Вычисление N-го знака числа Пи без вычисления предыдущих

/*  
    This program implements the BBP algorithm to generate a few hexadecimal
    digits beginning immediately after a given position id, or in other words
    beginning at position id + 1.  On most systems using IEEE 64-bit floating-
    point arithmetic, this code works correctly so long as d is less than
    approximately 1.18 x 10^7.  If 80-bit arithmetic can be employed, this limit
    is significantly higher.  Whatever arithmetic is used, results for a given
    position id can be checked by repeating with id-1 or id+1, and verifying 
    that the hex digits perfectly overlap with an offset of one, except possibly
    for a few trailing digits.  The resulting fractions are typically accurate 
    to at least 11 decimal digits, and to at least 9 hex digits.  
*/
 
/*  David H. Bailey     2006-09-08 */
 
#include <stdio.h>
#include <math.h>
 
int main()
{
  double pid, s1, s2, s3, s4;
  double series (int m, int n);
  void ihex (double x, int m, char c[]);
  int id = 1000000;
#define NHX 16
  char chx[NHX];
 
/*  id is the digit position.  Digits generated follow immediately after id. */
 
  s1 = series (1, id);
  s2 = series (4, id);
  s3 = series (5, id);
  s4 = series (6, id);
  pid = 4. * s1 - 2. * s2 - s3 - s4;
  pid = pid - (int) pid + 1.;
  ihex (pid, NHX, chx);
  printf (" position = %i\n fraction = %.15f \n hex digits =  %10.10s\n",
  id, pid, chx);
}
 
void ihex (double x, int nhx, char chx[])
 
/*  This returns, in chx, the first nhx hex digits of the fraction of x. */
 
{
  int i;
  double y;
  char hx[] = "0123456789ABCDEF";
 
  y = fabs (x);
 
  for (i = 0; i < nhx; i++){
    y = 16. * (y - floor (y));
    chx[i] = hx[(int) y];
  }
}
 
double series (int m, int id)
 
/*  This routine evaluates the series  sum_k 16^(id-k)/(8*k+m) 
    using the modular exponentiation technique. */
 
{
  int k;
  double ak, eps, p, s, t;
  double expm (double x, double y);
#define eps 1e-17
 
  s = 0.;
 
/*  Sum the series up to id. */
 
  for (k = 0; k < id; k++){
    ak = 8 * k + m;
    p = id - k;
    t = expm (p, ak);
    s = s + t / ak;
    s = s - (int) s;
  }
 
/*  Compute a few terms where k >= id. */
 
  for (k = id; k <= id + 100; k++){
    ak = 8 * k + m;
    t = pow (16., (double) (id - k)) / ak;
    if (t < eps) break;
    s = s + t;
    s = s - (int) s;
  }
  return s;
}
 
double expm (double p, double ak)
 
/*  expm = 16^p mod ak.  This routine uses the left-to-right binary 
    exponentiation scheme. */
 
{
  int i, j;
  double p1, pt, r;
#define ntp 25
  static double tp[ntp];
  static int tp1 = 0;
 
/*  If this is the first call to expm, fill the power of two table tp. */
 
  if (tp1 == 0) {
    tp1 = 1;
    tp[0] = 1.;
 
    for (i = 1; i < ntp; i++) tp[i] = 2. * tp[i-1];
  }
 
  if (ak == 1.) return 0.;
 
/*  Find the greatest power of two less than or equal to p. */
 
  for (i = 0; i < ntp; i++) if (tp[i] > p) break;
 
  pt = tp[i-1];
  p1 = p;
  r = 1.;
 
/*  Perform binary exponentiation algorithm modulo ak. */
 
  for (j = 1; j <= i; j++){
    if (p1 >= pt){
      r = 16. * r;
      r = r - (int) (r / ak) * ak;
      p1 = p1 - pt;
    }
    pt = 0.5 * pt;
    if (pt >= 1.){
      r = r * r;
      r = r - (int) (r / ak) * ak;
    }
  }
 
  return r;
}