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  • (HDOJ 1012)u Calculate e

    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
     

    Source
     

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     AC code:

    #include <stdio.h>
    #include
    <math.h>

    long int f(int n)
    {
        
    if(n==0||n==1)
         
    return 1;
        
    else
         
    return n*f(n-1);
    }

    double sum(int n)
    {
        
    int i;
        
    double s=0.000000000;
        
    for(i=0; i<=n;i++)
        {
            s
    +=(double)1/f(i);
        }
        
    return s;
    }

    int main()
    {
        
    int i;
       printf(
    "n e\n");
       printf(
    "- -----------\n");
       
    for(i=0;i<=9;i++)
       {
         
    if(i==0)
         {
             printf(
    "%d %d\n",i,1);
            }
         
         
    else if(i==1)
         {
             printf(
    "%d %d\n",i,2);
            }
            
    else if(i==2)
         {
             printf(
    "%d %.1f\n",i,2.5);
            }
            
    else 
            {
         printf(
    "%d %.9lf\n",i,sum(i));
         }
        }

        
    return 0;

    } 

    作者:cpoint
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
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  • 原文地址:https://www.cnblogs.com/cpoint/p/2015282.html
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