HDU 2674-- N！Again【技巧】

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4012    Accepted Submission(s): 2154

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
   
   
   
   

 

因为2009 = 41 * 7 * 7，所以对于大于41的数，它的阶乘会到达肯定可以用 2009  *  某数表示 ，然后取余2009 = 0。

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