ACM HDU 2674 N! Again(數論)

繼續數論。。

Problem Description

WhereIsHeroFrom:            Zty,what are you doing ?
Zty:                                    Iwant to calculate N!......
WhereIsHeroFrom:            Soeasy! How big N is ?
Zty:                                    1<=N <=1000000000000000000000000000000000000000000000…
WhereIsHeroFrom:            Oh! Youmust 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 linewill contain one integer N(0 <= N<=10^9). Process to end of file.

 

Output

For eachcase, output N! mod 2009

 

Sample Input

4 

5

 

 

Sample Output

24

120

/*****************************************************

數據規模  1<N<10^9，妥妥的要找規律了。。

從  N = 1  到N = 40  時還都能正常算出，(正常規律  num[i] = num[i-1] * i %2009;  (   1<N<10^9)，，，N 爲40時，num[N] = 245，可以發現  245 * 41 = 10045 = 2009 * 5，

所以，就可以知道了，N>40時，輸出全部爲  0。。。

***********************************************************/

#include<stdio.h>
#include<iostream>
using namespace std;
int num[42];
void cal()
{
	num[0] = 1,num[1] = 1;
	for(int i = 2;i<42;i++)
		num[i] = num[i-1]*i%2009;
}
int main()
{

	int n;
	cal();
	while(cin>>n)
    {
        if(n>41)
            printf("%d\n",0);
        else
            printf("%d\n",num[n]);
    }
	return 0;
}