hide handkerchief hdu 2104

最新推荐文章于 2025-08-09 15:58:26 发布

转载最新推荐文章于 2025-08-09 15:58:26 发布 · 126 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/tangcong/archive/2011/08/16/2141523.html

文章标签：

#java

本文介绍了一种基于儿童游戏“藏手帕”的算法问题，通过分析游戏规则，提出了解决方案。当玩家Haha在N个人形成的圆圈中寻找藏在某人背后的盒子中的手帕，每次跳过M-1个盒子进行搜索时，文章给出了判断Haha是否能通过特定次数找到手帕的条件。通过编程验证，发现当N和M的最大公约数等于1时，Haha能找到手帕。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

hide handkerchief

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1107 Accepted Submission(s): 530

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

Source

这题我写了两个程序，一个暴力来规律，最后发现只要gcd(N,M) == 1,就输出YES，哈哈，一个来AC。。。

 1 //暴力找规律的
 2 #include <stdio.h>
 3 #include <stdlib.h>
 4 #include <string.h>
 5 
 6 bool f[100000010];

15 int fun(int N, int M )
16 {
17   int count = 0, flag = 1;
18   int i = 1;
19   for( i = 1; i <= N; ) {
20           if (f[i] == 1)
21               break;
22           f[i] = 1, count++,printf("%d %d\n",i, i + M);
23           i += M;
24           if (i > N)
25               i = i % N;
26           
27   }
28   if (count == N)
29     return 1;
30    return 0;
31 }
32         
33 
34 
35 
36 int main( )
37 {
38     int N, M;
39     while (scanf("%d%d", &N, &M), N != -1 && M !=-1)
40     {    
41         memset(f, 0, sizeof(f));
42         if(fun(N, M))
43             puts("YES");
44         else
45             puts("NO");
46     }
47     return 0;
48 }

//AC 代码
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int gcd(int n, int m)
{
return m ? gcd(m, n %m) : n;
}

int main( )
{
int N, M;
while(scanf("%d%d",&N, &M), N != -1 && M != -1)
    {
if(gcd(N,M) == 1)
            puts("YES");
else
            puts("POOR Haha");
    }
return 0;
}