您的位置：首页 > 其它

ACM--steps--2.2.4--求最大公约数

2015-02-02 11:47 246 查看

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2020 Accepted Submission(s): 1145

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

[align=left]Sample Input[/align]

2
1 2
2 2

[align=left]Sample Output[/align]

NO
YES

[align=left]Author[/align]
weigang Lee

一开始想到Joseph。。。后来看了看，原来就是最大公约数的问题。

#include<iostream>
using namespace std;
bool dyx(int m,int n)
{
if(m>n)
{
m=m^n;
n=m^n;
m=m^n;
}
while(n%m)
{
int wyx=n%m;
n=m;
m=wyx;
}
return m==1?false:true;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int m,n;
cin>>m>>n;
if(dyx(m,n))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return  0;
}<span style="color:#66ff99;">
</span>

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航