Joseph(P1012)

本文探讨约瑟夫环问题的一个变种,即找到最小的m值,使得所有坏人在第一个好人被选中之前被选出。通过预先计算并列出关键数值结果,提供了一个高效的解决方案。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

直接模拟出会严重超时,反正只有几个数,打个表就行了,



int x[] = 27530169441187276321740933134599011358657,  2504881 };



#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
using namespace std;


int main()
{
	//freopen("fuck.txt","r",stdin);
	int i,j,k;
	
	int n;
	while (cin>>n,n)
	{
		for (j=n+1;;j++)
		{
			//cout<<j<<endl;
			int vist[30]={0};
			//memset(vist,0,sizeof(vist));
			k=n;

			int step=1;
			i=0;

			while (k)
			{
				i=(i+1)%(2*n);
				if (!vist[i])
				{
					step++;
				}
				if (step==j)
				{
					step=0;
					vist[i]=1;
					k--;
				}
			}
			for (i=0;i<n;i++)
				if (vist[i])
					break;
			//cout<<i<<endl;
			if (i==n)
				break;
		}
		cout<<j<<endl;
	}

	return 1;
}
Joseph
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 39656 Accepted: 14827

Description

The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.

Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.

Input

The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.

Output

The output file will consist of separate lines containing m corresponding to k in the input file.

Sample Input

3
4
0

Sample Output

5
30

Source



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值