hoj1017 Joseph's problem II

Joseph's problem II

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 Source : ACM ICPC Central European Regional 1995 Time limit : 2 sec Memory limit : 32 M

Submitted : 1048, Accepted : 509

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

约瑟夫问题

一共有2*k个人，前k个人不能死，问在杀死最后一个后k个人的时候，m最小是多少

由于，k是小于14的，可以先打表。。

从m开始模拟，如果在区间前k内，则m++，维护当前的前k区间，注意x'=(x-m)%i，此时x'可能小于0，那就i 知道满足为止

code：

#include <iostream>#include <queue>#include <stdio.h>#include <stack>#include <algorithm>#include <string.h>#include <map>#include <stdlib.h>using namespace std;bool solve(int k,int m){    int start = 0,end = k - 1;    bool flag = true;    for(int i=2*k;i>k;i--)    {        int kill = (m-1)%i;        if(kill>=start && kill<=end)        {            flag = false;            break;        }        start = ((start - m)%i+i)%i;        end = ((end - m)%i+i)%i;    }    return flag;}int kk[15];int main(){int k;for(int i=1;i<14;i++){for(int m=i+1;;m++)if(solve(i,m)){kk[i]=m;break;}}while(cin>>k){if(k==0) break;cout<<kk[k]<<endl;}return 0;}