hide handkerchief |
| Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 6646 Accepted Submission(s): 2173 |
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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". |
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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.
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Output
For each input case, you should only the result that Haha can find the handkerchief or not.
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Sample Input
3 2 |
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Sample Output
YES |
最一开始并没有发现这道题的规律，只是盲目地将前几种情况列了出来。
发现了一些“端倪”。。。然并卵，还是看了别人的博客
http://blog.sina.com.cn/s/blog_86a6befd01012bqi.html
发现了解法应该是利用辗转相除求解N和M-1的公约数
#include <iostream>
using namespace std;
int main(){
int n,m,t;
while () {
scanf("%d%d",&n,&m);
if(n==-&&m==-)break;
while (m!=) {
t=n%m;
n=m;
m=t;
}
if(n==)printf("YES\n");
else printf("POOR Haha\n");
}
return ;
}
总结：1.无限循环tips-->while(1){}