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
   
   
   
   
   
   
   
   
 
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

 

题意：

就是求e的值，而e的值是根据上面的那个公式求出来的n越大则e的值越精确，题中已经给了5个输出案例，按照上面的形式输出n=0-9时e的值。

代码：

 

#include<stdio.h>
int g(int a);
int main()
{
    int i;
    double sum=0;
    for(i=0;i<9;i++)
    {
        sum=sum+1.0/g(i);
        printf("%.9lf\n",sum);
    }
    return 0;
}
int g(int a)
{
    if(a==0||a==1)
    return 1;
    else
    return g(a-1)*a;
}

 

分析：

果断的暴力输出- -！

先把所有的数算出来然后输出。。