（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

Greater New York 2000

 

Recommend

JGShining

 

 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 ;

}