time limit per test
1 second
memory limit per test
256 megabytes
Pasha loves prime numbers∗! Once again, in his attempts to find a new way to generate prime numbers, he became interested in an algorithm he found on the internet:
- To obtain a new number y, repeat k times the decimal representation of the number x (without leading zeros).
For example, for x=52 and k=3, we get y=525252, and for x=6 and k=7, we get y=6666666.
Pasha really wants the resulting number y to be prime, but he doesn't yet know how to check the primality of numbers generated by this algorithm. Help Pasha and tell him whether y is prime!
∗An integer x is considered prime if it has exactly 2 distinct divisors: 1 and x. For example, 13 is prime because it has only 2 divisors: 1 and 13. Note that the number 1 is not prime, as it has only one divisor.
Input
Each test consists of several sets of input data. The first line contains a single integer t (1≤t≤100) --- the number of sets of input data. The following lines describe the sets of input data.
The first and only line of each data set contains two integers: x and k (1≤x≤109, 1≤k≤7).
Output
For each set of input data, output <<YES>> (without quotes) if the resulting number y will be prime, and <<NO>> otherwise.
You may output <<Yes>> and <<No>> in any case (for example, the strings <<yES>>, <<yes>>, and <<Yes>> will be recognized as positive answers).
Example
Input
Copy
4
52 3
6 7
7 1
1 7
Output
Copy
NO
NO
YES
NO解题说明：此题是一道数学题，题意就是将一个数字x重复k次，判断产生的数是否是质数，直接判断即可。
cpp
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
long long int n;
long int l, m, h, x, s;
l = 1;
scanf("%lld %ld", &n, &x);
if (x >= 2 && n != 1)
{
printf("NO\n");
}
else if (n == 1 && x != 2)
{
printf("NO\n");
}
else if (n == 1 && x == 2)
{
printf("YES\n");
}
else
{
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
{
printf("NO\n");
l = 0;
break;
}
}
if (l)
{
printf("YES\n");
}
}
}
return 0;
}