POJ 2244 Eeny Meeny Moo 约瑟夫

题意：约瑟夫环问题，给定人数n，求一个步长m使得最后剩下的是2.

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

int num[200];

int Josephus ( int n, int m, int s )
{
    if ( m == 1 )
        return ( s + n - 1 ) % n;

    for ( int i = 1; i <= n; i++ )
    {
        s = ( s + m ) % i;
        if ( i == n ) break;

        if ( s + m < i )
        {
            int x = (i-s) / (m-1); // s + m * x <= i + x;
            if ( i + x < n )
            {
                i = i + x;
                s = ( s + m * x ) % i;
            }
            else { s = (s + m * (n-i)) % n; break; }
        }
    }
    return s; 
}


int main()
{
    int n, i;
    memset(num,0,sizeof(num));
    while ( scanf("%d",&n) && n )
    {
        if ( num[n] )
        {
            printf("%d\n",num[n]); continue;
        }

        for ( i = 1; ; i++ )
        {
            if ( Josephus ( n - 1, i, 0 ) == 0  )
            {
                num[n] = i;
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}