HDU 1222 Wolf and Rabbit（数学，找规律）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9848    Accepted Submission(s): 5011

Problem 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

 

Author

weigang Lee

 

Source

杭州电子科技大学第三届程序设计大赛

 

分析：

    
    一个山围绕着n个洞，编号依次为0,1,2,...n-1，
    狼先从第一个洞开始寻找兔子，
    然后每隔m个洞寻找，依次循环，
    如果到最后每隔洞都被狼找了一遍，
    那么兔子是不安全的，要输出NO，否则输出YES。
    思路：
    求出n,m的最大公约数，如果两个数的最大公约数为1，
    那么每个洞都会被狼搜寻到，应该输出NO，否则输出YES。

   

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int gcd(int a,int b)//最大公约数
{
    if (b==)
        return a;
    return gcd(b, a%b);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d %d",&m,&n);
        if(gcd(n,m)==)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return ;
}