How to solve hdu2647 problem using N! again?

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：N!计算问题

描述：给定一个整数N，你需要计算N的阶乘（N!）。

示例：N=5，则5!=5 × 4 × 3 × 2 × 1=120

时间限制：2000/1000 MS（Java/其他）内存限制：32768/32768 K（Java/其他）总提交次数：4781接受提交次数：2527

问题来源：WhereIsHeroFrom，作者：Zty问题内容：Zty问我：你在做什么？我回答：我想计算N!...

N！Again

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4781Accepted Submission(s): 2527

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

Author

WhereIsHeroFrom

Source

​​HDU女生专场公开赛——谁说女子不如男 ​​

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看到n的范围 脑子里就知道是规律。。。

其实把结果输出一下就看到了。如果某个值正好整除2009 那么 后面的都是0了

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sca=new Scanner(System.in);
while(sca.hasNext()){
int x=sca.nextInt();
if(x>41){
System.out.println(0);
continue;
}
BigInteger res=BigInteger.ONE;
for(int i=2;i<=x;i++){
res=res.multiply(BigInteger.valueOf(i));
res=res.mod(BigInteger.valueOf(2009));
}
System.out.println(res);
}
}

}