N！Again

最新推荐文章于 2019-05-08 14:52:30 发布

原创最新推荐文章于 2019-05-08 14:52:30 发布 · 536 阅读

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杭电地址：http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=2§ionid=2&problemid=13

N！Again

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

Total Submission(s): 641 Accepted Submission(s): 361

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

解题思路:2009=7*7*41，N>=41时，N!因式分解一定含7*7*41，即N!%2009＝0.所以只要计算0<=N<=40时的答案就OK。设N!=m+2009*n，N!%2009=m，(N+1)!%2009=[(N+1)*(m+2009*n)]%2009=[m*(N+1)]%2009，有了这个就可以轻易递推求解了。

代码：

#include <iostream>
using namespace std;
int main()
{
	int m;
	int ch[41];
	ch[0]=1;
	for(int i=1;i<=41;i++)
	{
		ch[i]=(ch[i-1]*i)%2009;
	}
	while(cin>>m)
	{
		if(m>=41)
			cout<<0<<endl;
		else
			cout<<ch[m]<<endl;
	}
	return 0;

}