'9'])
# Sun Jan 09 2022 08:29:38 GMT+0000 (Coordinated Universal Time) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 92 | 307,200 | 4816627 | number = input()
int_part = ''
i=0
while number[i] != '.':
int_part += number[i]
i=i+1
fraction_part = number[i:]
if int_part[-1] != '9' and float(fraction_part) < 0.5:
round_no = int_part
elif int_part[-1] != '9' and float(fraction_part) >= 0.5:
round_no = int_part[0:-1] + str(int(int_part[-1])+1)
el... | 150 | 92 | 0 | 144267071 | s = input()
if s[s.index(".")-1] == '9': print("GOTO Vasilisa.")
else:
if int(s[s.index(".")+1]) < 5:
print(s[:s.index(".")])
else:
print(int(s[:s.index(".")])+1) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 92 | 0 | 225653365 | x, y = map(str, input(). split("."))
if x[-1] == "9" or x == "9":
print("GOTO Vasilisa.")
elif float("0." + y) < 0.5:
print(int(x))
else:
print(int(x)+1) | 150 | 92 | 0 | 145548415 | s = input()
nuqta = s.index(".")
butun_qism = int(s[0:nuqta])
kasr_qism = int(s[nuqta+1:])
if(butun_qism%10 == 9):
print("GOTO Vasilisa.")
else:
if(int(s[nuqta+1]) >= 5):
print(butun_qism+1)
else:
print(butun_qism) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 92 | 0 | 165454011 | i, d = tuple(input().split("."))
if i[-1] == "9":
print("GOTO Vasilisa.")
elif float(f"0.{d}") < 0.5:
print(int(i))
elif float(f"0.{d}") >= 0.5:
print(int(i) + 1) | 150 | 92 | 0 | 146844916 | a = input()
if int(a[a.index(".")-1]) == 9:
print("GOTO Vasilisa.")
elif int(a[a.index(".")+1]) < 5:
print(int(a[:a.index(".")]))
else:
print(int(a[:a.index(".")])+1) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 92 | 0 | 169353780 | n=input()
if "." not in n:
print(int(n))
else:
p=n.index(".")
if n[p-1]=="9":
print("GOTO Vasilisa.")
else:
s=n[:p]
x="0."+n[p+1:]
if float(x)>=0.5:
print(int(s)+1)
else:
print(int(s)) | 150 | 92 | 0 | 149005367 | s=list(input());r=s.index('.')
if s[r-1]=='9':print('GOTO Vasilisa.')
else:
if s[r+1]>='5':s[r-1]=str(int(s[r-1])+1)
print(''.join(s[:r])) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | PyPy 3 | TESTS | 134 | 218 | 0 | 114531778 | number = input()
integer, decimal = number.split('.')
if int(integer) % 10 == 9:
print('GOTO Vasilisa.')
else:
decimal = '0.' + decimal
if float(decimal) >= 0.5:
print(int(integer) + 1)
else:
print(int(integer)) | 150 | 92 | 0 | 160042145 | number = input().split('.')
if int(number[1][0]) >= 5 and int(number[0][-1]) < 9:
print(int(number[0]) + 1)
elif int(number[1][0]) < 5 and int(number[0][-1]) < 9:
print(int(number[0]))
else:
print('GOTO Vasilisa.') | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
99/A | 99 | A | Python 3 | TESTS | 134 | 92 | 0 | 170905104 | s=input()
k=s.split('.')
p=k[0]
p=p[::-1]
if p[0]=='9':
print('GOTO Vasilisa.')
else:
q='0.'+k[1]
if float(q)<0.5:
print(int(k[0]))
else:
print(int(k[0])+1) | 150 | 92 | 0 | 165274022 | n=input()
n=n.split('.')
if n[0][-1]=='9':
print("GOTO Vasilisa.")
elif n[1][0]<'5':
print(n[0])
else:
print(int(n[0])+1) | Codeforces Beta Round 78 (Div. 2 Only) | CF | 2,011 | 2 | 256 | Help Far Away Kingdom | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Ki... | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empt... | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | null | null | [{"input": "0.0", "output": "0"}, {"input": "1.49", "output": "1"}, {"input": "1.50", "output": "2"}, {"input": "2.71828182845904523536", "output": "3"}, {"input": "3.14159265358979323846", "output": "3"}, {"input": "12345678901234567890.1", "output": "12345678901234567890"}, {"input": "123456789123456789.999", "output... | 800 | ["strings"] | 150 | [{"input": "0.0\r\n", "output": "0"}, {"input": "1.49\r\n", "output": "1"}, {"input": "1.50\r\n", "output": "2"}, {"input": "2.71828182845904523536\r\n", "output": "3"}, {"input": "3.14159265358979323846\r\n", "output": "3"}, {"input": "12345678901234567890.1\r\n", "output": "12345678901234567890"}, {"input": "12345678... | false | stdio | null | true |
750/B | 750 | B | Python 3 | TESTS | 8 | 46 | 0 | 211954607 | c=True
x=20000
n= int(input())
for i in range(n):
t,d=map(str,input().split())
if d != 'South' and i==0:
c=False
elif d != 'North' and i == (n-1):
c=False
else:
if d == 'South' and x != 0:
x-=int(t)
elif x==20000 and d=='North':
c=False
eli... | 140 | 62 | 0 | 232602852 | ll=[]
x=int(input())
n,s=20000,0
for _ in range(x):
i,j=input().split()
i=int(i)
if (n==20000 or s==20000) and (j=="West"or j=="East"):
print("NO")
exit()
elif (j=="South" and i+s>20000) or (j=="North" and i+n>20000):
print("NO")
exit()
else:
if j=="South":
... | Good Bye 2016 | CF | 2,016 | 2 | 256 | New Year and North Pole | In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the N... | The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, $${ dir } _ { i } \in \{ \mathrm { N o r t h, ~ S o u t h, ~ W e s t, ~ E a s t } \}$$) — the length and the direction of the i-th part of the journey, according to the... | Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes. | null | Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole. | [{"input": "5\n7500 South\n10000 East\n3500 North\n4444 West\n4000 North", "output": "YES"}, {"input": "2\n15000 South\n4000 East", "output": "NO"}, {"input": "5\n20000 South\n1000 North\n1000000 West\n9000 North\n10000 North", "output": "YES"}, {"input": "3\n20000 South\n10 East\n20000 North", "output": "NO"}, {"input... | 1,300 | ["geometry", "implementation"] | 140 | [{"input": "5\r\n7500 South\r\n10000 East\r\n3500 North\r\n4444 West\r\n4000 North\r\n", "output": "YES\r\n"}, {"input": "2\r\n15000 South\r\n4000 East\r\n", "output": "NO\r\n"}, {"input": "5\r\n20000 South\r\n1000 North\r\n1000000 West\r\n9000 North\r\n10000 North\r\n", "output": "YES\r\n"}, {"input": "3\r\n20000 Sout... | false | stdio | null | true |
750/B | 750 | B | Python 3 | TESTS | 1 | 30 | 0 | 210218600 | n = int(input())
journey_parts = []
for _ in range(n):
ti, diri = input().split()
journey_parts.append((int(ti), diri))
current_latitude = 0
for ti, diri in journey_parts:
if diri == "North":
current_latitude -= ti
elif diri == "South":
current_latitude += ti
if current_latitude ... | 140 | 62 | 4,608,000 | 23437510 | import math
n = int(input())
fi = 0
theta = 0
flag = True
for n in range(n):
row = [x for x in input().split()]
dist = int(row[0])
napr = row[1]
if (fi == 0 and napr != 'South') or fi < 0:
flag = False
break
if (fi == 20000 and napr != 'North') or fi > 20000:
flag = Fals... | Good Bye 2016 | CF | 2,016 | 2 | 256 | New Year and North Pole | In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the N... | The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, $${ dir } _ { i } \in \{ \mathrm { N o r t h, ~ S o u t h, ~ W e s t, ~ E a s t } \}$$) — the length and the direction of the i-th part of the journey, according to the... | Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes. | null | Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole. | [{"input": "5\n7500 South\n10000 East\n3500 North\n4444 West\n4000 North", "output": "YES"}, {"input": "2\n15000 South\n4000 East", "output": "NO"}, {"input": "5\n20000 South\n1000 North\n1000000 West\n9000 North\n10000 North", "output": "YES"}, {"input": "3\n20000 South\n10 East\n20000 North", "output": "NO"}, {"input... | 1,300 | ["geometry", "implementation"] | 140 | [{"input": "5\r\n7500 South\r\n10000 East\r\n3500 North\r\n4444 West\r\n4000 North\r\n", "output": "YES\r\n"}, {"input": "2\r\n15000 South\r\n4000 East\r\n", "output": "NO\r\n"}, {"input": "5\r\n20000 South\r\n1000 North\r\n1000000 West\r\n9000 North\r\n10000 North\r\n", "output": "YES\r\n"}, {"input": "3\r\n20000 Sout... | false | stdio | null | true |
979/B | 979 | B | Python 3 | TESTS | 137 | 187 | 7,782,400 | 38270585 | n = int(input())
s1 = input()
s2 = input()
s3 = input()
s = len(s1)
d1, d2, d3 = dict(), dict(), dict()
for i in s1:
if i not in d1:
d1[i] = 1
else:
d1[i] += 1
for i in s2:
if i not in d2:
d2[i] = 1
else:
d2[i] += 1
for i in s3:
if i not in d3:
d3[i] = 1
else:
d3[i] += 1
max1, max2, max3 = 0, 0, 0
for... | 184 | 93 | 2,969,600 | 217007184 | from collections import Counter
n = int(input())
total = [0] * 3
for i in range(3):
s = input()
a = Counter(s)
MAX = max(a.values())
if MAX == len(s) and n == 1:
MAX = len(s) - 1
else:
MAX += n
MAX = min(MAX, len(s))
total[i] = MAX
max = max(total)
if total.count(max) >... | Codeforces Round 482 (Div. 2) | CF | 2,018 | 1 | 256 | Treasure Hunt | After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination... | The first line contains an integer $$$n$$$ ($$$0 \leq n \leq 10^{9}$$$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $$$10^{5}$$$ uppercase and lowercase Latin letters and is not empty. It is guaranteed th... | Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw". | null | In the first example, after $$$3$$$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $$$5$$$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $$$4$$$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon int... | [{"input": "3\nKuroo\nShiro\nKatie", "output": "Kuro"}, {"input": "7\ntreasurehunt\nthreefriends\nhiCodeforces", "output": "Shiro"}, {"input": "1\nabcabc\ncbabac\nababca", "output": "Katie"}, {"input": "15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE", "output": "Draw"}] | 1,800 | ["greedy"] | 184 | [{"input": "3\r\nKuroo\r\nShiro\r\nKatie\r\n", "output": "Kuro\r\n"}, {"input": "7\r\ntreasurehunt\r\nthreefriends\r\nhiCodeforces\r\n", "output": "Shiro\r\n"}, {"input": "1\r\nabcabc\r\ncbabac\r\nababca\r\n", "output": "Katie\r\n"}, {"input": "15\r\nfoPaErcvJ\r\nmZaxowpbt\r\nmkuOlaHRE\r\n", "output": "Draw\r\n"}, {"in... | false | stdio | null | true |
979/B | 979 | B | Python 3 | TESTS | 98 | 124 | 8,601,600 | 38261709 | from collections import Counter
names = ["Kuro", "Shiro", "Katie"]
def solve(s, n):
d = Counter(s)
if len(d) == 1 and n == 1:
return n - 1
m = max(d.values())
return min(m+n,len(s))
n = int(input())
p = []
for _ in range(3):
p.append(input())
l = sorted([ (solve(p[i],n),i) for i in range... | 184 | 93 | 4,505,600 | 176881870 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
x = [0] * 3
for i in range(3):
s = list(input().rstrip())
cnt = [0] * 130
for j in s:
cnt[j] += 1
m = max(cnt)
c = min(m + n, len(s))
if m == len(s) and n == 1:
c = m - 1
x[i] = ... | Codeforces Round 482 (Div. 2) | CF | 2,018 | 1 | 256 | Treasure Hunt | After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination... | The first line contains an integer $$$n$$$ ($$$0 \leq n \leq 10^{9}$$$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $$$10^{5}$$$ uppercase and lowercase Latin letters and is not empty. It is guaranteed th... | Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw". | null | In the first example, after $$$3$$$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $$$5$$$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $$$4$$$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon int... | [{"input": "3\nKuroo\nShiro\nKatie", "output": "Kuro"}, {"input": "7\ntreasurehunt\nthreefriends\nhiCodeforces", "output": "Shiro"}, {"input": "1\nabcabc\ncbabac\nababca", "output": "Katie"}, {"input": "15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE", "output": "Draw"}] | 1,800 | ["greedy"] | 184 | [{"input": "3\r\nKuroo\r\nShiro\r\nKatie\r\n", "output": "Kuro\r\n"}, {"input": "7\r\ntreasurehunt\r\nthreefriends\r\nhiCodeforces\r\n", "output": "Shiro\r\n"}, {"input": "1\r\nabcabc\r\ncbabac\r\nababca\r\n", "output": "Katie\r\n"}, {"input": "15\r\nfoPaErcvJ\r\nmZaxowpbt\r\nmkuOlaHRE\r\n", "output": "Draw\r\n"}, {"in... | false | stdio | null | true |
797/B | 797 | B | PyPy 3 | TESTS | 16 | 124 | 20,172,800 | 79682834 | a=int(input())
z=list(map(int,input().split()))
oddp=[]
eve=[]
oddn=[]
for i in range(len(z)):
if(z[i]%2==0 and z[i]>0):
eve.append(z[i])
elif(z[i]%2==1):
if(z[i]>0):
oddp.append(z[i])
else:
oddn.append(z[i])
if(len(oddp)==0):
print(max(oddn)+sum(eve))
else:
... | 126 | 78 | 10,547,200 | 228848341 | input()
total_sum, min_odd = 0, 10**4
for num in map(int, input().split()):
if num > 0:
total_sum += num
if num % 2 != 0:
min_odd = min(abs(num), min_odd)
if total_sum % 2 != 1:
total_sum -= min_odd
print(total_sum) | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Odd sum | You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deletin... | The first line contains integer number n (1 ≤ n ≤ 105).
The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | null | In the first example sum of the second and the fourth elements is 3. | [{"input": "4\n-2 2 -3 1", "output": "3"}, {"input": "3\n2 -5 -3", "output": "-1"}] | 1,400 | ["dp", "greedy", "implementation"] | 126 | [{"input": "4\r\n-2 2 -3 1\r\n", "output": "3\r\n"}, {"input": "3\r\n2 -5 -3\r\n", "output": "-1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n-1\r\n", "output": "-1\r\n"}, {"input": "15\r\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046\r\n", "output": "53507\r\n"}, {... | false | stdio | null | true |
815/A | 815 | A | PyPy 3 | TESTS | 16 | 155 | 5,324,800 | 112037896 | n, m = map(int, input().split())
g = [list(map(int, input().split())) for _ in range(n)]
a = [0] * n
b = [0] * m
a[0] = min(g[0])
if a[0] > 0:
for j in range(m):
g[0][j] -= a[0]
b[j] = g[0][j]
good = True
for i in range(n):
for j in range(m):
g[i][j] -= b[j]
if min(g[i]) < 0 or ... | 177 | 108 | 3,788,800 | 27849257 | import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
n,m=map(int,input().split())
realg=[list(map(int,input().split())) for _ in range(n)]
g=[[0]*m for _ in range(n)]
ans=[]
# get differences
f1=min([realg[0][i] for i in range(m)])
for i in range(m):
for _ in range(realg[0][i]-f1):
ans.append... | Codeforces Round 419 (Div. 1) | CF | 2,017 | 2 | 512 | Karen and Game | On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To ... | The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500). | If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.
The next k lines should each contain one of the following, describing the moves in the order they must b... | null | In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required... | [{"input": "3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1", "output": "4\nrow 1\nrow 1\ncol 4\nrow 3"}, {"input": "3 3\n0 0 0\n0 1 0\n0 0 0", "output": "-1"}, {"input": "3 3\n1 1 1\n1 1 1\n1 1 1", "output": "3\nrow 1\nrow 2\nrow 3"}] | 1,700 | ["brute force", "greedy", "implementation"] | 177 | [{"input": "3 5\r\n2 2 2 3 2\r\n0 0 0 1 0\r\n1 1 1 2 1\r\n", "output": "4\r\nrow 1\r\nrow 1\r\ncol 4\r\nrow 3\r\n"}, {"input": "3 3\r\n0 0 0\r\n0 1 0\r\n0 0 0\r\n", "output": "-1\r\n"}, {"input": "3 3\r\n1 1 1\r\n1 1 1\r\n1 1 1\r\n", "output": "3\r\nrow 1\r\nrow 2\r\nrow 3\r\n"}, {"input": "3 5\r\n2 4 2 2 3\r\n0 2 0 0 ... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
n, m = map(int, f.readline().split())
grid = [list(map(int, f.readline().split())) for _ in range(n)]
possible = True
g1_1 = grid[0][0] if n... | true |
797/B | 797 | B | Python 3 | TESTS | 16 | 124 | 409,600 | 77743956 | from collections import Counter,defaultdict,deque
import heapq as hq
from itertools import count, islice
#alph = 'abcdefghijklmnopqrstuvwxyz'
#from math import factorial as fact
import math
import sys
input=sys.stdin.readline
#print=sys.stdout.write
#tt = int(input())
#total=0
#n = int(input())
#n,m,k = [int(x) for x ... | 126 | 92 | 13,721,600 | 219660805 | def main():
n = int(input())
a = list(map(int, input().split()))
sum_ = 0
for i in range(n):
if a[i] > 0:
sum_ += a[i]
if sum_ % 2 == 1:
return sum_
min_pos = sum_
max_neg = sum_
for i in range(n):
if a[i] > 0 and a[i] % 2 == 1:
min_pos = m... | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Odd sum | You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deletin... | The first line contains integer number n (1 ≤ n ≤ 105).
The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | null | In the first example sum of the second and the fourth elements is 3. | [{"input": "4\n-2 2 -3 1", "output": "3"}, {"input": "3\n2 -5 -3", "output": "-1"}] | 1,400 | ["dp", "greedy", "implementation"] | 126 | [{"input": "4\r\n-2 2 -3 1\r\n", "output": "3\r\n"}, {"input": "3\r\n2 -5 -3\r\n", "output": "-1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n-1\r\n", "output": "-1\r\n"}, {"input": "15\r\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046\r\n", "output": "53507\r\n"}, {... | false | stdio | null | true |
797/B | 797 | B | PyPy 3 | TESTS | 37 | 233 | 10,752,000 | 84147165 | n=int(input())
l=sorted([int(i) for i in input().split()])
s=0
c=-1
c1=0
c2=0
for i in range(n):
if(l[i]>=0):
s+=l[i]
else:
c=i
if(c<n-1):
if(s%2==1):
print(s)
else:
for i in range(c,-1,-1):
if(l[i]%2==1):
k=l[i]
c1=1
... | 126 | 93 | 6,758,400 | 160946235 | n = int(input())
a = list(map(int, input().split()))
maximal = sum(x for x in a if x > 0)
if maximal % 2 == 1:
print(maximal)
else:
INF = 1e9
print(max(
maximal - min((x for x in a if x > 0 and x & 1), default=INF),
maximal + max((x for x in a if x < 0 and x & 1), default=-INF),
)) | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Odd sum | You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deletin... | The first line contains integer number n (1 ≤ n ≤ 105).
The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | null | In the first example sum of the second and the fourth elements is 3. | [{"input": "4\n-2 2 -3 1", "output": "3"}, {"input": "3\n2 -5 -3", "output": "-1"}] | 1,400 | ["dp", "greedy", "implementation"] | 126 | [{"input": "4\r\n-2 2 -3 1\r\n", "output": "3\r\n"}, {"input": "3\r\n2 -5 -3\r\n", "output": "-1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n-1\r\n", "output": "-1\r\n"}, {"input": "15\r\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046\r\n", "output": "53507\r\n"}, {... | false | stdio | null | true |
797/B | 797 | B | PyPy 3 | TESTS | 37 | 170 | 11,878,400 | 160947060 | n = int(input())
nums = [int(x) for x in input().split()]
posi = []
nega = []
odd = []
sum = 0
for x in nums:
if x >= 0:
posi.append(x)
else:
nega.append(x)
sort = sorted(posi, reverse=True) + sorted(nega, reverse = True)
ans = []
for x in range(len(sort)):
sum += sort[x]
ans.append(sum)... | 126 | 93 | 7,270,400 | 132354899 | def solution(n, a):
oddMin = 10**4
isOdd = False
sum = 0
for i in range(n):
if (a[i] > 0):
sum = sum+a[i]
if (a[i]%2 !=0):
isOdd = True
if (oddMin > abs(a[i])):
oddMin = abs(a[i])
if (isOdd == False):
return -1
if (sum... | Educational Codeforces Round 19 | ICPC | 2,017 | 1 | 256 | Odd sum | You are given sequence a1, a2, ..., an of integer numbers of length n. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deletin... | The first line contains integer number n (1 ≤ n ≤ 105).
The second line contains n integer numbers a1, a2, ..., an ( - 104 ≤ ai ≤ 104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | null | In the first example sum of the second and the fourth elements is 3. | [{"input": "4\n-2 2 -3 1", "output": "3"}, {"input": "3\n2 -5 -3", "output": "-1"}] | 1,400 | ["dp", "greedy", "implementation"] | 126 | [{"input": "4\r\n-2 2 -3 1\r\n", "output": "3\r\n"}, {"input": "3\r\n2 -5 -3\r\n", "output": "-1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n-1\r\n", "output": "-1\r\n"}, {"input": "15\r\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046\r\n", "output": "53507\r\n"}, {... | false | stdio | null | true |
815/A | 815 | A | PyPy 3 | TESTS | 11 | 171 | 4,812,800 | 58700504 | a,b=map(int,input().split())
m=[]
for i in range(a):
m.append(list(map(int,input().split())))
r=[]
if a <= b:
for i in range(a):
mn=min(m[i])
for j in range(b):
m[i][j]-=mn
for j in range(mn):
r.append('row ' + str(i+1))
for i in range(b):
mn=502
... | 177 | 124 | 1,843,200 | 48908067 | n,m=map(int,input().split())
L=[list(map(int,input().split())) for i in range(n)]
s="row"
s1="col"
if n>m :
L1=[[0 for i in range(n)] for j in range(m)]
for i in range(n) :
for j in range(m) :
L1[j][i]=L[i][j]
L=L1
n,m=m,n
s,s1=s1,s
w=[0 for i in range(n)]
w1=[0 for ... | Codeforces Round 419 (Div. 1) | CF | 2,017 | 2 | 512 | Karen and Game | On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To ... | The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500). | If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.
The next k lines should each contain one of the following, describing the moves in the order they must b... | null | In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required... | [{"input": "3 5\n2 2 2 3 2\n0 0 0 1 0\n1 1 1 2 1", "output": "4\nrow 1\nrow 1\ncol 4\nrow 3"}, {"input": "3 3\n0 0 0\n0 1 0\n0 0 0", "output": "-1"}, {"input": "3 3\n1 1 1\n1 1 1\n1 1 1", "output": "3\nrow 1\nrow 2\nrow 3"}] | 1,700 | ["brute force", "greedy", "implementation"] | 177 | [{"input": "3 5\r\n2 2 2 3 2\r\n0 0 0 1 0\r\n1 1 1 2 1\r\n", "output": "4\r\nrow 1\r\nrow 1\r\ncol 4\r\nrow 3\r\n"}, {"input": "3 3\r\n0 0 0\r\n0 1 0\r\n0 0 0\r\n", "output": "-1\r\n"}, {"input": "3 3\r\n1 1 1\r\n1 1 1\r\n1 1 1\r\n", "output": "3\r\nrow 1\r\nrow 2\r\nrow 3\r\n"}, {"input": "3 5\r\n2 4 2 2 3\r\n0 2 0 0 ... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
n, m = map(int, f.readline().split())
grid = [list(map(int, f.readline().split())) for _ in range(n)]
possible = True
g1_1 = grid[0][0] if n... | true |
370/D | 370 | D | Python 3 | TESTS | 40 | 390 | 52,019,200 | 5414654 | #!/usr/bin/python3
def readln(): return list(map(int, input().split()))
import sys
def exit():
print(-1)
sys.exit()
n, m = readln()
mon = [list(input()) for _ in range(n)]
hor = [i for i in range(n) if mon[i] != ['.'] * m]
rmon = list(zip(*mon))
ver = [j for j in range(m) if rmon[j] != ('.',) * n]
mini = hor[... | 125 | 1,465 | 35,123,200 | 139630574 | from itertools import chain
def draw_square(ans, square_a, ymin, xmin) -> list:
for i in range(square_a + 1):
if(ans[ymin][xmin + i] != 'w'):
ans[ymin] = ans[ymin][:xmin + i] + '+' + ans[ymin][xmin + i + 1:]
if(ans[ymin + square_a][xmin + i] != 'w'):
ans[ymin + square_a] = an... | Codeforces Round 217 (Div. 2) | CF | 2,013 | 2 | 256 | Broken Monitor | Innocentius has a problem — his computer monitor has broken. Now some of the pixels are "dead", that is, they are always black. As consequence, Innocentius can't play the usual computer games. He is recently playing the following game with his younger brother Polycarpus.
Innocentius is touch-typing a program that pain... | The first line contains the resolution of the monitor as a pair of integers n, m (1 ≤ n, m ≤ 2000). The next n lines contain exactly m characters each — the state of the monitor pixels at the moment of the game. Character "." (period, ASCII code 46) corresponds to the black pixel, and character "w" (lowercase English l... | Print the monitor screen. Represent the sought frame by characters "+" (the "plus" character). The pixels that has become white during the game mustn't be changed. Print them as "w". If there are multiple possible ways to position the frame of the minimum size, print any of them.
If the required frame doesn't exist, t... | null | In the first sample the required size of the optimal frame equals 4. In the second sample the size of the optimal frame equals 3. In the third sample, the size of the optimal frame is 1. In the fourth sample, the required frame doesn't exist. | [{"input": "4 8\n..w..w..\n........\n........\n..w..w..", "output": "..w++w..\n..+..+..\n..+..+..\n..w++w.."}, {"input": "5 6\n......\n.w....\n......\n..w...\n......", "output": "......\n+w+...\n+.+...\n++w...\n......"}, {"input": "2 4\n....\n.w..", "output": "....\n.w.."}, {"input": "2 6\nw..w.w\n...w..", "output": "-... | 2,100 | ["brute force", "constructive algorithms", "greedy", "implementation"] | 125 | [{"input": "4 8\r\n..w..w..\r\n........\r\n........\r\n..w..w..\r\n", "output": "..w++w..\r\n..+..+..\r\n..+..+..\r\n..w++w..\r\n"}, {"input": "5 6\r\n......\r\n.w....\r\n......\r\n..w...\r\n......\r\n", "output": "......\r\n+w+...\r\n+.+...\r\n++w...\r\n......\r\n"}, {"input": "2 4\r\n....\r\n.w..\r\n", "output": "...... | false | stdio | import sys
def read_grid(file_path, n):
with open(file_path) as f:
return [list(line.strip()) for line in f.readlines()[:n]]
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, m = map(int, f.readline().split())
input_grid = [list(line.strip()) for line i... | true |
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