【约瑟夫问题】——poj1012

                                                                                                            

                                                                                                            Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 51775		Accepted: 19698

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

题意：小船上有2*k个人，前一半为好人，后一半为坏人，要求杀死所有坏人，剩下所有好人的报数周期m。

注意：（从0开始编号）
     可以用从1开始报数的约瑟夫公式：people[i] = (people[i - 1] + m - 1) % (n - i + 1);
          （从0开始报数是：people[i] = (people[i - 1] + m ) % (n - i );

技巧：for循环里套用重置循环变量i。

下面贴上代码：

#include <stdio.h>

int main()
{
    int people[50] = {0}, k, Joseph[14] = {0};//Joseph用于打表，不然超时

    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个人的编号为k - 1,所以这里是<而不是<=
            {
                i = 0 ;
                m++;
            }
        }
        Joseph[k] = m;
        printf("%d\n", m);
    }
    return 0;
}