hdoj 2647 N！Again

N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4016    Accepted Submission(s): 2157

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

 

此题仍然用同余定理：前边文章中已经做出详细解释这里就不解释了

 此题还有一个技巧就是41（包括41）之后的所有数据结果都是0

因为40求出的结果是245==49*5      2009==49*41

下一步对41求阶乘并对2009取模就等于（41%2009*245%2009）%2009==（41*245）%2009==0

所以之后的每一步都等0

AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
	int n,m,j,i,s,t;
	while(scanf("%d",&n)!=EOF)
	{
		if(n>=41)
		printf("0
");
		else
		{	
		    s=1;
		    for(i=2;i<=n;i++)
		    {
			    s=s%2009*i%2009;
			    s=s%2009;
		    }
		    printf("%d
",s);
	    }
	}
	return 0;
}