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

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int i,sum;
    double a[100];
    sum=1;
    for(i=0;i<=9;i++)
    {
           if(i==0)
               a[i]=1;
           else
        {
                  sum=sum*i;
            a[i]=a[i-1]+1.0/sum;
        }
    }
    printf("n e\n- -----------\n");
    for(i=0;i<10;i++)
    {
        cout<<i<<" ";
        if(i<2)
            printf("%.0lf\n",a[i]);//小数点后保留0位有效数值
        else if(i==2)
            printf("%.1lf\n",a[i]);//小数点后保留1位有效数值
        else
            printf("%.09lf\n",a[i]);//小数点后保留9位有效数值
    }
    return 0;
}