/*
 * Computation of the n'th decimal digit of pi with very little memory.
 * Written by Fabrice Bellard on February 26, 1997.
 * 
 * We use a slightly modified version of the method described by Simon
 * Plouffe in "On the Computation of the n'th decimal digit of various
 * transcendental numbers" (November 1996). We have modified the algorithm
 * to get a running time of O(n^2) instead of O(n^3log(n)^3).
 *
 * This program uses a variation of the formula found by Gosper in 1974 :
 * 
 * pi = sum( (25*n-3)/(binomial(3*n,n)*2^(n-1)), n=0..infinity);
 *
 * This program uses mostly integer arithmetic. It may be slow on some
 * hardwares where integer multiplications and divisons must be done by
 * software. We have supposed that 'int' has a size of at least 32 bits. If
 * your compiler supports 'long long' integers of 64 bits, you may use the
 * integer version of 'mul_mod' (see HAS_LONG_LONG).  
 */

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

/* uncomment the following line to use 'long long' integers */
/* #define HAS_LONG_LONG */

#ifdef HAS_LONG_LONG
#define mul_mod(a,b,m) (( (long long) (a) * (long long) (b) ) % (m))
#else
#define mul_mod(a,b,m) fmod( (double) a * (double) b, m)
#endif

/* return the inverse of x mod y */
int inv_mod(int x,int y) {
  int q,u,v,a,c,t;

  u=x;
  v=y;
  c=1;
  a=0;
  do {
    q=v/u;
    
    t=c;
    c=a-q*c;
    a=t;
    
    t=u;
    u=v-q*u;
    v=t;
  } while (u!=0);
  a=a%y;
  if (a<0) a=y+a;
  return a;
}

/* return the inverse of u mod v, if v is odd */
int inv_mod2(int u,int v) {
  int u1,u3,v1,v3,t1,t3;
  
  u1=1;
  u3=u;
  
  v1=v;
  v3=v;
  
  if ((u&1)!=0) {
    t1=0;
    t3=-v;
    goto Y4;
  } else {
    t1=1;
    t3=u;
  }
  
  do {
    
    do {
      if ((t1&1)==0) {
	t1=t1>>1;
	t3=t3>>1;
      } else {
	t1=(t1+v)>>1;
	t3=t3>>1;
      }
      Y4:
    } while ((t3&1)==0);
    
    if (t3>=0) {
      u1=t1;
      u3=t3;
    } else {
      v1=v-t1;
      v3=-t3;
    }
    t1=u1-v1;
    t3=u3-v3;
    if (t1<0) {
      t1=t1+v;
    }
  } while (t3 != 0);
  return u1;
}


/* return (a^b) mod m */
int pow_mod(int a,int b,int m)
{
  int r,aa;
   
  r=1;
  aa=a;
  while (1) {
    if (b&1) r=mul_mod(r,aa,m);
    b=b>>1;
    if (b == 0) break;
    aa=mul_mod(aa,aa,m);
  }
  return r;
}
      
/* return true if n is prime */
int is_prime(int n)
{
   int r,i;
   if ((n % 2) == 0) return 0;

   r=(int)(sqrt(n));
   for(i=3;i<=r;i+=2) if ((n % i) == 0) return 0;
   return 1;
}

/* return the prime number immediatly after n */
int next_prime(int n)
{
   do {
      n++;
   } while (!is_prime(n));
   return n;
}

#define DIVN(t,a,v,vinc,kq,kqinc)		\
{						\
  kq+=kqinc;					\
  if (kq >= a) {				\
    do { kq-=a; } while (kq>=a);		\
    if (kq == 0) {				\
      do {					\
	t=t/a;					\
	v+=vinc;				\
      } while ((t % a) == 0);			\
    }						\
  }						\
}

int main(int argc,char *argv[])
{
  int av,a,vmax,N,n,num,den,k,kq1,kq2,kq3,kq4,t,v,s,i,t1;
  double sum;

  if (argc<2 || (n=atoi(argv[1])) <= 0) {
    printf("This program computes the n'th decimal digit of pi\n"
	   "usage: pi n , where n is the digit you want\n"
	   );
    exit(1);
  }
  
  N=(int)((n+20)*log(10)/log(13.5));
  sum=0;

  for(a=2;a<=(3*N);a=next_prime(a)) {
    vmax=(int)(log(3*N)/log(a));
    if (a==2) {
      vmax=vmax+(N-n);
      if (vmax<=0) continue;
    }
    av=1;
    for(i=0;i<vmax;i++) av=av*a;

    s=0;
    den=1;
    kq1=0;
    kq2=-1;
    kq3=-3;
    kq4=-2;
    if (a==2) {
      num=1;
      v=-n; 
    } else {
      num=pow_mod(2,n,av);
      v=0;
    }

    for(k=1;k<=N;k++) {

      t=2*k;
      DIVN(t,a,v,-1,kq1,2);
      num=mul_mod(num,t,av);
      
      t=2*k-1;
      DIVN(t,a,v,-1,kq2,2);
      num=mul_mod(num,t,av);

      t=3*(3*k-1);
      DIVN(t,a,v,1,kq3,9);
      den=mul_mod(den,t,av);

      t=(3*k-2);
      DIVN(t,a,v,1,kq4,3);
      if (a!=2) t=t*2; else v++;
      den=mul_mod(den,t,av);
      
      if (v > 0) {
	if (a!=2) t=inv_mod2(den,av);
	else t=inv_mod(den,av);
	t=mul_mod(t,num,av);
	for(i=v;i<vmax;i++) t=mul_mod(t,a,av);
	t1=(25*k-3);
	t=mul_mod(t,t1,av);
	s+=t;
	if (s>=av) s-=av;
      }
    }

    t=pow_mod(5,n-1,av);
    s=mul_mod(s,t,av);
    sum=fmod(sum+(double) s/ (double) av,1.0);
  }
  printf("Decimal digits of pi at position %d: %09d\n",n,(int)(sum*1e9));
  return 0;
}