狼和兔子 数论

数论Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

There is a hill with n holes around. The holes are signed from 0 to n-1.



A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n(0<m,n<2147483648).

Output

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

Sample Input

2
1 2
2 2

Sample Output

NO
YES

题目大意：：一只狼一只兔，狼围绕一个均匀分布着n个兔子洞的山转圈，狼每经过m个洞口，就会进入洞中，那么这个洞就是不安全的。
问n个洞中，是否存在安全的洞让兔子藏身。


解题思路：这题其实是找n和m有没有公约数，如果有公约数那么狼找完一圈之后他就会一直找这么几个洞，如果没有公约数，那么狼就总会找到的只是时间问题

具体代码如下：

#include <stdio.h>
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    int T;
    scanf("%d",&T);
     while(T--)
     {
         int m,n;
        scanf("%d%d",&m,&n);
         int j=gcd(n,m);
         if(j==1)
             printf("NO
");
         else
             printf("YES
");
     }
     return 0;
 }