u Calculate e-优快云博客

本文介绍了一种通过数学公式计算自然常数e的近似值的方法，并提供了一个C语言程序示例，该程序展示了如何使用从0到9的阶乘倒数之和来逐步逼近e的精确值。

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Problem Description

A simple mathematical formula for e is
e = sum((1/i!) i is from 0 to n

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

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答案：

#include <stdio.h>
#include <stdlib.h>
#include <string>

void main()
{
int i, j, n, k;
double e;
int v, s = 0;
int decimal, sign;
char* buffer;
int precision = 11;
char out[12];

printf("n e/n");
printf("- -----------/n");
for (n = 0; n <= 9; ++n)
{
  e = 0;
  for (i = 0; i <= n; ++i)
  {
   if (i == 0)
    v = 1;
   else
    v *= i;
   e += (double) 1.0 / v;
  }
  buffer = _ecvt(e, precision, &decimal, &sign);
  for (j = 10; j >= 0; --j)
  {
   if (buffer[j] != '0')
    break;
  }
  if (j != 0)
  {
   int add = 0, v;
   if (j == 10)
   {
    add = buffer[j]-'0'>=5?1:0;
    j = j-1;
   }
   else
    add = 0;

   for (k = j, i = k+1; i >= 0; --i)
   {
    if (i == decimal)
     out[i] = '.';
    else
    {
     if (add != 0)
     {
      v = buffer[k--]-'0' + add;
      out[i] = v%10 + '0';
      add = v/10;
     }
     else
      out[i] = buffer[k--];
    }
   }
   out[j+2] = '/0';
  }
  else
  {
   out[0] = buffer[0];
   out[1] = '/0';
  }

  printf("%d %s/n", n, out);
}
}