Wolf and Rabbit（最大公约数）

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
    
    

解题思路：

题目大意是给n个洞，收尾相接，标号为0 ~ n - 1，狼从第0个洞开始每m个洞走一步，可能有些洞永远无法走到，兔子藏在其中一个洞中，问是否安全。

这题列一下方程就很明显，将设狼走了s步，兔子藏在编号为a的洞中，则m * s - a = k * n。移项后：m * s + k * n = a。要使s和k为整数则a能被m和n的最大公约数整除。

而a是不确定的，所以只有最大公约数为1才行。

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;
int Gcd(int a, int b)
{
    return b ? Gcd(b, a % b) : a;
}
int main()
{
    int p, m, n;
    scanf("%d", &p);
    while(p--)
    {
        scanf("%d%d", &m, &n);
        if(Gcd(m, n) == 1)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}