HDU 1222 Wolf and Rabbit

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7022    Accepted Submission(s): 3502

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

 

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题意：有山有n个洞。标志为从0到n - 1。一只兔子必须藏在一个洞。一只狼搜索兔子以逆时针顺序。他进入的第一个洞是一个0。然后他将进入每隔开m个洞，再进入下一个洞。例如,m = 2和n = 6,狼将进入洞为0、2、4,0。如果兔子藏在洞1,3或5,她将生存。所以我们称这些漏洞的安全漏洞，给出n,m

问是否存在安全洞

思路：n,m互质就不存在，否则存在。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;__int64 init(__int64 a,__int64 b){__int64 t,m=1;if(a<b){t=a;a=b;b=t; }  while(m) { m=a%b; a=b; b=m; } return a;}int main(){int t;__int64 m,n;scanf("%d",&t);while(t--){scanf("%I64d%I64d",&m,&n);if(init(m,n)==1)printf("NO\n");elseprintf("YES\n");}return 0;}