POJ 1012 Joseph

Joseph

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52102		Accepted: 19840

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
自己写的枚举超时的不得了
借鉴了wmy大神的代码（虽然也超时，但是快多了）：

/*------------------------------分割线----------------------------------*/
#include<iostream>
using namespace std;
#include<cstdio>
int n,k,m;
int main()
{
    while(scanf("%d",&k)==1)
    {
        if(k==0) break;
        for(m=k+1;;m++)
        {
            n=2*k;
            int p=0;
            bool flag=true;
            for(int i=1;i<=k;++i)
            {/*我在这里用了暴力，挨个挨个的数*/
                p=(p+m-1)%n;/*最巧妙的在于这句，p表示删除人在之前被删的序列（不是原序列）的编号，这个应该很好理解，每次人数减少，所以总人数n也--*/
                if(p<k)/*如果我们删除的对的话，那么前k个人的编号是不变的，所以p始终是不能等于p的*/
                {
                    flag=false;
                    break;
                }
                n--;
            }
            if(flag)
            {
                printf("%d
",m);
                break;
            }
        }
    }
    
    return 0;
}
/*------------------------------分割线----------------------------------*/
打表代码：
#include<iostream>
using namespace std;
int pp[15]={0,2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};
int main()
{
    int k;
    while(cin>>k)
    {
        if(k==0) break;
        else cout<<pp[k]<<endl;
    }
    return 0;;
}