hide handkerchief
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6917 Accepted Submission(s): 3110
Problem Description
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
开始看了半天没看懂题目，发现是N个人围成一个圈，定义M，每隔M个人搜索一次，类似于约瑟夫环但又不去掉盒子的搜索，当M与N互质是必然搜到手帕。用辗转相除法求得二者的公约数为1是成立。
AC代码如下：
#include <iostream>
using namespace std;
int main()
{
int a ,b ,r ;
while (cin >> a >> b )
{
if(a==-1&&b==-1) break;
else
while(b!=0 )
{
r = a%b;
a = b;
b = r;
}
if(a==1)
{
cout<<"YES"<<endl;
}
else
cout<<"POOR Haha"<<endl;
}
return 0 ;
}