(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;

} 

posted @ 2011-04-13 21:04  cpoint  阅读(467)  评论(0编辑  收藏  举报
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