打表法解 1012 -- Joseph

Joseph

Time Limit: 1000MS
Memory Limit: 10000K

Total Submissions: 16411
Accepted: 6092

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

340

Sample Output

530

没有想到好的解题办法，并且Time Limit: 1000MS，Memory Limit: 10000K，还有0 < k < 14. 

那就用打表的方法啦，第一次用。

程序如下：注释掉部分为计算程序。

// poj1012Joseph.cpp : Defines the entry point for the console application.
//

#include
using namespace std;

int main()
{
    int data[14]={2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881,13482720};
    int n=0;
    while(cin>>n && n)
        cout<    return 0;
}
//
//int main()
//{
//    int n=0;
//    bool data[28]={};
//    int flg1=0;
//    long int flg2=1;
//    int bad;
//    while(cin>>n && n)
//    {
//        flg2=1;
//        flg1=n+n;
//        bad=n;
//        for(int k=0;k//        int i=1,j=0;//j is the real position oa array
//        while(bad)
//        {
//            if(j==flg1)
//                j=0;//-----again from the begin.
//
//
//            if(i==flg2)//-----------get one position.
//            {
//                if(j//                {
//                    flg2++;//--increase the step;
//                    j=0;
//                    bad=n;
//                    i=1;
//                    for(int k=0;k//                }
//                else if(data[j]==1)//----position in the bad guys'
//                {
//                    data[j]=0;
//                    bad--;
//                    j++;
//                    i=1;
//                }
//                else if(data[j]==0) j++;
//           
//            }
//            else if(data[j]==1)//------in one real position,so i++;
//            {
//                i++;
//                j++;
//            }
//            else
//            {
//                j++;
//            }
//        }
//        cout<//
//    }
//    return 0;
//}
//
//