c++ - 找到尽可能多的 2 平方根的数字

Question

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
 double a = sqrt(2);
 cout << a << endl;
}

嗨，这是找到 2 的 sqrt 的程序，它在输出中只打印 1.41421 如何以这样的方式实现它，它将在小数点后打印 200000 位

1.41421..........upto 200 000 digits

有没有办法打印成这样？

Answer 1

It can be shown那个

sqrt(2) = (239/169)*1/sqrt(1-1/57122)

并且 1/sqrt(1-1/57122) 可以使用泰勒级数展开有效地计算:

1/sqrt(1-x) = 1 + (1/2)x + (1.3)/(2.4)x^2 + (1.3.5)/(2.4.6)x^3 + ...

There's also a C program available that uses this method (我稍微重新格式化并更正了它):

/*
** Pascal Sebah : July 1999
**
** Subject:
**
**    A very easy program to compute sqrt(2) with many digits.
**    No optimisations, no tricks, just a basic program to learn how
**    to compute in multiprecision.
**
** Formula:
**
**    sqrt(2) = (239/169)*1/sqrt(1-1/57122)
**
** Data:
**
**    A big real (or multiprecision real) is defined in base B as:
**      X = x(0) + x(1)/B^1 + ... + x(n-1)/B^(n-1)
**      where 0<=x(i)<B
**
** Results: (PentiumII, 450Mhz)
**
**    1000   decimals :   0.02seconds
**    10000  decimals :   1.7s
**    100000 decimals : 176.0s
**
** With a little work it's possible to reduce those computation
** times by a factor of 3 and more.
*/

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

long B = 10000; /* Working base */
long LB = 4;    /* Log10(base)  */

/*
** Set the big real x to the small integer Integer
*/
void SetToInteger(long n, long* x, long Integer)
{
  long i;
  for (i = 1; i < n; i++)
    x[i] = 0;
  x[0] = Integer;
}

/*
** Is the big real x equal to zero ?
*/
long IsZero(long n, long* x)
{
  long i;
  for (i = 0; i < n; i++)
    if (x[i])
      return 0;
  return 1;
}

/*
** Addition of big reals : x += y
**  Like school addition with carry management
*/
void Add(long n, long* x, long* y)
{
  long carry = 0, i;
  for (i = n - 1; i >= 0; i--)
  {
    x[i] += y[i] + carry;
    if (x[i] < B)
      carry = 0;
    else
    {
      carry = 1;
      x[i] -= B;
    }
  }
}

/*
** Multiplication of the big real x by the integer q
*/
void Mul(long n, long* x, long q)
{
  long carry = 0, xi, i;
  for (i = n - 1; i >= 0; i--)
  {
    xi = x[i] * q;
    xi += carry;
    if (xi >= B)
    {
      carry = xi / B;
      xi -= carry * B;
    }
    else
      carry = 0;
    x[i] = xi;
  }
}

/*
** Division of the big real x by the integer d
**  Like school division with carry management
*/
void Div(long n, long* x, long d)
{
  long carry = 0, xi, q, i;
  for (i = 0; i < n; i++)
  {
    xi    = x[i] + carry * B;
    q     = xi / d;
    carry = xi - q * d;
    x[i]  = q;
  }  
}

/*
** Print the big real x
*/
void Print(long n, long* x)
{
  long i;
  printf("%ld.", x[0]);
  for (i = 1; i < n; i++)
    printf("%04ld", x[i]);
  printf("\n");
}

/*
** Computation of the constant sqrt(2)
*/
int main(void)
{
  long NbDigits = 200000, size = 1 + NbDigits / LB;
  long* r2 = malloc(size * sizeof(long));
  long* uk = malloc(size * sizeof(long));
  long k = 1;
  /*
  ** Formula used:
  **    sqrt(2) = (239/169)*1/sqrt(1-1/57122)
  ** and
  **   1/sqrt(1-x) = 1+(1/2)x+(1.3)/(2.4)x^2+(1.3.5)/(2.4.6)x^3+...
  */
  SetToInteger(size, r2, 1); /* r2 = 1 */
  SetToInteger(size, uk, 1); /* uk = 1 */
  while (!IsZero(size, uk))
  {
    Div(size, uk, 57122); /* uk = u(k-1)/57122 * (2k-1)/(2k) */
    Div(size, uk, 2 * k);
    Mul(size, uk, 2 * k - 1);
    Add(size, r2, uk);    /* r2 = r2+uk */
    k++;
  }
  Mul(size, r2, 239);
  Div(size, r2, 169);  /* r2 = (239/169)*r2 */

  Print(size, r2);     /* Print out of sqrt(2) */

  free(r2);
  free(uk);

  return 0;
}

计算 200,000 位的 sqrt(2) 大约需要一分钟。

但是请注意，在 200,000 位时，由于累积的舍入误差，最后生成的 11 位数字不正确，如果您想要 200,000 位正确的数字，则需要运行 200,012 位。