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

Sample Output

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\n");

		else

		{

		    s=1;

		    for(i=2;i<=n;i++)

		    {

			    s=s%2009*i%2009;

			    s=s%2009;

		    }

		    printf("%d\n",s);

	    }

	}

	return 0;

}

巴特西

hdoj 2647 N！Again

N！Again

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