UC知道wolf and rabbit
来源：百度知道 编辑：UC知道 时间：2024/09/22 07:34:16
Wolf and Rabbit
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 78 Accepted Submission(s) : 17
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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 exis
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 78 Accepted Submission(s) : 17
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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 exis
#include<iostream>
#include<cmath>
using namespace std;
int main()
{ int T;
int i,j,n;
int m;
int work;
cin>>T;
while(T--)
{
work=0;
cin>>m>>n;
for(i=2;i<=min(m,n);i++)
if(m%i==0&&n%i==0)
{
cout<<"YES"<<endl;
work=1;
break;
}
if(work==0)
cout<<"NO"<<endl;
}
return 0;
}
25