poj 1012(约瑟夫环)

int g[15]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};

打表比较好吧，13的数据量。

如果用循环链表模拟暴力要跑很多时间才能搞定。 根据这题只看前k个和后k个。  第一次到s%2*k，  那么第二次则到  (s+s%(2*k-1))%(2*k-1) ，如此递推只需k的时间就可以A了。

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43659		Accepted: 16425

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

Central Europe 1995

#include <stdio.h>
#include <string.h>

int g[15]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
/*
int mark[30];
int l[30],r[30];

int main()
{
    for(int k=1;k<=13;k++)
    {
        for(int j=k; ;j++)
        {
            int s=0,ts;
            int flag=0;
            for(int i=2*k;i>k;i--)
            {
                s=(s+(j%i))%i;
                if(s<=k-1) 
                {
                    flag=1;
                    break;
                }
            }

            if(flag==0)
            {
                printf("%d,",j+1);
                break;
            }
        }

    }
    return 0;
}*/

int main()
{
    int n;
    while(scanf("%d",&n)&&n)
        printf("%d\n",g[n]);
    return 0;
}