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：

给定e的计算公式，按格式输出n为0-9e的结果。

题目解析：

简单递归可以计算出e的值，要注意输出格式。

具体代码：

#include<iostream>
#include<algorithm>
using namespace std;
int getnum(int n){
	if(n==0||n==1)
		return 1;
	else
		return n*getnum(n-1);
}
int main()
{
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    printf("1 2\n");
    printf("2 2.5\n");
    printf("3 2.666666667\n");
    printf("4 2.708333333\n");
    double sum=0; 
    for(int i=0;i<=9;i++){
    	int fm=getnum(i);
    	sum+=(double)1/fm;
    	if(i>4)
    		printf("%d %.9f\n",i,sum);
	}
    return 0;
}