利用数学公式求开方

最近突然想起来数学中的如下规律：

1   = 1;

4   = 1 + 3;

9   = 1 + 3 +5;

16 = 1 + 3 +5 + 7;

25 = 1 + 3 +5 + 7 + 9;

36 = 1 + 3 +5 + 7 + 9 + 11;

49 = 1 + 3 +5 + 7 + 9 + 11 +13;

...

利用该公式可以计算开方，而且进行特殊处理便可以小数点后面几位。不过限于整数的长度，计算精度和范围有限。

闲时自己动书编写了如下程序：

#include <stdio.h>
float sqrt(float num, int precision)
{
  int m;
  int n;
  int data, j;
  float value;
  for(m = 1, j = 1; j <= precision; j++)
  {
    if(num * m < 0xFFFFFFFF/100) 
    {
      m *= 100;
    }
    else
    {
      printf ("the number or precision is beyond the scope\n");
      return 0.0;
    }
  }

  data = (int) (num * m + 0.5);  

  for(n = 0, j = 0; j < data; n++)
    j += (2 * n + 1);

  if(j - data >= n )
    n = n - 1;

  value = (float)n;

  for(j = 1; j <= precision; j++)
    value /= 10;
  
  return value;
}
int main(void)
{
  printf ("sqrt(2, 0) = %.0f\n",sqrt((float)2, 0) );
  printf ("sqrt(2, 1) = %.1f\n",sqrt((float)2, 1) );
  printf ("sqrt(2, 2) = %.2f\n",sqrt((float)2, 2) );
  printf ("sqrt(2, 3) = %.3f\n",sqrt((float)2, 3) );
  printf ("sqrt(2, 4) = %.4f\n",sqrt((float)2, 4) );
  printf ("sqrt(2, 5) = %.5f\n",sqrt((float)2, 5) );
  printf ("sqrt(2.5, 1) = %.1f\n",sqrt((float)2.5, 1) );
  printf ("sqrt(16, 0) = %.0f\n",sqrt((float)16, 0) );
  printf ("sqrt(625, 3) = %.3f\n",sqrt((float)625, 3) );
  printf ("sqrt(6000, 3) = %.3f\n",sqrt((float)6000, 3) );
  printf ("sqrt(7000, 2) = %.2f\n",sqrt((float)7000, 2) );
 
  return 0;
}

输出结果如下：

sqrt(2, 0) = 1
sqrt(2, 1) = 1.4
sqrt(2, 2) = 1.41
sqrt(2, 3) = 1.414
sqrt(2, 4) = 1.4142
the number or precision is beyond the scope
sqrt(2, 5) = 0.00000
sqrt(2.5, 1) = 1.6
sqrt(16, 0) = 4
sqrt(625, 3) = 25.000
the number or precision is beyond the scope
sqrt(6000, 3) = 0.000
sqrt(7000, 2) = 83.67