Joseph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3001 Accepted Submission(s): 1694 

Problem 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

题目大意：一共有2k个人，前k个好人和后k个坏人，每隔m个人挑一个人出来，

知道最后一个坏人被挑出来为止，这中间不能有好人被挑出

数据量较小，可以通过暴力枚举

#include <iostream>#include<string.h>using namespace std;int main(){    int a[20],ans[20];    int i,n,k,m;    memset(a,0,sizeof(a));    for(i=1;i<=14;i++)    {        int mm=2*i;        m=1;        for(k=1;k<=i;k++)        {            a[k]=(a[k-1]+m-1)%(mm+1-k);//从第一个人开始数            if(a[k]<i)            {                k=0;                m++;            }        }        ans[i]=m;    }    int nnn;    while(cin>>n)    {        if(n==0)break;        cout<<ans[n]<<endl;    }    return 0;}