杭电_ACM_Wolf and Rabbit

最新推荐文章于 2021-05-13 16:25:24 发布

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最新推荐文章于 2021-05-13 16:25:24 发布

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原文链接：http://www.cnblogs.com/chuanlong/archive/2012/10/30/2745787.html

本文通过一个趣味性的故事背景，深入探讨了寻找安全洞穴的问题，利用数学中的最大公约数（GCD）概念解决实际问题。详细解释了如何判断是否存在安全洞穴，并通过代码实现了解决方案。此过程不仅锻炼了读者的数学思维，还展示了算法在解决实际问题中的应用。

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

View Code

 1 #include <stdio.h>
 2 //get the greatest common divisor
 3 int gcd(int a, int b)
 4 {
 5     int r;
 6     while (b)
 7     {
 8         r = a % b;
 9         a = b;
10         b = r;
11     }
12     return a;
13 }
14 int main()
15 {
16     int N, m, n;
17     scanf("%d", &N);
18     while(N--)
19     {
20         scanf("%d %d", &n, &m);
21         if(gcd(n, m) == 1)
22             printf("NO\n");
23         else
24             printf("YES\n");
25     }
26     return 0;
27 }