POJ - 1012 Joseph

题意：约瑟夫环的问题，前K个人是好人，后K个人是坏人，求最小的M，枚举M，套用公式ans[i]=(ans[i-1]+m-1)%(n-i+1),报数是从1开始的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main(){
   int Joseph[14] = {0};
   int k;
   while (scanf("%d",&k) != EOF && k){
       if (Joseph[k]){
           printf("%d\n",Joseph[k]);
           continue;
       }
       int n = 2*k;
       int ans[30] = {0};
       int m = 1;
       for (int i = 1; i <= k; i++){
           ans[i] = (ans[i-1]+m-1)%(n-i+1);
           if (ans[i] < k){
               i = 0;
               m++;
           }
       }
       Joseph[k] = m;
       printf("%d\n",m);
   } 
   return 0;
}