HDU N！Again

題目

Problem Description

WhereIsHeroFrom: Zty, what are you doing ?

Zty: I want to calculate N!..

WhereIsHeroFrom: So easy! How big N is ?

Zty: 1 <=N <=1000000000000000000000000000000000000000000000…

WhereIsHeroFrom: Oh! You must be crazy! Are you Fa Shao?

Zty: No. I haven’s finished my saying. I just said I want to calculate N! mod 2009

Hint : 0! = 1, N! = N*(N-1)!

Input

Each line will contain one integer N(0 <= N<=10^9). Process to end of file.

Output

For each case, output N! mod 2009

Sample Input

4

5

Sample Output

24

120

一、分析

2009=4177，有因為41的階乘裡含有41，7，14（2*7）是以41和41之後的階乘都能整除2009，是以當n>=41時直接輸出0即可

二、代碼

代碼如下（示例）：

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

/*
2009=41*7*7，有因為41的階乘裡含有41，7，14（2*7）是以41和41之後的階乘都能整除2009
*/
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        long long sum=1;
        if(n==0) printf("1\n");
        else if(n>=41) printf("0\n");
        else
        {
            for(int i=1;i<=n;i++)
            {
                sum=(sum*i)%2009;
            }
               printf("%lld\n",sum);
        }

    }
    return 0;
}