求自然数的近似值

   1:  
            /* e≌1+1/1！+1/2！+1/3！+...+1/n！，要求直到最后一项小于10-6为止 */

   2:   

   3:  #include "Stdio.h"

   4:  #include "Conio.h"

   5:   

   6:  
            int main(void)

   7:  {

   8:   float e=1,n=1; /*e是结果,n是每项的阶层*/

   9:   int i=1;

  10:   while(1/n>0.0000001)

  11:   {

  12:     n*=i;

  13:     e+=(1/n);

  14:     i++;

  15:    }

  16:   printf("%f",e);

  17:    getch();

  18:    return 0;

  19:  }