E - Wolf and Rabbit

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. 

InputThe 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). 
OutputFor 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
找互质的数

 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 int TM(int a,int b)
16 {
17     if(a%b==0)
18         return b;
19     return TM(b,a%b);
20 }
21 int main()
22 {
23     int n,m;
24     int t;
25     cin>>t;
26     while(t--)
27     {
28         cin>>n>>m;
29         if(n==1)
30         {
31             cout<<"NO"<<endl;
32             continue;
33         }
34         if(TM(m,n)!=1)
35             cout<<"YES"<<endl;
36         else
37             cout<<"NO"<<endl;
38     }
39 
40     return 0;
41 }

View Code

 

转载于:https://www.cnblogs.com/dulute/p/7272963.html