HDU 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 11 22 2.53 2.6666666674 2.708333333
思路：简单的计算自然底数e的精确值问题：因为n==0||n==1||n==2时小数位数不为9，所以可以按照特殊情况对待。从3到9看为计算的普通情况
代码如下：
#include<stdio.h>int main(){    printf("n e\n");    printf("- -----------\n");    printf("0 1\n1 2\n2 2.5\n");    int i,j,k;    for(i=3;i<=9;i++)   //从3到9    {        double sum=1;                for(j=1;j<=i;j++)          //计算  从3到9时某个数 阶乘和。        {            int ans=1;            for(k=2;k<=j;k++)           //计算在某个数下的每个数的阶乘。            ans=ans*k;            sum=sum+1.0/ans;                //结果                }             printf("%d %.9lf\n",i,sum);              //%.9lf  小数点后9位    }    return 0;}