ACM 辗转相除法（求最大公约数 ）

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：

题目就是让我们求最大公约数，如果两个数的最大公约数等于1，就输出NO，否则输出YES。我们用辗转相除法求最大公约数

程序代码：

#include <iostream>
using namespace std;
int gcd(int a,int b)
{
     return b==0 ? a : gcd(b,a%b);
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int m,n;
		cin>>m>>n;
		int p=gcd(m,n);
		if(p==1)
			cout<<"NO"<<endl;
		else
			cout<<"YES"<<endl;
	}
	return 0;
}