message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000 | instruction | 0 | 31,158 | 22 | 62,316 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
x,y,z = map(int,input().split())
if x==y==z:
print('YES')
print(x,y,z)
elif x==y and x>z:
print('YES')
print(x,1,z)
elif x==z and x>y:
print('YES')
print(1,x,y)
elif y==z and y>x:
print('YES')
print(x,1,z)
else:
print('NO')
``` | output | 1 | 31,158 | 22 | 62,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000 | instruction | 0 | 31,159 | 22 | 62,318 |
Tags: math
Correct Solution:
```
t = int(input())
while(t>0):
t = t-1
a = input()
A = list(map(int,list(a.split())))
m = max(A)
d = {}
for i in range(3):
if A[i] in d:
d[A[i]] = d[A[i]]+1
else:
d[A[i]] = 1
if d[m]>=2:
a = m
if d[m] == 3:
b = m
c = m
else:
for i in d:
if i != m:
b = i
c = i
break
print("YES")
print(a,end=" ")
print(b,end=" ")
print(c)
else:
print("NO")
``` | output | 1 | 31,159 | 22 | 62,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000 | instruction | 0 | 31,160 | 22 | 62,320 |
Tags: math
Correct Solution:
```
t = int(input())
k = [[] for i in range(t)]
for i in range(t):
x, y, z = map(int, input().split())
a = b = c = 0
if max(x, y) == max(y, z) and max(y, z) == max(x, z):
k[i].append(min(x, y))
k[i].append(min(y, z))
k[i].append(min(z, x))
else:
k[i].append(-1)
for i in k:
if i[0] == -1:
print("NO")
else:
print("YES")
print(*i)
``` | output | 1 | 31,160 | 22 | 62,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000 | instruction | 0 | 31,161 | 22 | 62,322 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a=[int(x) for x in input().split()]
a.sort()
if a[1]!=a[2]:
print("NO")
else:
print("YES")
print(a[0],a[0],a[2])
``` | output | 1 | 31,161 | 22 | 62,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given three positive (i.e. strictly greater than zero) integers x, y and z.
Your task is to find positive integers a, b and c such that x = max(a, b), y = max(a, c) and z = max(b, c), or determine that it is impossible to find such a, b and c.
You have to answer t independent test cases. Print required a, b and c in any (arbitrary) order.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains three integers x, y, and z (1 ≤ x, y, z ≤ 10^9).
Output
For each test case, print the answer:
* "NO" in the only line of the output if a solution doesn't exist;
* or "YES" in the first line and any valid triple of positive integers a, b and c (1 ≤ a, b, c ≤ 10^9) in the second line. You can print a, b and c in any order.
Example
Input
5
3 2 3
100 100 100
50 49 49
10 30 20
1 1000000000 1000000000
Output
YES
3 2 1
YES
100 100 100
NO
NO
YES
1 1 1000000000 | instruction | 0 | 31,162 | 22 | 62,324 |
Tags: math
Correct Solution:
```
import math
LI = lambda: list(map(int,input().split()) )
MI = lambda: map(int,input().split())
SI = lambda: input()
II = lambda: int(input())
t = II()
for q in range(t):
x,y,z = MI()
m = max(x,y,z)
if x == y == m or x == z == m:
print("YES")
print(m,min(x,y,z),min(x,y,z))
elif y == z == m:
print("YES")
print(min(x,y,z),min(x,y,z),m)
else:
print("NO")
``` | output | 1 | 31,162 | 22 | 62,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input
The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).
Output
Output a single integer — the required minimal sum.
Examples
Input
2
1 2
Output
2
Input
3
2 4 6
Output
6
Input
2
12 18
Output
12
Input
5
45 12 27 30 18
Output
15
Note
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1. | instruction | 0 | 31,316 | 22 | 62,632 |
Tags: greedy, math
Correct Solution:
```
def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
n=int(input())
arr=list(map(int , input().split()))
ans=arr[0]
for i in range(1,n):
ans=find_gcd(ans,arr[i])
print(ans*n)
``` | output | 1 | 31,316 | 22 | 62,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,884 | 22 | 63,768 |
Tags: binary search, math
Correct Solution:
```
from collections import *
import sys
# "". join(strings)
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
t =ri()
for _ in range(t):
n,k =rl()
q=(k-1)//(n-1)
print(k+q)
``` | output | 1 | 31,884 | 22 | 63,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,885 | 22 | 63,770 |
Tags: binary search, math
Correct Solution:
```
import math
import sys
input = sys.stdin.readline
t=int(input())
for i in range(t):
n,k=map(int,input().split())
q=k//(n-1)
r=k%(n-1)
if r==0:
print(n*q-1)
else:
print(n*q+r)
``` | output | 1 | 31,885 | 22 | 63,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,886 | 22 | 63,772 |
Tags: binary search, math
Correct Solution:
```
m = int(input())
for i in range(m):
n, k = map(int, input().split())
p = int(k*n/(n-1))
for x in range(p-2, p+2):
if x == k + x//n:
print(x)
break
``` | output | 1 | 31,886 | 22 | 63,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,887 | 22 | 63,774 |
Tags: binary search, math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Mon Nov 2 16:05:33 2020
@author: 章斯岚
"""
import math
t=int(input())
for i in range(t):
n,k=map(int,input().split())
print(math.ceil(k*(n/(n-1)))-1)
``` | output | 1 | 31,887 | 22 | 63,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,888 | 22 | 63,776 |
Tags: binary search, math
Correct Solution:
```
for i in range(int(input())):
n,k=map(int,input().split())
a=n-1
b=k//a
b=n*b
c=k%a
if(c==0):
print(b-1)
else:
print(b+c)
``` | output | 1 | 31,888 | 22 | 63,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,889 | 22 | 63,778 |
Tags: binary search, math
Correct Solution:
```
def solution():
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
x = m // n
y = (m % n)
result = m - y
m = x + y
while m >= n :
x = m // n
y = (m % n)
result += m - y
m = x + y
print(result + m)
solution()
``` | output | 1 | 31,889 | 22 | 63,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,890 | 22 | 63,780 |
Tags: binary search, math
Correct Solution:
```
import sys
t=int(sys.stdin.readline())
for _ in range(t):
n, k=map(int, sys.stdin.readline().split())
cnt=0
last=0
if k%(n-1)==0:
cnt=k//(n-1)
print(cnt*n-1)
else:
cnt=k//(n-1)+1
last=cnt*n-1
print(last-(cnt*(n-1)-k))
``` | output | 1 | 31,890 | 22 | 63,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1 | instruction | 0 | 31,891 | 22 | 63,782 |
Tags: binary search, math
Correct Solution:
```
import math
def function(n, k):
x = int(math.floor((k) / (n - 1)))
if (k + x) % n == 0:
return ((k + x) - 1)
else:
return k + x
if __name__ == '__main__':
t = int(input())
for k1 in range(t):
n, k = map(int, input().rstrip().split())
print(function(n, k))
``` | output | 1 | 31,891 | 22 | 63,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,k = list(map(int,input().split()))
s = int((k-1)/(n-1))
print(k + s)
``` | instruction | 0 | 31,892 | 22 | 63,784 |
Yes | output | 1 | 31,892 | 22 | 63,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t = int(input())
while t > 0:
n, k = [int(x) for x in input().split()]
lower = k // (n - 1)
upper = lower + 1
low = n * lower
up = n * upper
st = (n * lower) - lower
if st == k:
print (low - 1)
else:
print (low + (k - st))
t -= 1
``` | instruction | 0 | 31,893 | 22 | 63,786 |
Yes | output | 1 | 31,893 | 22 | 63,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t=int(input())
while(t>0):
t-=1
n,k=map(int,input().split())
print(k + (k-1)//(n-1))
``` | instruction | 0 | 31,894 | 22 | 63,788 |
Yes | output | 1 | 31,894 | 22 | 63,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t=int(input())
import math
while t>0:
t-=1
n,k=[int(x) for x in input().split()]
numsets=math.ceil(k/(n-1))
if k%(n-1)==0:
print((numsets-1)*n+n-1)
else:
print((numsets-1)*n+k%(n-1))
``` | instruction | 0 | 31,895 | 22 | 63,790 |
Yes | output | 1 | 31,895 | 22 | 63,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t = int(input())
for j in range(t):
n, k = map(int, input().split())
g = k % n
h = (k // n) + g
print(h + k)
``` | instruction | 0 | 31,896 | 22 | 63,792 |
No | output | 1 | 31,896 | 22 | 63,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
# Python3 implementation to find
# the K'th non-divisible
# number of N
import math
# Function to find the Kth
# not divisible by N
def kthNonDivisible(N, K):
return K + math.floor((K - 1) / (N - 1))
# Driver Code
N = 3
K = 6
# Function Call
print(kthNonDivisible(N, K))
# This code is contributed by ishayadav181
``` | instruction | 0 | 31,897 | 22 | 63,794 |
No | output | 1 | 31,897 | 22 | 63,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
def main():
t = int(input())
l = []
for i in range(t):
n, k = map(int, input().split())
maxi = 10 ** 9 * 2
mini = 1
while maxi != mini:
srednji = (maxi + mini) // 2
djeljivi = srednji // n
if srednji - djeljivi == k:
l.append(srednji)
break
if srednji - djeljivi > k:
maxi = srednji - 1
else:
mini = srednji + 1
for i in l:
print(i)
if __name__ == '__main__':
main()
``` | instruction | 0 | 31,898 | 22 | 63,796 |
No | output | 1 | 31,898 | 22 | 63,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.
For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).
Output
For each test case print the k-th positive integer that is not divisible by n.
Example
Input
6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1
Output
10
15
1999999999
113
1000000001
1
Submitted Solution:
```
t = int(input())
a = []
for i in range(t):
n = str(input())
a.append(n)
for i in a:
b = i.split(' ')
n = int(b[0])
k = int(b[1])
if -k % (1/n - 1) == 0:
x = round(-k // (1/n - 1) - 1)
else:
x = round(-k // (1/n - 1))
print(x)
``` | instruction | 0 | 31,899 | 22 | 63,798 |
No | output | 1 | 31,899 | 22 | 63,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,988 | 22 | 63,976 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
def primeFactors(n):
c=0
while n % 2 == 0:
c+=1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
c+=1
n = n // i
if n > 2:
c+=1
return c
for _ in range(int(input())):
a,b,k=list(map(int,input().split()))
if a<b:
a,b=b,a
if a==b and k==0:
print("YES")
elif k==1 and a!=b and a%b==0 :
print("YES")
else:
x=primeFactors(a)
y=primeFactors(b)
if k>=2 and k<=(x+y):
print("YES")
else:
print("NO")
``` | output | 1 | 31,988 | 22 | 63,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,989 | 22 | 63,978 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def d(n):
a=0;d=3
while~n&1:n>>=1;a+=1
while d*d<=n:
while n%d<1:n//=d;a+=1
d+=2
return a+(n>1)
for s in[*open(0)][1:]:a,b,k=map(int,s.split());print("YNEOS"[[k>d(a)+d(b),0<max(a,b)%min(a,b)or a==b][k<2]::2])
``` | output | 1 | 31,989 | 22 | 63,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,990 | 22 | 63,980 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from math import *
from collections import defaultdict as dt
from sys import stdin
inp = lambda : stdin.readline().strip()
I = lambda : int(inp())
M = lambda : map(int,inp().split())
L = lambda : list(M())
mod = 1000000007
inf = 100000000000000000000
ss = "abcdefghijklmnopqrstuvwxyz"
############## All the best start coding #############
def prime(n):
prime = [True for i in range(n+1)]
p = 2
while(p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
c = []
for p in range(2, n+1):
if prime[p]:
c.append(p)
return c
primes=prime(100000)
def solve():
a1,a2,k=M()
if((a1==a2 and k==1) or (a1==a2 and k==0)):
print("No")
return
s1=s2=0
a3=gcd(a1,a2)
q=a1//a3
w=a2//a3
if(q!=1 or w!=1):
if(q!=1 and w!=1):
s1+=2
else:
s1+=1
c=0
n=a1
for x in primes:
while n%x==0:
n=n//x
c=c+1
if(n!=1):
c=c+1;
q=c
c=0
n=a2
for x in primes:
while n%x==0:
n=n//x
c=c+1
if(n!=1):
c=c+1
w=c
s2=0
s2+=(q+w)
if(s1<=k and k<=s2):
print("Yes")
else:
print("No")
##################### Submit now #####################
tt=1
tt=I()
for _ in range(tt):
solve()
``` | output | 1 | 31,990 | 22 | 63,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,991 | 22 | 63,982 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import *
from collections import defaultdict, deque, Counter
import math, string
from heapq import *
from operator import add
from itertools import accumulate
BUFSIZE = 8192
sys.setrecursionlimit(10 ** 5)
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
inf = float("inf")
en = lambda x: list(enumerate(x))
ceil_ = lambda a, b: (a + b - 1) // b
ii = lambda: int(input())
r = lambda: map(int, input().split())
rr = lambda: list(r())
# --------------------------
pp = []
prime = []
def seve(n):
global pp, prime
prime = [True] * (n + 1)
p = 2
while p * p <= n:
if prime[p] == True:
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
for p in range(2, n + 1):
if prime[p]:
pp += (p,)
def solve():
a, b, k = r()
da, db = a, b
x = y = 0
i = 0
n = len(pp)
# print(pp[:10])
while i < n and a > 1:
if a % pp[i] == 0:
x += 1
a //= pp[i]
continue
i += 1
if a > 1:
x += 1
i = 0
while i < n and b > 1:
if b % pp[i] == 0:
y += 1
b //= pp[i]
# print('b' , b)
continue
i += 1
# print(x , y)
if b > 1:
y += 1
if k == 1:
if (da % db == 0 or db % da == 0) and (da != db):
print("YES")
else:
print("NO")
return
if x + y >= k:
print("YES")
else:
print("NO")
seve(10 ** 5)
# print(len(pp))
for _ in " " * ii():
solve()
``` | output | 1 | 31,991 | 22 | 63,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,992 | 22 | 63,984 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def factorize(n):
a = 0
while n & 1 == 0:
a += 1
n >>= 1
f = 3
while f*f<=n:
if n%f==0:
a += 1
n//=f
else:
f+=2
if n!=1:
a += 1
return a
import sys
input=sys.stdin.readline
t = int(input())
ans = [-1] * t
for iiii in range(t):
a, b, c = map(int, input().split())
if c == 1:
if a == b:
ans[iiii] = "No"
else:
if max(a, b) % min(a, b):
ans[iiii] = "No"
else:
ans[iiii] = "Yes"
else:
if factorize(a) + factorize(b) >= c:
ans[iiii] = "Yes"
else:
ans[iiii] = "No"
print("\n".join(ans))
``` | output | 1 | 31,992 | 22 | 63,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,993 | 22 | 63,986 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right
import sys
def input():
return sys.stdin.readline().rstrip()
def getN():
return int(input())
def getNM():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def getArray(intn):
return [int(input()) for i in range(intn)]
mod = 10 ** 9 + 7
MOD = 998244353
INF = float('inf')
eps = 10 ** (-10)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]
#############
# Main Code #
#############
# aをaの約数のどれかにする
# bをbの約数のどれかにする
# ちょうどkターンでできるか 因数を何個かずつ落として
# 2回あれば必ず落とせる 両方1にするので
# kが大きいと厄介
# エラストテネスの篩
prime = [2]
max = 32000 # √10 ** 9
limit = int(math.sqrt(max))
data = [i + 1 for i in range(2, max, 2)]
while limit > data[0]:
prime.append(data[0])
data = [j for j in data if j % data[0] != 0]
prime = prime + data
# 素因数分解
def prime_factorize(n):
divisors = []
# 27(2 * 2 * 7)の7を出すためにtemp使う
temp = n
for i in prime:
if i > int(math.sqrt(n)):
break
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
# 素因数を見つけるたびにtempを割っていく
temp //= i
divisors.append([i, cnt])
if temp != 1:
divisors.append([temp, 1])
if divisors == []:
divisors.append([n, 1])
return divisors
T = getN()
for _ in range(T):
A, B, K = getNM()
g = math.gcd(A, B)
if K == 1:
if (A == g) ^ (B == g):
print('YES')
else:
print('NO')
else:
o1, o2 = 0, 0
for k, v in prime_factorize(A):
if k > 1:
o1 += v
for k, v in prime_factorize(B):
if k > 1:
o2 += v
if K <= o1 + o2:
print('YES')
else:
print('NO')
``` | output | 1 | 31,993 | 22 | 63,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,994 | 22 | 63,988 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def zip_sorted(a,b): return zip(*sorted(zip(a,b)))[0],zip(*sorted(zip(a,b)))[1]
def gcd(a,b): return math.gcd(a,b)
def lcm(a,b): return ((a*b)//math.gcd(a,b))
def ncr(n,r): return math.comb(n,r)
def npr(n,r): return (math.factorial(n)//math.factorial(n-r))
def decimal_to_binary(n): return bin(n).replace("0b", "")
def binary_to_decimal(s): return int(s,2)
import sys, os.path
import math
from collections import defaultdict,deque
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
S = lambda : list(map(str,input()))
import math
def primeFactors(n):
count = 0
while n % 2 == 0:
count+=1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
count+=1
n = n // i
if n > 2:
count+=1
return count
def main():
t,=I()
for t1 in range(t):
a,b,k = I()
if k==1:
if math.gcd(a,b)==min(a,b) and a!=b:
print("YES")
else:
print("NO")
else:
g1 = math.gcd(a,b)
c1 = primeFactors(a//g1)
c2 = primeFactors(b//g1)
c3 = primeFactors(g1)
if k>c1+c2+(2*c3):
print("NO")
else:
if a==1 or b==1:
if 1<=k and k<=max(c1,c2):
print("YES")
else:
print("NO")
else:
if 2<=k and k<=c1+c2:
print("YES")
else:
# k = k-(c1+c2)
# print(c3,k)
if c3==0:
print("NO")
else:
if 2<=k and k<=(c1+c2+2*c3):
print("YES")
else:
print("NO")
main()
``` | output | 1 | 31,994 | 22 | 63,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO | instruction | 0 | 31,995 | 22 | 63,990 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import math
primes = []
for i in range(2,math.floor(10**4.5)+1):
isprime = True
for j in range(2,math.floor(i**0.5)+1):
if i%j == 0:
isprime = False
break
if isprime:
primes.append(i)
T = int(input())
for t in range(T):
A,B,K = tuple(map(int,input().split()))
if K == 1:
if ((A%B == 0 or B%A == 0) and A != B):
print('YES')
else:
print('NO')
else:
if K > 58:
print('NO')
else:
s = 0
for p in primes:
#if p**2 > A or A == 1:
# break
while A%p == 0:
s += 1
A //= p
while B%p == 0:
s += 1
B //= p
#for p in primes:
# if p**2 > B or B == 1:
# break
s += (A>1)+(B>1)
if s >= K:
print('YES')
else:
print('NO')
``` | output | 1 | 31,995 | 22 | 63,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
import sys
import math
from math import factorial, inf, gcd, sqrt
from heapq import *
from functools import *
from itertools import *
from collections import *
from typing import *
from bisect import *
import random
from sys import stdin, stdout
sys.setrecursionlimit(10**5)
def inp():
return stdin.readline().strip()
def mp():
return map(int, inp().split())
def cntFac(n):
c = 0
j = 2
while n % j == 0:
c += 1
n //= j
j = 3
while j * j <= n:
while n % j == 0:
c += 1
n //= j
j += 2
return c + 1 if n > 1 else c
for _ in range(int(inp())):
a, b, k = mp()
x1, x2 = cntFac(a), cntFac(b)
if k == 1:
print("YES" if (a % b == 0 or b % a == 0) and a != b else "NO")
else:
print("YES" if k <= x1 + x2 else "NO")
``` | instruction | 0 | 31,996 | 22 | 63,992 |
Yes | output | 1 | 31,996 | 22 | 63,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
ans=0
b=[]
vis=[]
tot=0
def dfs(root):
global tot,vis,gp
for child in gp[root]:
if vis[child]==0:
tot+=1
vis[child]=1
dfs(child)
pre=[[] for i in range(3)]
pp=[1]*(31625)
pp[0]=0
pp[1]=0
for i in range(2,31625):
if pp[i]==1:
for j in range(i*i,31625,i):
pp[j]=0
prime=[]
for i in range(2,31625):
if pp[i]:
prime.append(i)
def func(a):
x=a
tot=0
for i in prime:
if x%i==0:
while(x%i==0):
tot+=1
x//=i
if x>1:
tot+=1
return tot
def hnbhai(tc):
a,b,k=sd()
if k==1:
if (a%b==0)^(b%a==0):
print("YES")
return
print("NO")
return
if func(a)+func(b)<k:
print("NO")
return
print("YES")
return
for _ in range(sb()):
hnbhai(_+1)
``` | instruction | 0 | 31,997 | 22 | 63,994 |
Yes | output | 1 | 31,997 | 22 | 63,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
def a1(arr, n, x, y):
l = 0
r = n-1
ans = 0
while (l < r):
temp = arr[l]+arr[r]
if(temp>= x and temp<=y):
ans+=(r-l)
l+=1
elif (temp < x):
l+=1
elif(temp > y):
r-=1
return ans
t = int(input())
yn = ["NO", "YES"]
for _ in range(t):
a, b, k = mapin()
x = primeFactors(a)
y = primeFactors(b)
# print(x, y)
if(k == 1):
if(a == 1 and b == 1):
print("No")
elif((b%a == 0 or a%b == 0)and b!=a):
print("Yes")
else:
if(len(x) >= 1 and len(y) >= 1):
print("No")
else:
print("Yes")
else:
if(len(x)+len(y)>= k):
print("Yes")
else:
print("No")
``` | instruction | 0 | 31,998 | 22 | 63,996 |
Yes | output | 1 | 31,998 | 22 | 63,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
import sys,os,io
import math
from collections import defaultdict
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def primeFactors(n):
cnt = defaultdict(lambda : 0)
if n==1:
return cnt
while not n%2:
cnt[2]+=1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
cnt[i]+=1
n = n // i
if n > 2:
cnt[n]+=1
return cnt
for _ in range (int(input())):
a,b,k = [int(i) for i in input().split()]
ca = primeFactors(a)
cb = primeFactors(b)
common = 0
total = 0
temp = 0
for i in ca:
if ca[i]>cb[i]:
temp += 1
break
for i in cb:
if cb[i]>ca[i]:
temp += 1
break
for i in ca:
common += min(ca[i], cb[i])
total += ca[i]
for i in cb:
total += cb[i]
diff = total - common
if a==b and k==1:
print("NO")
continue
if k>total:
print("NO")
continue
if k==1 and temp==2:
print("NO")
continue
print("YES")
``` | instruction | 0 | 31,999 | 22 | 63,998 |
Yes | output | 1 | 31,999 | 22 | 63,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
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 : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
ap = lambda ab,bc,cd : ab[bc].append(cd)
li = lambda : list(input_())
pr = lambda x : print(x)
prinT = lambda x : print(x)
f = lambda : sys.stdout.flush()
inv =lambda x:pow(x,mod-2,mod)
mod = 10**9 + 7
mx = 4*10**4
adj = [1 for i in range (mx)]
p = 2
adj[0] = 0
adj[1] = 0
while (p*p < mx) :
if (adj[p] == 1) :
for i in range(p*p,mx,p) :
adj[i] = 0
p += 1
prime = []
for i in range (mx) :
if (adj[i] == 1) :
prime.append(i)
print(len(prime))
def cal (x) :
t = 0
for i in prime :
while (x%i == 0) :
t += 1
x = x//i
if (x == 1) :
break
if (x > 1) :
t += 1
return t
for _ in range (ii()) :
a,b,k = il()
t1 = cal(a)
t2 = cal(b)
if (a == b) :
if (k == 1) :
print("NO")
continue
if (k>=2 and k<= t1+t2) :
print("YES")
else :
print("NO")
continue
if (a == 1 or b == 1) :
if (k>=1 and k<= t1 + t2) :
print("YES")
else :
print("NO")
continue
if (gcd(a,b) != 1) :
if (gcd(a,b) == a or gcd(a,b) == b) :
if (k >= 1 and k<= t1+t2) :
print("YES")
else :
print("NO")
else :
if (k >= 2 and k<= t1+t2) :
print("YES")
else :
print("NO")
else :
if (k >= 2 and k <= t1 + t2) :
print("YES")
else :
print("NO")
``` | instruction | 0 | 32,000 | 22 | 64,000 |
No | output | 1 | 32,000 | 22 | 64,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
import sys
from sys import stdin, stdout
input, print = stdin.readline, stdout.write
from collections import OrderedDict, defaultdict
import bisect
from math import sqrt
prm = [1]*5005
p = []
for i in range(2, 5005):
if prm[i] == 1:
p.append(i)
for j in range(i + i, 5005, i):
prm[j] = 0
def pp(m):
global p
ret = 0
for x in p:
if x*x > m:
break
while m % x == 0:
m //= x
ret += 1
if m > 1: ret += 1
return ret
t = int(input())
while t > 0:
t -= 1
a, b, k = [int(x) for x in input().split()]
A = pp(a)
B = pp(b)
y = [A, B]
# print(str(y))
res = "NO"
if k == 1:
if (a % b == 0 or b % a == 0) and a != b:
res = "YES"
else:
if A + B >= k:
res = "YES"
print(str(res))
print('\n')
``` | instruction | 0 | 32,001 | 22 | 64,002 |
No | output | 1 | 32,001 | 22 | 64,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
import sys
INF = float('inf')
MOD = 10**9 + 7
MOD2 = 998244353
def solve():
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LC(): return list(input())
def IC(): return [int(c) for c in input()]
def MI(): return map(int, sys.stdin.readline().split())
T = II()
MAX = 100010
Prime = [True for i in range(MAX)]
Prime[0] = False
Prime[1] = False
for i in range(2, MAX):
if (Prime[i]):
for j in range(2 * i, MAX, i):
Prime[j] = False
PrimeNum = []
for i in range(3, MAX):
if(Prime[i]):
PrimeNum.append(i)
def prime_factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
Index = 2
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f = PrimeNum[Index]
Index+=1
if n != 1:
a.append(n)
return a
import math
for t in range(T):
A,B,K = MI()
if(K==1):
if((math.gcd(A,B) == A or math.gcd(A,B) == B) and not(A==B)):
print("YES")
else:
print("NO")
continue
if(A==B):
M=0
elif(math.gcd(A,B) == A or math.gcd(A,B) == B):
M=1
else:
M=2
N = len(prime_factorize(A)) + len(prime_factorize(B))
if(M<= K <= N):
print("YES")
else:
print("NO")
return
solve()
``` | instruction | 0 | 32,002 | 22 | 64,004 |
No | output | 1 | 32,002 | 22 | 64,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers a and b. In one turn, you can do one of the following operations:
* Take an integer c (c > 1 and a should be divisible by c) and replace a with a/c;
* Take an integer c (c > 1 and b should be divisible by c) and replace b with b/c.
Your goal is to make a equal to b using exactly k turns.
For example, the numbers a=36 and b=48 can be made equal in 4 moves:
* c=6, divide b by c ⇒ a=36, b=8;
* c=2, divide a by c ⇒ a=18, b=8;
* c=9, divide a by c ⇒ a=2, b=8;
* c=4, divide b by c ⇒ a=2, b=2.
For the given numbers a and b, determine whether it is possible to make them equal using exactly k turns.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
Each test case is contains three integers a, b and k (1 ≤ a, b, k ≤ 10^9).
Output
For each test case output:
* "Yes", if it is possible to make the numbers a and b equal in exactly k turns;
* "No" otherwise.
The strings "Yes" and "No" can be output in any case.
Example
Input
8
36 48 2
36 48 3
36 48 4
2 8 1
2 8 2
1000000000 1000000000 1000000000
1 2 1
2 2 1
Output
YES
YES
YES
YES
YES
NO
YES
NO
Submitted Solution:
```
prime = [0] * 31623
t = 4
while t < 31623:
prime[t] = 1
t += 2
for i in range(3, 178, 2):
if prime[i] == 0:
temp = i * i
while temp < 31623:
prime[temp] = 1
temp += 2 * i
check = [i for i in range(2, 31623) if not prime[i]]
def count_prime(x):
c = 0
for i in check:
if not x:
break
while x % i == 0 and x > 0:
x //= i
c += 1
return c
for i in range(int(input())):
a, b, k = map(int, input().split())
if a == b and k == 0:
print("YES")
elif a == b:
print('NO')
elif (a % b == 0 or b % a == 0) and k == 1:
print('YES')
else:
ca = max(1, count_prime(a))
if a == 1:
ca = 0
cb = max(1, count_prime(b))
if b == 1:
cb = 0
if 2 <= k <= (ca + cb):
print('YES')
else:
print("NO")
``` | instruction | 0 | 32,003 | 22 | 64,006 |
No | output | 1 | 32,003 | 22 | 64,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (<image>, <image>).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172 | instruction | 0 | 32,075 | 22 | 64,150 |
Tags: brute force, implementation, math, number theory
Correct Solution:
```
import sys
p,q = map(int,sys.stdin.readline().split())
def prime(n):
#print(int(n**0.5))
for div in range(2,int(n**0.5)+1):
if n%div==0:
return False
return True
def palindrom(n):
n = str(n)
for pos in range((len(n)+1)//2):
if n[pos]!=n[-1-pos]:
return False
return True
def findMaxN(p,q):
A = p/q
n = 1
pN = 0
rubN = 1
checkAgain = False
while True:
n+=1
if prime(n):
pN += 1
checkAgain = True
if palindrom(n):
rubN+=1
checkAgain = True
if checkAgain:
checkAgain = False
if pN>A*rubN:
#return n-1
break
good_n = n-1
check_to = n+10000
delta = 0
last_good = False
while n<check_to:
n+=1
delta+=1
if prime(n):
pN += 1
checkAgain = True
if palindrom(n):
rubN+=1
checkAgain = True
#if n == 172:
#print(n,pN,A*rubN)
if checkAgain:
checkAgain = False
if pN<=A*rubN:
#return n-1
good_n = n
check_to+=delta
delta = 0
last_good = True
else:
if last_good:
last_good = False
good_n = n-1
return good_n
def doTest():
assert findMaxN(1,1)==40
assert findMaxN(1,42)==1
assert findMaxN(6,4)==172
doTest()
'''
last = -1
pN = 0
rubN = 1
n = 1
for i in range(100000):
n+=1
tmp = pN<=6*rubN
if prime(n):
pN += 1
if palindrom(n):
rubN+=1
if tmp!=last:
print(n)
print(pN,rubN)
print()
last = tmp
'''
print(findMaxN(p,q))
``` | output | 1 | 32,075 | 22 | 64,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (<image>, <image>).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172
Submitted Solution:
```
def Prov(val):
val=str(val)
return val==val[::-1]
p,q= map(int,input().split(' '))
pmas = [1 for i in range(10**6)]
pmas[0]=0
pmas[1]=0
for i in range(2,10**6):
if pmas[i]==1:
for j in range(i+i,10**6,i):
pmas[j]=0
palmas=[1 if Prov(i) else 0
for i in range(10**6)]
su=0
for i in range(1,10**6):
if palmas[i]:
su+=p
palmas[i]=su
su=0
for i in range(1,10**6):
if pmas[i]:
su+=q
pmas[i]=su
su=0
for i in range(1,10**6):
if pmas[i]<=palmas[i]:
su=i
print(su)
``` | instruction | 0 | 32,076 | 22 | 64,152 |
No | output | 1 | 32,076 | 22 | 64,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (<image>, <image>).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172
Submitted Solution:
```
import sys
import math
MAXNUM = math.inf
MINNUM = -1 * math.inf
ASCIILOWER = 97
ASCIIUPPER = 65
def getInt():
return int(sys.stdin.readline().rstrip())
def getInts():
return map(int, sys.stdin.readline().rstrip().split(" "))
def getString():
return sys.stdin.readline().rstrip()
def printOutput(ans):
sys.stdout.write()
pass
def isPalindrome(string):
k = len(string)
for i in range(k // 2):
if string[i] != string[k - i - 1]:
return False
return True
def palnums(k):
"""brute force palnums up to k"""
count = [0 for _ in range(k)]
for i in range(1, k + 1):
if isPalindrome(str(i)):
count[i] = 1
for i in range(1, len(count)):
count[i] = count[i - 1] + count[i]
return count
def primes(k):
"""Sieve primes up to k"""
isprime = [1 for _ in range(k + 1)]
isprime[0] = 0
isprime[1] = 0
MAX = math.ceil(math.sqrt(k) + 1)
for i in range(2, MAX):
if isprime[i] == 1:
for j in range(i ** 2, k + 1, i):
isprime[j] = 0
for i in range(1, len(isprime)):
isprime[i] = isprime[i - 1] + isprime[i]
return isprime
def checkrep(primes):
primes2 = [1 for _ in range(1000)]
primes2[0] = 0
primes2[1] = 0
for i in range(2,1000):
for j in range(2, i):
if i % j == 0:
primes2[i] = 0
break
for i in range(1, len(primes2)):
primes2[i] = primes2[i - 1] + primes2[i]
for i in range(1000):
if primes[i] != primes2[i]:
print("ERROR AT", i)
assert 1 == 0
def compare(pal, prime, n, p, q):
#print()
#print("N", n)
#print("prime", prime[n], q * prime[n])
#print("pal", pal[n], p * pal[n])
return q * prime[n] <= p * pal[n]
def binsearch(pal, prime, p, q):
l = 0
r = 1000000
ans = None
while l <= r:
m = (l + r) // 2
if compare(pal, prime, m, p, q):
ans = m
l = m + 1
else:
r = m - 1
return ans
def solve(p, q):
palcount = palnums(1000000)
primecount = primes(1000000)
ans = None
for i in range(1, 1000000):
if compare(palcount, primecount, i, p ,q):
ans = i
if ans:
return ans
return "Palindromic tree is better than splay tree"
def readinput():
p, q = getInts()
print(solve(p, q))
readinput()
``` | instruction | 0 | 32,077 | 22 | 64,154 |
No | output | 1 | 32,077 | 22 | 64,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (<image>, <image>).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def ispali(k):
return str(k)==str(k)[::-1]
def main():
p,q=map(int, input().split())
isprime=[1 for i in range(1000001)]
isprime[1]=0
for i in range(2, 10001):
if isprime[i]:
for j in range(i*i,1000001, i ):
isprime[j]=0
pie=0
rub=0
ans='Palindromic tree is better than splay tree'
for i in range(1,1000001):
if isprime[i]:
pie+=1
if ispali(i):
rub+=1
if pie*q<=rub*p:
ans=i
print(ans)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 32,078 | 22 | 64,156 |
No | output | 1 | 32,078 | 22 | 64,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient A find the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (<image>, <image>).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def primeN(n):
prime = [True for i in range(n+1)]
prime[0]=False
prime[1]=False
p=2
while(p*p<=n):
if(prime[p]):
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
return prime
def isPalindrome(n):
n=str(n)
return n==n[::-1]
N=10**3
primes=primeN(N)
# palindrome=[0,1]
# prime=[0,0]
# for i in range(2,N+1):
# if(isPalindrome(i)):
# palindrome.append(palindrome[-1]+1)
# else:
# palindrome.append(palindrome[-1])
# if(primes[i]):
# prime.append(prime[-1]+1)
# else:
# prime.append(prime[-1])
# print(prime[40])
# print(palindrome[40])
ans="Palindromic tree is better than splay tree"
p,q=value()
prime=0
palindrome=0
for i in range(1,N+1):
if(primes[i]): prime+=1
if(isPalindrome(i)): palindrome+=1
if( prime*q <= p*palindrome):
ans=i
print(ans)
``` | instruction | 0 | 32,079 | 22 | 64,158 |
No | output | 1 | 32,079 | 22 | 64,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.
For example, the prairie partitions of 12, 17, 7 and 1 are:
12 = 1 + 2 + 4 + 5,
17 = 1 + 2 + 4 + 8 + 2,
7 = 1 + 2 + 4,
1 = 1.
Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.
Output
Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.
If there are no such values of m, output a single integer -1.
Examples
Input
8
1 1 2 2 3 4 5 8
Output
2
Input
6
1 1 1 2 2 2
Output
2 3
Input
5
1 2 4 4 4
Output
-1
Note
In the first example, Alice could get the input sequence from [6, 20] as the original sequence.
In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3]. | instruction | 0 | 32,134 | 22 | 64,268 |
Tags: binary search, constructive algorithms, greedy, math
Correct Solution:
```
from collections import Counter
from math import log2, ceil
MAX = ceil(log2(10 ** 12))
def can():
seqs_cp = Counter(seqs)
for num in set(nums):
cnt = nums[num]
while cnt != 0 and num < 2 ** 63:
dif = min(cnt, seqs_cp[num])
cnt -= dif
seqs_cp[num] -= dif
num *= 2
if cnt > 0:
return False
return True
n = int(input())
seq = list(map(int, input().split()))
cnt = Counter(seq)
nums = Counter(seq)
seqs = Counter()
cur_cnt = cnt[1]
cur_pow = 1
while cur_cnt != 0:
nums[cur_pow] -= cur_cnt
cur_pow *= 2
if cur_cnt > cnt[cur_pow]:
seqs[cur_pow // 2] = cur_cnt - cnt[cur_pow]
cur_cnt = cnt[cur_pow]
for num in set(nums):
addition = nums[num]
nums[num] = 0
nums[2 ** int(log2(num))] += addition
# remove elements with zero count
nums -= Counter()
seqs -= Counter()
cur_len = sum(seqs[num] for num in set(seqs))
res = []
cur_pow = 1
while can():
res.append(cur_len)
while seqs[cur_pow] == 0 and cur_pow <= 2 ** 63:
cur_pow *= 2
if cur_pow > 2 ** 63:
break
other_pow = 1
while other_pow <= cur_pow:
nums[other_pow] += 1
other_pow *= 2
seqs[cur_pow] -= 1
cur_len -= 1
print(-1 if len(res) == 0 else ' '.join(map(str, reversed(res))))
``` | output | 1 | 32,134 | 22 | 64,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x.
For example, the prairie partitions of 12, 17, 7 and 1 are:
12 = 1 + 2 + 4 + 5,
17 = 1 + 2 + 4 + 8 + 2,
7 = 1 + 2 + 4,
1 = 1.
Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice's original sequence could contain. Find all possible options!
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.
Output
Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.
If there are no such values of m, output a single integer -1.
Examples
Input
8
1 1 2 2 3 4 5 8
Output
2
Input
6
1 1 1 2 2 2
Output
2 3
Input
5
1 2 4 4 4
Output
-1
Note
In the first example, Alice could get the input sequence from [6, 20] as the original sequence.
In the second example, Alice's original sequence could be either [4, 5] or [3, 3, 3]. | instruction | 0 | 32,135 | 22 | 64,270 |
Tags: binary search, constructive algorithms, greedy, math
Correct Solution:
```
from collections import Counter
from math import log2
powers = [2 ** p for p in range(64)]
def can():
near = Counter()
for num in set(nums):
p = int(log2(num))
near[p] += nums[num]
seq_cp = seqs - Counter()
for num in set(near):
cnt = near[num]
while cnt != 0 and num < 64:
dif = min(cnt, seq_cp[powers[num]])
cnt -= dif
seq_cp[powers[num]] -= dif
num += 1
if cnt > 0:
return False
return True
n = int(input())
seq = list(map(int, input().split()))
cnt = Counter(seq)
nums = Counter(seq)
seqs = Counter()
cur_cnt = cnt[1]
cur_pow = 1
while cur_cnt != 0:
nums[cur_pow] -= cur_cnt
cur_pow *= 2
if cur_cnt > cnt[cur_pow]:
seqs[cur_pow // 2] = cur_cnt - cnt[cur_pow]
cur_cnt = cnt[cur_pow]
for num in set(nums):
addition = nums[num]
nums[num] = 0
nums[2 ** int(log2(num))] += addition
# remove elements with zero count
nums -= Counter()
seqs -= Counter()
cur_len = sum(seqs[num] for num in set(seqs))
res = []
cur_pow = 1
while can():
res.append(cur_len)
while seqs[cur_pow] == 0 and cur_pow <= 2 ** 63:
cur_pow *= 2
if cur_pow > 2 ** 63:
break
other_pow = 1
while other_pow <= cur_pow:
nums[other_pow] += 1
other_pow *= 2
seqs[cur_pow] -= 1
cur_len -= 1
print(-1 if len(res) == 0 else ' '.join(map(str, reversed(res))))
``` | output | 1 | 32,135 | 22 | 64,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set.
Input
The first line contains a single integer n (1 ≤ n ≤ 16).
The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.
The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.
Output
Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.
Examples
Input
3
2 3 5
7
Output
8
Input
5
3 7 11 13 31
17
Output
93
Note
The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:
(1, 2, 3, 4, 5, 6, 8, ...)
The seventh number in this list (1-indexed) is eight.
Submitted Solution:
```
n = int(input())
matrix = set(int(i) for i in input().split())
k = int(input())
print(matrix)
def is_prime(x):
if x > 3 and x % 2 == 0 or x <= 1:
return False
for i in range(3, int(x ** 0.5) + 1, 2):
if x % i == 0:
return False
return True
def right(m):
k = 1
i = 0
for el in m:
if (m[i] not in matrix):
k = 0
break
i += 1
return k
def deliteli(l):
a = []
for i in range(1, l + 1, 1):
if (l % i == 0 and is_prime(i)):
a.append(i)
return a
i = 1
b = 2
while (i != k):
m = deliteli(b)
if (right(m) == 1):
i += 1
if i != k:
b += 1
else:
b += 1
print(b)
``` | instruction | 0 | 32,206 | 22 | 64,412 |
No | output | 1 | 32,206 | 22 | 64,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set.
Input
The first line contains a single integer n (1 ≤ n ≤ 16).
The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.
The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.
Output
Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.
Examples
Input
3
2 3 5
7
Output
8
Input
5
3 7 11 13 31
17
Output
93
Note
The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:
(1, 2, 3, 4, 5, 6, 8, ...)
The seventh number in this list (1-indexed) is eight.
Submitted Solution:
```
from math import *
n = int(input())
primes = [int(i) for i in input().split()]
k = int(input())
elements = [1]
while len(elements) < k:
potential = []
for a in range(0, n):
prime = primes[a]
temp = 1
while elements[temp - 1] * prime <= elements[-1]:
if elements[temp - 1] * prime * prime <= elements[-1]:
temp *= prime
else:
temp += 1
potential.append(prime * elements[temp - 1])
elements.append(min(potential))
print(elements[-1])
``` | instruction | 0 | 32,207 | 22 | 64,414 |
No | output | 1 | 32,207 | 22 | 64,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set.
Input
The first line contains a single integer n (1 ≤ n ≤ 16).
The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.
The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.
Output
Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.
Examples
Input
3
2 3 5
7
Output
8
Input
5
3 7 11 13 31
17
Output
93
Note
The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:
(1, 2, 3, 4, 5, 6, 8, ...)
The seventh number in this list (1-indexed) is eight.
Submitted Solution:
```
from math import log
MAXVAL = 10**18
n = int(input())
primes = list(map(int, input().split()))
k = int(input())
possible_numbers = [1]
for prime in primes:
additional_numbers = []
for number in possible_numbers[:k]:
maxpow = int (1.01 + log(MAXVAL/number)/log(prime))
if maxpow>1:
additional_numbers.extend([number*(prime**pow) for pow in range(1, maxpow) if number*(prime**pow) <= MAXVAL])
else:
break
print('at prime', prime, 'got', additional_numbers[:10])
possible_numbers.extend(additional_numbers)
possible_numbers.sort()
print('up to', prime, 'total', len(possible_numbers), possible_numbers[:20])
print(sorted(possible_numbers)[k-1])
``` | instruction | 0 | 32,208 | 22 | 64,416 |
No | output | 1 | 32,208 | 22 | 64,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set.
Input
The first line contains a single integer n (1 ≤ n ≤ 16).
The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.
The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.
Output
Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.
Examples
Input
3
2 3 5
7
Output
8
Input
5
3 7 11 13 31
17
Output
93
Note
The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:
(1, 2, 3, 4, 5, 6, 8, ...)
The seventh number in this list (1-indexed) is eight.
Submitted Solution:
```
n = int(input())
arr = [int(i) for i in input().split()]
k = int(input())
num = []
def resh(delim, gen):
for i in gen:
if i % delim:
yield i
else:
num.append(i)
gen = resh(arr[0], range(10**4+1))
for i in arr[1:]:
gen = resh(i, gen)
for k, v in enumerate(gen):
pass
num.sort()
num[0] = 1
print(num[6])
``` | instruction | 0 | 32,209 | 22 | 64,418 |
No | output | 1 | 32,209 | 22 | 64,419 |
Provide a correct Python 3 solution for this coding contest problem.
Example
Input
3
Output
1 | instruction | 0 | 32,457 | 22 | 64,914 |
"Correct Solution:
```
N = int(input())
def gcd(m, n):
while n:
m, n = n, m % n
return m
def lcm(m, n):
return m // gcd(m, n) * n
def carmichael(x):
r = 1
b = 0
while x & 1 == 0:
b += 1
x >>= 1
if b > 1:
r = 2 if b == 2 else 2**(b-2)
y = 3
while y*y <= x:
if x % y == 0:
c = 0
while x % y == 0:
x //= y
c += 1
r = lcm(r, (y-1) * y**(c-1))
y += 1
if x > 1:
r = lcm(r, x-1)
return r
def solve(x, c):
if c == 1:
return 1
x = x % c
sq = int(c ** .5) + 1
mp = {}
y = 1
for i in range(sq):
mp[y] = i
y = (y * x) % c
if y == 1:
return i+1
if y in mp:
return -1
z = 1
for i in range(sq):
z = (z * y) % c
if z in mp:
return sq*(i+1) - mp[z]
return -1
c = carmichael(N)
k = solve(N, c)
if k == -1 or pow(N, k, c) != 1 % c:
print(-1)
else:
print(k)
``` | output | 1 | 32,457 | 22 | 64,915 |
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