HDU 1222-优快云博客

本博客介绍了一个基于最大公约数(GCD)的算法问题，通过判断两个正整数的最大公约数是否为1来确定兔子是否能在狼的搜索中生存。详细解析了输入输出格式，提供了C++代码实现，利用递归函数计算GCD。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目传送

题目描述:
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
```
题目思路：
只需要判断m和n的最大公约数是否等于1即可，不等于一则永远抓不到；
代码：

/*************************************************************************
    > File Name: HDU-1222.cpp
    > Author: sunowsir
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;

int gcd(int a,int b){
    return b==0? a:gcd(b , a%b);
}

int main(){
    int m,n,N;
    cin>>N;
    while(N--){
        cin>>m>>n;
        if(gcd(m,n)!=1)  cout<<"YES"<<endl;
        else  cout<<"NO"<<endl;
    }
    return 0;
}