message stringlengths 2 23.4k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 129 108k | cluster float64 6 6 | __index_level_0__ int64 258 216k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by chan... | instruction | 0 | 16,884 | 6 | 33,768 |
Tags: implementation, strings
Correct Solution:
```
s1 = input()
s2 = input()
vowel = list('aeiou')
if len(s1)!= len(s2):
print('No')
else:
done = False
for i in range(len(s1)):
if s1[i] in vowel and s2[i] not in vowel or s1[i] not in vowel and s2[i] in vowel:
done = True
print('No')
break
if not don... | output | 1 | 16,884 | 6 | 33,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by chan... | instruction | 0 | 16,885 | 6 | 33,770 |
Tags: implementation, strings
Correct Solution:
```
s = input()
t = input()
s = list(s)
t = list(t)
vowel = ['a','e','i','o','u']
if len(s)!=len(t):
print("No")
else:
for i in range(len(s)):
if s[i] in vowel and t[i] in vowel:
continue
elif s[i] not in vowel and t[i] not in vowel:
... | output | 1 | 16,885 | 6 | 33,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,157 | 6 | 34,314 |
Tags: *special, implementation
Correct Solution:
```
password = input()
if (any(c.isupper() for c in password) and len(password) > 4 and any(c.islower() for c in password) and any(c.isdigit() for c in password)):
print("Correct")
else:
print("Too weak")
``` | output | 1 | 17,157 | 6 | 34,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,158 | 6 | 34,316 |
Tags: *special, implementation
Correct Solution:
```
from math import*
def add_bit(n, k):
return n | (1 << k)
s = input().strip()
ans = 0
if len(s) >= 5:
ans = add_bit(ans, 0)
for i in range(len(s)):
if s[i] >= 'a' and s[i] <= 'z':
ans = add_bit(ans, 1)
if s[i] >= 'A' and s[i] <= 'Z':
... | output | 1 | 17,158 | 6 | 34,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,159 | 6 | 34,318 |
Tags: *special, implementation
Correct Solution:
```
line = input()
length = False
little = False
big = False
number = False
length = (len(line) > 4)
for symbol in line:
if '0' <= symbol <= '9':
number = True
elif 'a' <= symbol <= 'z':
little = True
elif 'A' <= symbol <= 'Z':
big = T... | output | 1 | 17,159 | 6 | 34,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,160 | 6 | 34,320 |
Tags: *special, implementation
Correct Solution:
```
import sys
s=str(input())
t2=0
t3=0
t4=0
if len(s)<5:
print("Too weak")
sys.exit(0)
for i in range(len(s)):
if 'a'<=s[i]<='z':
t2=1
if 'A'<=s[i]<='Z':
t3=1
if '0'<=s[i]<='9':
t4=1
if t2&t3&t4:
print("Correct")
else:
print("Too weak")
``` | output | 1 | 17,160 | 6 | 34,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,161 | 6 | 34,322 |
Tags: *special, implementation
Correct Solution:
```
big = False
low = False
digit = False
word = input()
for i in word:
if len(word) < 5:
print("Too weak")
exit()
if i.isupper():
big = True
if i.islower():
low = True
if i.isdigit():
digit = True
if big and low a... | output | 1 | 17,161 | 6 | 34,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,162 | 6 | 34,324 |
Tags: *special, implementation
Correct Solution:
```
import re
s=input()
print(["Too weak","Correct"][all([re.search(t,s) for t in ['.{5}','[a-z]','[A-Z]','\d']])])
``` | output | 1 | 17,162 | 6 | 34,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,163 | 6 | 34,326 |
Tags: *special, implementation
Correct Solution:
```
n = input()
print(["Too weak", "Correct"][len(n)>=5 and any([x for x in n if x in 'qwertyuiopasdfghjklzxcvbnm']) and any([x for x in n if x in 'QWERTYUIOPASDFGHJKLZXCVBNM']) and any([x for x in n if x in '1234567890'])])
``` | output | 1 | 17,163 | 6 | 34,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough... | instruction | 0 | 17,164 | 6 | 34,328 |
Tags: *special, implementation
Correct Solution:
```
s = input()
digit = 0
char = 0
large = 0
small = 0
if len(s) < 5:
print("Too weak")
exit()
for i in s:
if i.isdigit():
digit+=1
elif i.isupper():
large+=1
elif i.islower():
small+=1
else:
char+=1
if large >= 1 and small >= 1 and digit >= 1:
print("Corr... | output | 1 | 17,164 | 6 | 34,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,376 | 6 | 34,752 |
Tags: implementation, strings
Correct Solution:
```
possible = set(range(ord('a'), ord('z') + 1))
redundant_shocks = 0
N = int(input())
for _ in range(N - 1):
line = input()
if len(possible) == 1 and line[0] != ".":
redundant_shocks += 1
continue
description, string = line.split()
if... | output | 1 | 17,376 | 6 | 34,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,377 | 6 | 34,754 |
Tags: implementation, strings
Correct Solution:
```
n=int(input())
l=[]
for i in range(26):
l.append(chr(97+i))
#print(l)
ansi=-1
ansc='.'
ans=0
f=0
for i in range(n-1):
c,s=map(str,input().split())
#print(c,s)
if f==0 and len(l)==1:
f=1
ansi=i
ansc=l[0]
if c=='.':
fo... | output | 1 | 17,377 | 6 | 34,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,378 | 6 | 34,756 |
Tags: implementation, strings
Correct Solution:
```
number = int(input())
a = set()
for i in 'qwertyuiopasdfghjklzxcvbnm':
a.add(i)
n = set()
bs = set()
result = 0
for j in range(number - 1):
s = set()
z, s = input().split()
if z == '!':
if len(a) == 1:
result += 1
n = set()
... | output | 1 | 17,378 | 6 | 34,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,379 | 6 | 34,758 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
l = [1] * 26
ok = 1
u = 0
abc = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',]
for i in range(n):
s = list(input().split(' '))
if s[0] == '!':
if ok == 0:
u += 1
... | output | 1 | 17,379 | 6 | 34,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,380 | 6 | 34,760 |
Tags: implementation, strings
Correct Solution:
```
from sys import stdin, stdout
input = stdin.readline
n = int(input())
queries = list(input().split() for _ in range(n))
chars = [chr(ord('a')+i) for i in range(26)]
chars_set = set(list(chars))
res, cnt = '', 0
for c, f in queries:
if c == '.':
for q in... | output | 1 | 17,380 | 6 | 34,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,381 | 6 | 34,762 |
Tags: implementation, strings
Correct Solution:
```
# cook your dish here
n = int(input())
know=False
ans =0
let = set(['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'])
for i in range(n):
t,w = input().split()
w = set([l for l in w])
if know:
... | output | 1 | 17,381 | 6 | 34,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,382 | 6 | 34,764 |
Tags: implementation, strings
Correct Solution:
```
n = list(map(int, input().strip().split()))[0]
dp = dict()
for i in range(26):
dp[chr(ord('a') + i)] = 0
answer = 0
for i in range(n):
now = list(input('').strip().split())
if len(dp) == 1:
if now[0] == '!':
answer += 1
elif n... | output | 1 | 17,382 | 6 | 34,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receiv... | instruction | 0 | 17,383 | 6 | 34,766 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
S = set('abcdefghijklmnopqrstuvwxyz')
lines = [input().strip() for _ in range(n)]
ok, cnt = 0, 0
for line in lines[:-1]:
c, s = line.split()
if ok:
if c == '!' or c == '?':
cnt += 1
continue
if '!' == c:
... | output | 1 | 17,383 | 6 | 34,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,139 | 6 | 36,278 |
Tags: dp, implementation, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
s = input()[:-1]
n = len(s)
dp = [[False]*2 for _ in range(n+1)]
dp[n][0] = True
dp[n][1] = True
dp[n-2][0] = True
dp[n-3][1] = True
for i in range(n-4, -1, -1):
if (dp[i+2][0] and i+4<=n and s[... | output | 1 | 18,139 | 6 | 36,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,140 | 6 | 36,280 |
Tags: dp, implementation, strings
Correct Solution:
```
letters = input()
ll = len(letters)
dp = [[False for _ in range(ll)] for _ in range(2)]
dp[0][-2] = True
dp[1][-3] = True
for i in range(ll-4,4,-1):
if dp[0][i+3]==True or (dp[1][i+3]==True and letters[i:i+3] != letters[i+3:i+6]):
dp[1][i] = True
if dp[1][i... | output | 1 | 18,140 | 6 | 36,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,141 | 6 | 36,282 |
Tags: dp, implementation, strings
Correct Solution:
```
import sys
sys.setrecursionlimit(15000)
s = input()
s = s[5:] + " "
res = set()
aux = set()
def getWords(x,y):
if (x,y) in aux:
return
aux.add((x,y))
if x > 1 and s[x:y] != s[x-2:x]:
res.add(s[x-2:x])
getWords(x-2,x)
if x ... | output | 1 | 18,141 | 6 | 36,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,142 | 6 | 36,284 |
Tags: dp, implementation, strings
Correct Solution:
```
def main():
s = input()[5:]
n = len(s)
if n < 2:
print(0)
return
res2, res3 = set(), set()
dp2 = [False] * (n + 1)
dp3 = [False] * (n + 1)
dp2[-1] = dp3[-1] = True
for i in range(n, 1, -1):
if dp3[i] or dp2[i... | output | 1 | 18,142 | 6 | 36,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,143 | 6 | 36,286 |
Tags: dp, implementation, strings
Correct Solution:
```
a=input()
n=len(a)
sufiks = set()
suf_lens = [0] * (n+1)
suf_lens[n] = 3
def main(i, len_suf):
if (suf_lens[i] & len_suf - 1 and a[i : i + len_suf] != a[i-len_suf:i]) or suf_lens[i] & 4 - len_suf :
sufiks.add(a[i-len_suf : i])
suf_lens[i - len_... | output | 1 | 18,143 | 6 | 36,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,144 | 6 | 36,288 |
Tags: dp, implementation, strings
Correct Solution:
```
s = (input())
n = len(s)
x = set()
a = [[0 for i in range(2)] for j in range(n+3)]
b = [0 for i in range(n+3)]
b[n] = 1
a[n][0] = 1
a[n][1] = 1
for i in range(n - 1, 4, -1):
if b[i + 2]:
if a[i + 2][0] and (s[i:i + 2] != s[i + 2:i + 4]):
... | output | 1 | 18,144 | 6 | 36,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,145 | 6 | 36,290 |
Tags: dp, implementation, strings
Correct Solution:
```
s = input()
if len(s) <= 5:
print(0)
else:
words = set()
s = s[5:]
r = len(s)
if r == 4:
words.add(s[1:])
words.add(s[2:])
words.add(s[:2])
if r == 2:
words.add(s)
if r == 3:
words.add(s)
... | output | 1 | 18,145 | 6 | 36,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, ... | instruction | 0 | 18,146 | 6 | 36,292 |
Tags: dp, implementation, strings
Correct Solution:
```
#!/usr/bin/env python3
# 667C_ling.py - Codeforces.com/problemset/problem/667/C by Sergey 2016
import unittest
import sys
from collections import deque
###############################################################################
# Ling Class (Main Program)
##... | output | 1 | 18,146 | 6 | 36,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,147 | 6 | 36,294 |
Yes | output | 1 | 18,147 | 6 | 36,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,148 | 6 | 36,296 |
Yes | output | 1 | 18,148 | 6 | 36,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,149 | 6 | 36,298 |
Yes | output | 1 | 18,149 | 6 | 36,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,150 | 6 | 36,300 |
Yes | output | 1 | 18,150 | 6 | 36,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,151 | 6 | 36,302 |
No | output | 1 | 18,151 | 6 | 36,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,152 | 6 | 36,304 |
No | output | 1 | 18,152 | 6 | 36,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,153 | 6 | 36,306 |
No | output | 1 | 18,153 | 6 | 36,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to ... | instruction | 0 | 18,154 | 6 | 36,308 |
No | output | 1 | 18,154 | 6 | 36,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,789 | 6 | 37,578 |
Tags: implementation, strings
Correct Solution:
```
import sys
kek = ""
bol = 1
for s in sys.stdin:
ok = 0
for i in range(len(s)):
if s[i] == ' ':
continue
if s[i] == '#':
ok = 1
else:
ok = 0
break
kekw = (bol == 0 and ok == 0)
for j in... | output | 1 | 18,789 | 6 | 37,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,790 | 6 | 37,580 |
Tags: implementation, strings
Correct Solution:
```
import sys
a = sys.stdin.readlines()
def good(x):
c = ' '
for y in x:
if y != ' ':
c = y
break
return c == '#'
ans = []
now = None
for p in a:
if good(p):
if now != None:
ans.append(now)
... | output | 1 | 18,790 | 6 | 37,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,791 | 6 | 37,582 |
Tags: implementation, strings
Correct Solution:
```
import sys
result = ''
prev_amp = True
for line in sys.stdin:
if '#' in line:
i = line.index('#')
if list(line[:i]) == [' '] * i:
if not prev_amp:
result += '\n' + line
else:
result += line... | output | 1 | 18,791 | 6 | 37,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,792 | 6 | 37,584 |
Tags: implementation, strings
Correct Solution:
```
import sys
res = ''
has_endl = True
for line in sys.stdin:
if '#' in line:
i = line.index('#')
if list(line[:i]) == [' '] * i:
if not has_endl:
res += '\n'
res += line
has_endl = True
... | output | 1 | 18,792 | 6 | 37,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,793 | 6 | 37,586 |
Tags: implementation, strings
Correct Solution:
```
v = []
c = []
resp = ""
while True:
try:
v.append(input())
except:
break
i = 0
while i < len(v):
k = v[i].replace(" ","")
if(k == ""):
v[i] = k
c.append(False)
elif(k[0] == '#'):
c.append(True)
else:
v[i] = k
c.append(False)
i+=1
n = len(v)
i =... | output | 1 | 18,793 | 6 | 37,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,794 | 6 | 37,588 |
Tags: implementation, strings
Correct Solution:
```
import sys
result = ''
prev_amp = True
for line in sys.stdin:
if '#' in line:
i = line.index('#')
if list(line[:i]) == [' '] * i:
if not prev_amp:
result += '\n' + line
else:
result += line... | output | 1 | 18,794 | 6 | 37,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but only get to know the basic rules of writing spell... | instruction | 0 | 18,795 | 6 | 37,590 |
Tags: implementation, strings
Correct Solution:
```
from sys import stdin, stdout
def f(t):
for i in t:
if i != ' ': return i == '#'
return False
r, s = [], []
q = None
for t in stdin.readlines():
if f(t):
if q != None:
s.append(q)
q = None
s.append(t[: -1])
... | output | 1 | 18,795 | 6 | 37,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but on... | instruction | 0 | 18,796 | 6 | 37,592 |
No | output | 1 | 18,796 | 6 | 37,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but on... | instruction | 0 | 18,797 | 6 | 37,594 |
No | output | 1 | 18,797 | 6 | 37,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but on... | instruction | 0 | 18,798 | 6 | 37,596 |
No | output | 1 | 18,798 | 6 | 37,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's dive into one of the most interesting areas of magic β writing spells. Learning this exciting but challenging science is very troublesome, so now you will not learn the magic words, but on... | instruction | 0 | 18,799 | 6 | 37,598 |
No | output | 1 | 18,799 | 6 | 37,599 |
Provide a correct Python 3 solution for this coding contest problem.
JAG-channel
Nathan O. Davis operates an electronic bulletin board called JAG-channel. He is currently working on adding a new feature called Thread View.
Like many other electronic bulletin boards, JAG-channel is thread-based. Here, a thread refers... | instruction | 0 | 19,366 | 6 | 38,732 |
"Correct Solution:
```
while 1:
n=int(input())
if n==0:break
a=[list(input())for _ in [0]*n]
for i in range(n):
for j in range(len(a[i])):
if a[i][j]=='.':a[i][j]=' '
else:
if j>0:
a[i][j-1]='+'
x=i-1
... | output | 1 | 19,366 | 6 | 38,733 |
Provide a correct Python 3 solution for this coding contest problem.
JAG-channel
Nathan O. Davis operates an electronic bulletin board called JAG-channel. He is currently working on adding a new feature called Thread View.
Like many other electronic bulletin boards, JAG-channel is thread-based. Here, a thread refers... | instruction | 0 | 19,367 | 6 | 38,734 |
"Correct Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().... | output | 1 | 19,367 | 6 | 38,735 |
Provide a correct Python 3 solution for this coding contest problem.
JAG-channel
Nathan O. Davis operates an electronic bulletin board called JAG-channel. He is currently working on adding a new feature called Thread View.
Like many other electronic bulletin boards, JAG-channel is thread-based. Here, a thread refers... | instruction | 0 | 19,368 | 6 | 38,736 |
"Correct Solution:
```
def count_dot(lis):
c=0
for i in lis:
if not(i=='.'):
return c
c+=1
def solve(N):
a=[]
dot =[]
for i in range(N):
a.append([char for char in input()])
dot.append(count_dot(a[i]))
for i in range(N):
for j in range(len(a[... | output | 1 | 19,368 | 6 | 38,737 |
Provide a correct Python 3 solution for this coding contest problem.
JAG-channel
Nathan O. Davis operates an electronic bulletin board called JAG-channel. He is currently working on adding a new feature called Thread View.
Like many other electronic bulletin boards, JAG-channel is thread-based. Here, a thread refers... | instruction | 0 | 19,369 | 6 | 38,738 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.rea... | output | 1 | 19,369 | 6 | 38,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
JAG-channel
Nathan O. Davis operates an electronic bulletin board called JAG-channel. He is currently working on adding a new feature called Thread View.
Like many other electronic bulletin bo... | instruction | 0 | 19,370 | 6 | 38,740 |
No | output | 1 | 19,370 | 6 | 38,741 |
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