POJ 1012 Joseph

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44650		Accepted: 16837

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
题目大意：约瑟夫问题，总共有2 * k我的报数，前面k个是好人，后面k个是坏人，报道m的人要死去，问当m为什么值时，坏人所有死去以前不会有好人死去。
解题方法：从1开始枚举m的值，遇到不知足的则将m加一再次枚举，知道知足为止。

#include <stdio.h>

int main()
{
    int people[50] = {0}, k, Joseph[14] = {0};//Joseph用于打表，否则超时
    while(scanf("%d", &k) != EOF && k != 0)
    {
        if (Joseph[k] != 0)
        {
            printf("%d\n", Joseph[k]);
            continue;
        }
        int n = 2 * k;
        int m = 1;
        people[0] = 0;//people[0] = 0表示编号从0开始
        for (int i = 1; i <= k; i++)
        {
            //每次枚举完毕将剩下的人按从0到n - i编号，只要好人没有杀掉，则前面k - 1个编号是不变的
            people[i] = (people[i - 1] + m - 1) % (n - i + 1);
            if (people[i] < k)//第k我的的编号为k - 1,因此这里是<而不是<=
            {
                i = 0 ;
                m++;
            }
        }
        Joseph[k] = m;
        printf("%d\n", m);
    }
    return 0;
}