poj 2939 Flavius Josephus Reloaded

http://poj.org/problem?id=2939

题意：但某个人点到两次时这个人自杀，当某个人点到三次时，循环结束，求结束时剩余的人数；

思路：用哈希写，经典的题啊；（代码是参考其他队友的）

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define inf 1000009
using namespace std;
struct node
{
    long long data;
    int num;
    int next;
}node[inf];
int link[inf];
long long x = 0;
int a = 0;
int b = 0;
int m = 0;
int n = 0;
void init()
{
    for(int i = 0;i < inf; ++i)
    link[i] = 0;
    m = 1;
}
int find(long long q,int x)
{
    for(int i = link[x];i ; i = node[i].next)
    {
        if(node[i].data == q)
        {
        ++node[i].num;
        //printf("%d jj\n",node[i].num);
        //getchar();
        return node[i].num;
        }
    }
    return 0;
}
void add(long long q,int x)
{
    node[m].data = q;
    node[m].num = 1;
    node[m].next=link[x];
    link[x] = m++;
}
int main()
{
    while(scanf("%d",&n),n != 0)
    {
        init();
        scanf("%d%d",&a,&b);
        x = b%n;
        int s = 0;
        while(1)
        {
            int temp = find(x,x%inf);
            //printf("%d gg\n",temp);
            if( temp == 0)
            add(x,x%inf);
            else
            if(temp == 2)
            ++s;
            else
            if(temp == 3)
            {
                printf("%d\n",n - s);
                break;
            }
            x = (((a*x)%n*x)+b)%n;
        }
    }
    return 0;
}