HDOJ 2674 N！Again（找规律）

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
   
   
   
   

题目大意：求n!%2009.

直接求解必定超时,能不能缩小范围呢，我刚开始想n%=2009但是确实没有证据这么做果然WA。但是%2009显然当阶乘到2009

以后的时候每次必有2009为因子，故一定为0,但阶乘的值太大，那么就可以想到什么时候阶乘可以达到含有2009这个因子，

2009=41*7*7所以40阶乘后全为0.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define mod 2009
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
    LL n,m,j,k;
    while(~scanf("%lld",&n))
    {
        k=1;
        if(n>40)
        {
            printf("0\n");continue;
        }
        for(int i=1;i<=n;i++)
        {
            k=(k%mod*i%mod)%mod;
        }
        printf("%lld\n",k);
    }
}