HDU-1443 Joseph

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
3
4
0

Sample Output
5
30

题意：每个示例输入一个k，共有n=2*k人，求使得前k个出队的都是n人中后k个人里面的（即第k+1至2*k个）的最小m值。

AC代码：

#include <iostream>using namespace std;bool find(int k,int m){    int n=2*k;    int count=0,now=0;    while(count<k)    {        now = (now+m-1)%n;        if(now<k)//因为我们只需要前k个出队的是后k个，故不需要用队列来模拟            return false;        else        {            count++;            n--;            now%=n;        }    }    return true;}int main(){    int k;    int ans[14]={0};//用来保存已求过的答案,避免重复，也可以直接先打表,不保存会TLE    //第一次提交TLE了还以为是自己的算法问题，然后手动输入打表AC的，后来去查别人的解法才发现是因为自己没保存已求过的答案导致超时的    //int ans[]={2,7,5,30,169,441,1872,7632,1740,93313,459901,1358657,2504881};    while(cin>>k&&k)    {        if(ans[k-1]!=0)            cout<<ans[k-1]<<endl;        else        {            for(int i=k+1;;++i)            {                if(find(k,i))                {                    cout<<i<<endl;                    ans[k-1]=i;                    break;                }                if(i%k==0&&(i/k)%2==0)                {                    i+=k;                    continue;                }            }        }    }    return 0;}