POJ 1012：Joseph

Posted 2020-09-23 liguangsunls

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 50068		Accepted: 19020

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

非常小的时候就有的约瑟夫问题，就是一群人(人数为n)围成一桌，从1到n标上号，然后来一个数m，每次数到m的人就被淘汰，从下一个人開始再数m个数，数到m的再被淘汰，就这么淘汰去吧。

这题是有n个好人，n个坏人。

好人的标号是从1到n，坏人的标号是从n+1到2*n。题目要找一个m，把坏人都淘汰掉，好人一个都不淘汰。

这题的关键在于不要纠结与坏人的标号，不论人数还剩多少，好人的标号始终是1到n。坏人的标号始终在后面。淘汰一个坏人。仅仅需把剩余的人数减1，剩下的坏人把之前淘汰的坏人填补上，穿好他们的标号就好。所以举个样例

6个人：1 2 3 4 5 6

m=5

第一次从1開始数5位，淘汰5，剩余 1 2 3 4 5(6就往前移一位。穿上5的衣服，这样好人就还是标号1 2 3，坏人标号4 5。剩余5个人)

第二次从5開始数5位，淘汰4。剩余 1 2 3 4 (好人标号1 2 3,坏人标号4)

第三次从4開始数5位，淘汰4。剩余1 2 3 。游戏结束。

为什么不要纠结于坏人的标号呢？由于不easy得出公式啊，如今不计较坏人的标号的话，我得到的公式就是

kill_num=(kill_num+m-1)%rest

所以我记录一个kill的vector，仅仅要每次淘汰的标号大于n或是等于0，即符合标准，我就把它扔进去，什么时候kill的人数等于n了。说明找到的m是正确的。否则就m++，再找。

(找m)代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

int people[50];
vector <int> kill;

int main()
{
	int n,k=0;
	while(cin>>n)
	{
		int result=n+1,rest=2*n,kill_num=1;
		int n2=2*n;
		
		memset(people,0,sizeof(people));
		kill.clear();
		while(1)
		{
			if(kill.size()==n)
				break;
			if((result+kill_num-1)%rest==0)
			{
				kill_num=rest;
				rest--;
				kill.push_back(rest);
			}
			else if((result+kill_num-1)%rest<=n)
			{
				kill_num=1;
				kill.clear();
				rest=n2;
				result++;
			}
			else
			{
				kill_num=(result+kill_num-1)%rest;
				rest--;
				kill.push_back(kill_num);
			}
		}
		cout<<result<<endl;
	}

	return 0;
}

终于打表代码：

#include <iostream>
using namespace std;

int main()
{
	int result[16];
	int n;
	
	result[1] = 2;
	result[2] = 7;
	result[3] = 5;
	result[4] = 30;
	result[5] = 169;
	result[6] = 441;
	result[7] = 1872;
	result[8] = 7632;
	result[9] = 1740;
	result[10] = 93313;
	result[11] = 459901;
	result[12] = 1358657;
	result[13] = 2504881;
	result[14] = 13482720;
	
	while(cin>>n && n)
	{
		cout<<result[n]<<endl;
	}
	return 0;
}