A - Wolf and Rabbit_rpg rabbit-优快云博客

本文链接：https://blog.youkuaiyun.com/DNMTOOBA/article/details/75675459

本文探讨了一种关于狼寻找兔子的游戏算法，通过分析输入参数m和n的关系来判断是否存在安全洞穴供兔子躲避。该算法考虑了不同情况下洞穴的安全性，并提供了相应的代码实现。

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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

题解：
无论n==1或m==1，结果都是NO;
当n>m时，如果n是m的倍数，那么只有固定的部分洞会被遍历；
当n<=m时，如果m是n的倍数，那么也只有固定的部分洞会被遍历；

代码如下：

#include<cstdio>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int m,n;
        scanf("%d%d",&m,&n);
        if(n>m){
            if(n==1||m==1){
                printf("NO\n");
            }else if(n%m==0){
                printf("YES\n");
            }else{
                printf("NO\n");
            }
        }else{
            if(n==1||m==1){
                printf("NO\n");
            }else if(m%n==0){
                printf("YES\n");
            }else{
                printf("NO\n");
            }
        }

    }
    return 0;
 }