F - N！Again

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

分析：
光是用同余定理还不行，n太大，所以结合同余定理和规律；


代码：

#include<cstdio>
int main()
{long long n,i;
int mod=2009;
while(scanf("%lld",&n)!=EOF)
{long long sum=1;
if(n<=50)
		for(i=1;i<=n;i++)
	 sum=sum%mod*i%mod;
	 else sum=0;
	printf("%lld\n",sum);
	}
	return 0;
}