hdoj 2674 N！Again 【好题】

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4008    Accepted Submission(s): 2152

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

 

因为2009 = 41 * 7 * 7，所以对于大于或者等于41的数，它的阶乘会到达2009，然后2009取余2009 = 0。。。

#include <cstdio>
#define LL long long
int main()
{
	int n;
	while(scanf("%d", &n) != EOF)
	{
		if(n >= 41)
		{
			printf("0\n");
			continue;
		}
		LL ans = 1;
		for(int i = 1; i <= n; i++) ans = (ans * i) % 2009;
		printf("%lld\n", ans);
	}
	return 0;
}