# u Calculate e
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 37341    Accepted Submission(s): 16897  

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
题目大意为：求出0-9的情况下这个公式求出来的e的值是多少，从结果可以看出来，0和1的时候保留整数部分，2的时候保留一位小数，3,4以及以后的时候保留九位小数。

import java.text.DecimalFormat;

public class Main{

	public static void main(String[] args) {
		double sum = 0;
		System.out.println("n e");
		System.out.println("- -----------");
		for(int i = 0; i < 10; i++)
		{
			sum += 1.0 / functionMuti(i);
			if( i == 0 || i == 1)
			{
				DecimalFormat decimalForm = new DecimalFormat("0");
				System.out.println(i + " " + decimalForm.format(sum));
			}
			else if(i == 2)
			{
				DecimalFormat decimalForm = new DecimalFormat("0.0");
				System.out.println(i + " " + decimalForm.format(sum));
			}
			else 
			{
				DecimalFormat decimalForm = new DecimalFormat("0.000000000");
				System.out.println(i + " " + decimalForm.format(sum));
			}
		}
		
	}

	private static int functionMuti(int i) {
		if(i == 1)
			return 1;
		else if(i == 0)
			return 1;
		else
		{
			return  i * functionMuti(--i);
		}
	}
}