N！Again

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

Total Submission(s): 3440    Accepted Submission(s): 1865

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=7*7*41， 所以41及往后的阶乘对2009取余都为0.

#include <stdio.h>

int arr[42] = {1, 1};

int main()
{
	int i;
	for(i = 2; i < 42; ++i)
		arr[i] = (arr[i-1] * i) % 2009;
	while(scanf("%d", &i) == 1){
		if(i < 42) printf("%d\n", arr[i]);
		else printf("0\n");
	}
	return 0;
}

HDU2674 N！Again 【数学】