HDOJ 1222 Wolf and Rabbit ( GCD 和思维 )

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7346    Accepted Submission(s): 3677

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

21 22 2

 

Sample Output

NOYES

 

Author

weigang Lee

 

Source

杭州电子科技大学第三届程序设计大赛

思路：如果两个数字是互质的，就会被捉到。因为，两数的最大公约数是 1 ，这就意味着，每次寻找都会是隔着旁边的洞穴寻找，还要注意特殊情况，当两者相等或者一者为 1 的时候。

代码：

#include<stdio.h>#include<stdlib.h>#include<math.h>#include<string.h>#include<algorithm>#define MYDD 1103using namespace std;int GCD(int x,int y) {//最大公约数if(y==0)return x;return GCD(y,x%y);//辗转相除法符合递归}int LCM(int x,int y) {//最小公倍数return x*y/GCD(x,y);}int main() {int t,n,m;scanf("%d",&t);while(t--) {scanf("%d %d",&m,&n);if(m==1) {puts("NO");//找到，不安全} else {if(m==n) {puts("YES");//找不到} else {if(GCD(m,n)==1)puts("NO");//找到，不安全elseputs("YES");//找不到}}}return 0;}