message stringlengths 2 59.7k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the ... | instruction | 0 | 89,894 | 20 | 179,788 |
No | output | 1 | 89,894 | 20 | 179,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.
Let f(n) be the ... | instruction | 0 | 89,895 | 20 | 179,790 |
No | output | 1 | 89,895 | 20 | 179,791 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,026 | 20 | 180,052 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0162
"""
import sys
from sys import stdin
from bisect import bisect_right, bisect_left
from math import ceil, log
input = stdin.readline
def main(args):
hammings = []
temp = set()
for i in range(c... | output | 1 | 90,026 | 20 | 180,053 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,027 | 20 | 180,054 |
"Correct Solution:
```
while 1:
n=list(map(int,input().split()))
if n[0]==0:break
a=0
for i in range(n[0],n[1]+1):
b=i
while b%5==0: b/=5
while b%3==0: b/=3
while b%2==0: b/=2
if b==1:a+=1
print(a)
``` | output | 1 | 90,027 | 20 | 180,055 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,028 | 20 | 180,056 |
"Correct Solution:
```
while True:
A = list(map(int, input().split()))
if A[0] == 0:
break
n, m = A[0], A[1]
ans = 0
for i in range(n, m+1):
b = i
while b % 5 == 0:
b //= 5
while b % 3 == 0:
b //= 3
while b % 2 == 0:
b //=... | output | 1 | 90,028 | 20 | 180,057 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,029 | 20 | 180,058 |
"Correct Solution:
```
MAX = 1000000
hamming_list = [False] * (MAX + 1)
hamming_list[0] = False
hamming_list[1] = True
for index in range(2, MAX + 1):
if index % 2 == 0:
if hamming_list[index // 2]:
hamming_list[index] = True
elif index % 3 == 0:
if hamming_list[index // 3]:
... | output | 1 | 90,029 | 20 | 180,059 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,030 | 20 | 180,060 |
"Correct Solution:
```
MAX = 1000000
hamming_list = [False] * (MAX + 1)
hamming_list[0] = False
hamming_list[1] = True
for index in range(2, MAX + 1):
if index / 2 % 1 == 0:
if hamming_list[index // 2]:
hamming_list[index] = True
elif index / 3 % 1 == 0:
if hamming_list[index // ... | output | 1 | 90,030 | 20 | 180,061 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,031 | 20 | 180,062 |
"Correct Solution:
```
H = [False for i in range(1000001)]
H[1] = True
for i in range(20):
for j in range(13):
for k in range(9):
if (2 ** i) * (3 ** j) * (5 ** k) < 1000001:
H[(2 ** i) * (3 ** j) * (5 ** k)] = True
else:
break
while True:
L = in... | output | 1 | 90,031 | 20 | 180,063 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,032 | 20 | 180,064 |
"Correct Solution:
```
while 1:
n=list(map(int,input().split()))
if n[0]==0:break
a=0
for i in range(n[0],n[1]+1):
b=i
while b%2==0: b/=2
while b%3==0: b/=3
while b%5==0: b/=5
if b==1:a+=1
print(a)
``` | output | 1 | 90,032 | 20 | 180,065 |
Provide a correct Python 3 solution for this coding contest problem.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming numbers, but 11, 13, 14 etc. are not humming numbers.
A... | instruction | 0 | 90,033 | 20 | 180,066 |
"Correct Solution:
```
while 1:
datas = list(map(int, input().split()))
if datas[0] == 0:
break
n, m = datas[0], datas[1]
cnt = 0
for i in range(n, m+1):
b = i
while b % 5 == 0:
b //= 5
while b % 3 == 0:
b //= 3
while b % 2 == 0:
... | output | 1 | 90,033 | 20 | 180,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming ... | instruction | 0 | 90,034 | 20 | 180,068 |
Yes | output | 1 | 90,034 | 20 | 180,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming ... | instruction | 0 | 90,036 | 20 | 180,072 |
Yes | output | 1 | 90,036 | 20 | 180,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The number obtained by multiplying 1 by 2, 3, 5 several times (0 or more times) is called the Hamming numbers. For example
* 1
* 1 x 2 x 2 = 4
* 1 x 2 x 2 x 3 x 5 x 5 = 300
Etc. are humming ... | instruction | 0 | 90,038 | 20 | 180,076 |
No | output | 1 | 90,038 | 20 | 180,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,419 | 20 | 180,838 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
from itertools import product
n = int(input().strip())
if n<102:
print(n)
else:
res = set()
combination = (str(i)+str(j) for i in range(10) for j in range(i+1,10))
for it in (product(comb,repeat=l) for comb in combination for l in range(... | output | 1 | 90,419 | 20 | 180,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,420 | 20 | 180,840 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
def p(k):
if 0 < k <= n:
s.add(k)
k *= 10
p(k + x)
p(k + y)
n = int(input())
s = set()
for x in range(10):
for y in range(10):
p(x)
print(len(s))
``` | output | 1 | 90,420 | 20 | 180,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,421 | 20 | 180,842 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
# https://codeforces.com/contest/244/problem/B
def gen(digit, x, length, S):
S.append(x)
if length == 10:
return
if len(digit) == 1:
for i in range(0, 10):
next_x = x * 10 + i
i... | output | 1 | 90,421 | 20 | 180,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,422 | 20 | 180,844 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
#CF Round 150. Div II Prob. A - Dividing Orange
import sys
dp = [[[-1 for j in range(3)] for i in range (1 << 10)] for k in range(11)]
In = sys.stdin
n = In.readline().strip()
def go (idx, mask, equal):
if dp[idx][mask][equal] != -1:
retu... | output | 1 | 90,422 | 20 | 180,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,423 | 20 | 180,846 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
def dfs(k):
if 0 < k <= n:
s.add(k)
k *= 10
dfs(k + x)
dfs(k + y)
n = int(input())
s = set()
for x in range(10):
for y in range(10):
dfs(x)
print(len(s))
``` | output | 1 | 90,423 | 20 | 180,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,424 | 20 | 180,848 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
def findNum(num):
if 0 < num <= n:
s.add(num)
num*=10
findNum(num+x)
findNum(num+y)
n = int(input())
s = set()
for x in range(10):
for y in range(10):
findNum(x)
print(len(s))
``` | output | 1 | 90,424 | 20 | 180,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,425 | 20 | 180,850 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
n=int(input())
def bruteforce(number):
res=0
liste=[]
while number>0:
integer=number%10
if integer not in liste:
liste.append(integer)
number=number//10
return len(liste)<=2
answer=[0]
def ... | output | 1 | 90,425 | 20 | 180,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example, if x = 4, and y = 7, then numbers 47, 744, 4 ... | instruction | 0 | 90,426 | 20 | 180,852 |
Tags: bitmasks, brute force, dfs and similar
Correct Solution:
```
ii=lambda:int(input())
kk=lambda:map(int, input().split())
ll=lambda:list(kk())
q =[]
s=set()
n=ii()
for x in range(10):
for y in range(10):
q.append(x)
while q:
q2 = []
for item in q:
if item > 0 and item <= n:
s.add(item)
item... | output | 1 | 90,426 | 20 | 180,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,427 | 20 | 180,854 |
Yes | output | 1 | 90,427 | 20 | 180,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,428 | 20 | 180,856 |
Yes | output | 1 | 90,428 | 20 | 180,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,429 | 20 | 180,858 |
Yes | output | 1 | 90,429 | 20 | 180,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,430 | 20 | 180,860 |
Yes | output | 1 | 90,430 | 20 | 180,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,431 | 20 | 180,862 |
No | output | 1 | 90,431 | 20 | 180,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,432 | 20 | 180,864 |
No | output | 1 | 90,432 | 20 | 180,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,433 | 20 | 180,866 |
No | output | 1 | 90,433 | 20 | 180,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarpus loves lucky numbers. Everybody knows that lucky numbers are positive integers, whose decimal representation (without leading zeroes) contain only the lucky digits x and y. For example... | instruction | 0 | 90,434 | 20 | 180,868 |
No | output | 1 | 90,434 | 20 | 180,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,642 | 20 | 181,284 |
Tags: math, number theory
Correct Solution:
```
a, b, c = map(int, input().split())
pos = -1
for i in range(1, 309):
if int((a * pow(10, i) // b)) % 10 == c:
pos = i
break
print(pos)
#128 7 0
``` | output | 1 | 90,642 | 20 | 181,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,643 | 20 | 181,286 |
Tags: math, number theory
Correct Solution:
```
import decimal
a, b, c = map(int, input().split())
decimal.getcontext().prec = 3 * b + 100
s = str(decimal.Decimal(a) / decimal.Decimal(b)).ljust(3 * b + 100, '0')
r = s.find(str(c), 2, 2 * b + 8) - 1
if r == -2:
print(-1)
else:
print(r)
``` | output | 1 | 90,643 | 20 | 181,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,644 | 20 | 181,288 |
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 300
split_strings = input().split()
[a, b, c] = map(lambda x: int(x), split_strings)
a = Decimal(a)
b = Decimal(b)
decimal = (a / b) * 10
for i in range(1, 300):
ones = int(decimal)
fractional = decimal % 1
if o... | output | 1 | 90,644 | 20 | 181,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,645 | 20 | 181,290 |
Tags: math, number theory
Correct Solution:
```
a,b,c = map(int,input().split(" "))
a = a%b
i = 1
while i <= b :
if a==0:
if c==0 :
print(i)
exit()
else :
print(-1)
exit()
if a<b : a *= 10
while(a<b) :
if c==0 :
print(i)
... | output | 1 | 90,645 | 20 | 181,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,646 | 20 | 181,292 |
Tags: math, number theory
Correct Solution:
```
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
num = []
cnt = 0
while True:
cnt += 1
num.append( a )
n = a*10//b
if n == c:
break
a = a*10 - n*b
if a in num:
cnt = -1
break
print( cnt )
``` | output | 1 | 90,646 | 20 | 181,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,647 | 20 | 181,294 |
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 10000
a, b, c = map(int, input().split())
d = Decimal(a)/Decimal(b)
# print(d)
s = str(d)[2:]
if len(s) > 900:
s = s[:-1]
else:
s += '0'
# print(s)
try:
ans = s.index(str(c))
except ValueError:
print(-1)
else:
... | output | 1 | 90,647 | 20 | 181,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,648 | 20 | 181,296 |
Tags: math, number theory
Correct Solution:
```
a,b,c=map(int,input().split())
r=a%b
t=b
cnt=1
f=0
while(t!=0):
r*=10
k=r//b
r=r%b
if(k==c):
f=1
break
cnt+=1
t-=1
if(f):
print(cnt)
else:
print(-1)
``` | output | 1 | 90,648 | 20 | 181,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).
Output
... | instruction | 0 | 90,649 | 20 | 181,298 |
Tags: math, number theory
Correct Solution:
```
from decimal import *
getcontext().prec = 2048
getcontext().rounding = ROUND_DOWN
num, denom, target = [int(x) for x in input().split()]
a = Decimal(num)/Decimal(denom)
a = str(a)
a = a[a.find('.') : ]
if len(a) < 2047:
a = a + '0'
print(a.find(str(target)))
``` | output | 1 | 90,649 | 20 | 181,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,650 | 20 | 181,300 |
Yes | output | 1 | 90,650 | 20 | 181,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,651 | 20 | 181,302 |
Yes | output | 1 | 90,651 | 20 | 181,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,652 | 20 | 181,304 |
Yes | output | 1 | 90,652 | 20 | 181,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,653 | 20 | 181,306 |
Yes | output | 1 | 90,653 | 20 | 181,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,654 | 20 | 181,308 |
No | output | 1 | 90,654 | 20 | 181,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,655 | 20 | 181,310 |
No | output | 1 | 90,655 | 20 | 181,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,656 | 20 | 181,312 |
No | output | 1 | 90,656 | 20 | 181,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers... | instruction | 0 | 90,657 | 20 | 181,314 |
No | output | 1 | 90,657 | 20 | 181,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as ... | instruction | 0 | 90,694 | 20 | 181,388 |
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit((10**9))
n = int(input())
s = set()
arr = [1,5,10,50]
def comb(num, digit):
global n
if digit == n:
s.add(num)
else:
for i in range(4):
comb(num+arr[i], di... | output | 1 | 90,694 | 20 | 181,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as ... | instruction | 0 | 90,695 | 20 | 181,390 |
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
n = int(input())
if n < 12:
print([1, 4, 10, 20, 35, 56, 83, 116, 155, 198, 244, 292][n])
else:
print(n*50-n-247)
``` | output | 1 | 90,695 | 20 | 181,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as ... | instruction | 0 | 90,696 | 20 | 181,392 |
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
def check(n, b, c):
for tb in range(0, 9):
for tc in range(0, min(n - tb + 1, 9)):
s1 = 9 * b + 4 * c
s2 = 9 * tb + 4 * tc
if s1 > s2 and (s1 - s2) % 49 == 0:
return False
return... | output | 1 | 90,696 | 20 | 181,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as ... | instruction | 0 | 90,697 | 20 | 181,394 |
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
a = [0, 4, 10, 20, 35, 56, 83, 116, 155, 198, 244, 292]
n = int(input())
if n < len(a):
print (a[n])
else:
print (a[11] + 49*(n-11))
``` | output | 1 | 90,697 | 20 | 181,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as ... | instruction | 0 | 90,698 | 20 | 181,396 |
Tags: brute force, combinatorics, dp, greedy, math
Correct Solution:
```
def ceildiv(a, b):
return -(-a // b)
def test_ceildiv():
assert ceildiv(6, 3) == 2
assert ceildiv(7, 3) == 3
assert ceildiv(5, 3) == 2
def brute_force(n, m):
"""Count numbers in range [0; m] equal to 4a + 9b + 49c
...f... | output | 1 | 90,698 | 20 | 181,397 |
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