HDU 2104 - hide handkerchief(数论)

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本文介绍了一种基于儿童游戏“藏手帕”的寻宝算法。该算法通过设定人数N和步长M，来判断是否能遍历所有人形成的一个环形队列。核心在于判断M与N是否互质以确保遍历的完整性。

hide handkerchief

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4216 Accepted Submission(s): 2008

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
3 2
-1 -1

Sample Output
YES

题意:
给出N个人,让这N个人坐成环,然后给出步数M,每M步就标记,看能否使得N个人都被标记.

解题思路:
这N个人看能否构成模N加法循环群,即使得群中元素为1-N,如果可以构成,那么标志着所有的人都被标记了.

如果是N阶加法循环群,那么他的生成元和N是互质的.

AC代码:

#include<stdio.h>
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int main()
{
    int n;
    int m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n == -1 && m == -1)  return 0;
        if(gcd(n,m) == 1)
        {
            printf("YES\n");
        }
        else
        {
            printf("POOR Haha\n");
        }
    }
    return 0;
}