The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.
Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .
Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place
of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A.
So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha".
Input
There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.
Output
For each input case, you should only the result that Haha can find the handkerchief or not.
Sample Input
3 2
-1 -1
Sample Output
YES
Answer
#include <iostream>
using namespace std;
int main()
{
int m,n,temp,r; // haha 在m个人中 每n个找一次 所以只要m和n互质 就可以搜遍每个盒子 所以此题是求最大公约数问题
while(cin>>m>>n,m!=-1 || n!=-1)
{
if(m<n)
{
temp=m;
m=n;
n=temp;
}
r=n; // 求最大公约数
while(r!=0)
{
r=m%n;
m=n;
n=r;
}
if(m==1)
cout<<"YES"<<endl;
else
cout<<"POOR Haha"<<endl;
}
return 0;
}
本文回忆了作者儿时玩过的‘寻找手帕’游戏,并介绍了游戏规则和玩法。通过输入人数N和搜索间隔M,计算Haha是否能找到隐藏的手帕。文章详细解释了如何判断Haha能否找到手帕,包括使用最大公约数的概念解决实际问题。

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