E - Wolf and Rabbit ( 狼和羊 )

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. 

InputThe 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). 
OutputFor 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

题目意思:有 n 个山洞 , 从 0 , 1, 2,----,n-1,编号; 狼从0号山洞开始,每走m步进入一次山洞,无限循环,当有山洞狼没有进入的话,说明羊是安全的;

解法:数学题,找规律  

#include <iostream>

using namespace std;

int DP(int a,int b)
{
    if(a==1)
        return 0;
    int n = 1;
    while(1)
    {
         n = b % a;
        if(n==0)
            return 1;
        if(n==1)
            return 0;
        b = a;
        a = n;
    }

}

int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int a,b;
        cin>>a>>b;
       if(DP(a,b))
        cout<<"YES"<<endl;
       else
        cout<<"NO"<<endl;
    }

    return 0;
}