Wolf and Rabbit

http://acm.hdu.edu.cn/showproblem.php?pid=1222

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6052    Accepted Submission(s): 3032


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
 

 题解:就是求n和m 是否互质,求gcd(n,m)是否等于1即可

代码:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 int gcd (int a , int b)
 6 {
 7     return b==0? a :gcd(b,a%b);
 8 }
 9 int main()
10 {
11     int n , m ;
12     int t ;
13     cin>>t;
14     while(t--)
15     {
16         scanf("%d%d",&n,&m);
17         if(gcd(n,m)==1)
18             cout<<"NO"<<endl;
19         else cout<<"YES"<<endl;
20     }
21     return 0;
22 }
原文地址:https://www.cnblogs.com/shanyr/p/4675865.html