Wolf and Rabbit

标签：nes   pos   first   signed   amp   integer   str   无法   技术   

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

题意 ： 狼可以每次走m步，而这里有n个洞，问是否存在狼无法走到的洞让兔子藏身。
分析 ： 只要n和m存在着最大公约数，那么狼就会一直在0，最大公约数，m的位置上循环，不会走到其他的洞上去。
这里简单讲一下最大公约数和最小公倍数的算法：
　　　　最大公约数

int gcd(int i,int j)
{
    if(j==0)
        return i;
    else
        gcd(j,i%j);
}

　　　　最小公倍数 = i*j/最大公约数；

代码：

#include <iostream>

using namespace std;

int gcd(int i,int j)
{
    if(j==0)
        return i;
    else
        gcd(j,i%j);
}

int main()
{
   int T;
   while(cin >> T)
   {
       while(T--)
       {
            int a,b;
            cin >> a >> b;
            if(gcd(a,b)==1)
                cout << "NO";
            else
                cout << "YES";
            cout << endl;
       }
   }
}

Wolf and Rabbit

标签：nes   pos   first   signed   amp   integer   str   无法   技术   

原文地址：http://www.cnblogs.com/7750-13/p/7270114.html