/*
 * Computation of the n'th decimal digit of \pi with very little memory.
 * Written by Fabrice Bellard on January 8, 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 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 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 (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;
}

int main(int argc, char *argv[])
{
    int av, a, vmax, N, n, num, den, k, kq, kq2, t, v, s, i;
    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(2));

    sum = 0;

    for (a = 3; a <= (2 * N); a = next_prime(a)) {

	vmax = (int) (log(2 * N) / log(a));
	av = 1;
	for (i = 0; i < vmax; i++)
	    av = av * a;

	s = 0;
	num = 1;
	den = 1;
	v = 0;
	kq = 1;
	kq2 = 1;

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

	    t = k;
	    if (kq >= a) {
		do {
		    t = t / a;
		    v--;
		} while ((t % a) == 0);
		kq = 0;
	    }
	    kq++;
	    num = mul_mod(num, t, av);

	    t = (2 * k - 1);
	    if (kq2 >= a) {
		if (kq2 == a) {
		    do {
			t = t / a;
			v++;
		    } while ((t % a) == 0);
		}
		kq2 -= a;
	    }
	    den = mul_mod(den, t, av);
	    kq2 += 2;

	    if (v > 0) {
		t = inv_mod(den, av);
		t = mul_mod(t, num, av);
		t = mul_mod(t, k, av);
		for (i = v; i < vmax; i++)
		    t = mul_mod(t, a, av);
		s += t;
		if (s >= av)
		    s -= av;
	    }

	}

	t = pow_mod(10, 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;
}