/*
     * 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).  
     */

     /* Adapted to FemtoRV32 (Bruno Levy Feb. 2021) */

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

    // Bruno TODO: find a way of:
    // [x] get rid of sqrtf()
    // [ ] implementing mul_mod() using int32 arithmetics 
    // [ ] replace log() and fmod() by int32 arithmetics


//#define RV32_FASTCODE __attribute((section(".fastcode")))
#define RV32_FASTCODE

/* 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) RV32_FASTCODE;
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) RV32_FASTCODE;
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) RV32_FASTCODE;
int is_prime(int n)
{
    int r, i;
    if ((n % 2) == 0)
    return 0;

    // initial program was doing that:
    // r = (int) (sqrt(n));
    // for (i = 3; i <= r; i += 2)

    // I think it is more efficient to do that:
    for(i=3; i*i<=n; i += 2)
    if ((n % i) == 0)
        return 0;
    return 1;
}

/* return the prime number immediatly after n */
int next_prime(int n) RV32_FASTCODE;
int next_prime(int n)
{
    do {
    n++;
    } while (!is_prime(n));
    return n;
}

int digits(int n) RV32_FASTCODE;
int digits(int n) {
    int av, a, vmax, N, num, den, k, kq, kq2, t, v, s, i;
    double sum;

    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);
    }
    return (int) (sum * 1e9);
}


void main() {
    printf("\npi = 3.");
    for(int n=1; ;n+=9) {
       int D = digits(n);
       
       // Stop at Feynman point (comment out to get more decimals)
       if(D == 998372978) {
	  printf("99 etc...\n");
	  break;
       }
       
       printf("%d",D);
       fflush(stdout);
    }
}