HDU 1008 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

 

 

简单题

 #include<stdio.h>
 int fact(int n)
 {
     int res=;
     while(n>=)
     {
         res*=n;
         n--;
     }
     return res;
 }
 int main()
 {
     printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n3 2.666666667\n4 2.708333333\n");
     int n,i;
     double e;
     for(n=;n<=;n++)
     {
         e=0.0;
         for(i=;i<=n;i++)
         {
             e+=1.0/double(fact(i));
         }
         printf("%d %.9lf\n",n,e);
     }
     return ;
 }