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 1
1 2
2 2.5
3 2.666666667
4 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; }