HDOJ 1012 u Calculate e

Problem Description
A simple mathematical formula for e is

where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output
n e

0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

简单的阶乘运算。
对于小数大于9位的，保留9位小数，四舍五入。

public class Main {
    public static void main(String[] args) {
        //打表输出：
        System.out.println("n e");
        System.out.println("- -----------");
        System.out.println("0 1");
        System.out.println("1 2");
        System.out.println("2 2.5");

        //3-9 的数：
        for(int i=3;i<10;i++){
            double a=0;
            for(int k=0;k<=i;k++){
                a = a+fact(a,k);
            }
            System.out.print(i+" ");
            //默认为四舍五入
            System.out.printf("%.9f",a);
            System.out.println();
        }

    }

    //返回数为i的阶乘分之一
    private static double fact(double a, int i) {
        double e = 1;
        if(i==0){
            return 1;
        }
        for(int j=1;j<=i;j++){
            e = e*j;
        }
        return 1/e;
    }

}