A - Wolf and Rabbit

最新推荐文章于 2021-05-23 19:07:16 发布

DTL66

最新推荐文章于 2021-05-23 19:07:16 发布

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分类专栏： 3.GCD+LCM

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本文介绍了一道经典的狼追兔问题，通过数学方法判断兔子在循环路径中是否能避开狼的搜索。文章提供了完整的C++代码实现，并利用最大公约数算法来判断安全洞穴是否存在。

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A - Wolf and Rabbit

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

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

1.题意：

2.思路：

3.失误：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
 
__int64 GCD(__int64 a,__int64 b)
{
    if(a%b==0)
     return b;
     return GCD(b,a%b); 
}

int main()
{
    __int64 m,i,n,t;
	cin>>t;
	while(t--) 
    {
    	cin>>m>>n;
    	if(GCD(m,n)==1)
    	{
    		cout<<"NO"<<endl;
		}
		else
		{
			cout<<"YES"<<endl;
		}
	}
    return 0;
    
}