hdoj 1222 Wolf and Rabbit

最新推荐文章于 2019-04-07 21:46:48 发布

原创最新推荐文章于 2019-04-07 21:46:48 发布 · 701 阅读

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本文介绍了一种经典的狼追兔问题，通过分析狼在n个洞中按特定规律寻找兔子的过程，探讨了兔子如何选择安全洞穴以躲避狼的捕捉。文章提供了判断存在安全洞穴的条件，并附带实现代码。

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Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7065 Accepted Submission(s): 3532

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

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

解题思路：只要n和m互质时-.-才可遍历所有的洞。。

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int gcd(int nn,int mm)
{
    if (nn%mm==0) return mm;
    else
    return gcd(mm,nn%mm);
}
int main()
{
    int n,m,lp,t;scanf("%d",&t);
    while (t--)
    {
    	scanf("%d%d",&n,&m);
        if (n>m)
        lp=gcd(n,m);
        else
        lp=gcd(m,n);
        if (lp==1)
        printf("NO\n");
        else
        printf("YES\n");
    }
     return 0;
}