HDUOJ1042N!(Java大数)

Problem Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

Input
One N in one line, process to the end of file.

Output
For each N, output N! in one line.

第一次写了一个在while外面的数组就直接存好了范围内每个数的阶乘但是wa了
然后就换了一个方法
AC代码1

import java.io.BufferedInputStream;
import java.util.*;
import java.math.*;
public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        int n;
//      BigInteger a[] = new BigInteger[10010];
//      a[0] = BigInteger.valueOf(1);
//      for(int i = 1; i <= 10000; i++)
//      {
//          a[i] = a[i-1].multiply(BigInteger.valueOf(i)); 
//      }
        while(in.hasNext())
        {
            n = in.nextInt();
            BigDecimal ans = fun(n);
            System.out.println(ans);

        }
    }
    public static BigDecimal fun(int n)
    {
        BigDecimal ans = BigDecimal.valueOf(1);
        for(int i = 1; i <= n; i++)
        {
            ans = ans.multiply(BigDecimal.valueOf(i));
        }
        return ans;
    }

}

emmmmmm又交了一次
第一个方法就把BigInteger改成BigDecimal就OK了
AC代码2

import java.io.BufferedInputStream;
import java.util.*;
import java.math.*;
public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        int n;
        BigDecimal a[] = new BigDecimal[10010];
        a[0] = BigDecimal.valueOf(1);
        for(int i = 1; i <= 10000; i++)
        {
            a[i] = a[i-1].multiply(BigDecimal.valueOf(i)); 
        }
        while(in.hasNext())
        {
            n = in.nextInt();
            //BigDecimal ans = fun(n);
            BigDecimal ans = a[n];
            System.out.println(ans);
        }
    }
}