HDU ACM 1222 Wolf and Rabbit

最新推荐文章于 2018-09-25 09:20:22 发布

金甲虫Scarb

最新推荐文章于 2018-09-25 09:20:22 发布

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分类专栏： ----------1.编程语言---------- ---ACM-递归文章标签： acm

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本文探讨了一道数学问题，通过编程实现来解决寻找兔子在狼搜索路径中可能生存的洞穴的问题。利用最大公因数的概念，判断是否存在安全洞穴，提供了解题思路与参考代码。

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原题描述

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

解题思路

试验得出当m和n互质时，不存在安全的洞。所以本题可以转化为求最大公因数。
当最大公因数为1，则m和n互质，不存在安全洞；当最大公因数大于1，存在安全洞。

参考代码

/*
    HDU ACM 1222 Wolf and Rabbit
    Author: Scarb
    2016-3-15
*/
#include <iostream>
using namespace std;
__int64 gcd(__int64 a, __int64 b)   // a > b
{
    if (a < b)return gcd(b, a);
    if (!b)return a;
    return gcd(b, a%b);
}
int main()
{
    int p;
    __int64 m, n;
    cin >> p;
    while(p--)
    {
        cin >> m >> n;
        if (gcd(m, n) == 1)cout << "NO\n";
        else cout << "YES\n";
    }
}