HDOJ 2674 N！Again（同余定理）

N！Again

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

Total Submission(s): 4044 Accepted Submission(s): 2177



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

赛后清题，组队赛第一场，跪了！！！



这一题考查的是同余定理，但是不能硬做会超时，要剪枝。



注：此题运用的同余定理公式为：(n*m)%c=(n%c*m%c)%c



一，题目是求n的阶乘对2009取余，即可知所有大于等于2009的数的阶乘对2009取余都等于0，所以有以下程序：

#include<cstdio>
int main()
{
	int n,i,ans;
	while(scanf("%d",&n)!=EOF)
	{
		ans=1;
		if(n>=2009)
		  printf("0\n");
		else
		{
			for(i=2;i<=n;i++)
				ans=(ans*(i%2009))%2009;
			printf("%d\n",ans);
		}
	}
	return 0;
}

二，2009=41*7*7，即可知所有大于等于41的数的阶乘对2009取余都等于零，时间复杂度再次减少，代码如下：

#include<cstdio>
int main()
{
	int n,i,ans;
	while(scanf("%d",&n)!=EOF)
	{
		ans=1;
		if(n>=41)
		  printf("0\n");
		else
		{
			for(i=2;i<=n;i++)
				ans=(ans*(i%2009))%2009;
			printf("%d\n",ans);
		}
	}
	return 0;
}