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;    }}