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/*

Redistribution and use in source and binary forms, with or without

modification, are permitted provided that the following conditions are met:

    * Redistributions of source code must retain the above copyright

    notice, this list of conditions and the following disclaimer.

    * Redistributions in binary form must reproduce the above copyright

    notice, this list of conditions and the following disclaimer in the

    documentation and/or other materials provided with the distribution.

    * Neither the name of "The Computer Language Benchmarks Game" nor the

    name of "The Computer Language Shootout Benchmarks" nor the names of

    its contributors may be used to endorse or promote products derived

    from this software without specific prior written permission.

THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"

AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE

IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE

ARE DISCLAIMED.  IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE

LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR

CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF

SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS

INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN

CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)

ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE

POSSIBILITY OF SUCH DAMAGE.

*/

/* The Computer Language Benchmarks Game

  http://shootout.alioth.debian.org/

  contributed by Paolo Bonzini & Sean Bartlett

  modified by Michael Mellor

*/

#include <stdio.h>

#include <stdlib.h>

#include <gmp.h>

static mpz_t numer, accum, denom, tmp1, tmp2;

static int extract_digit()

{

  if (mpz_cmp(numer, accum) > 0)

    return -1;

  /* Compute (numer * 3 + accum) / denom */

  mpz_mul_2exp(tmp1, numer, 1);

  mpz_add(tmp1, tmp1, numer);

  mpz_add(tmp1, tmp1, accum);

  mpz_fdiv_qr(tmp1, tmp2, tmp1, denom);

  /* Now, if (numer * 4 + accum) % denom... */

  mpz_add(tmp2, tmp2, numer);

  /* ... is normalized, then the two divisions have the same result.  */

  if (mpz_cmp(tmp2, denom) >= 0)

    return -1;

  return mpz_get_ui(tmp1);

}

static void next_term(unsigned int k)

{

  unsigned int y2 = k*2 + 1;

  mpz_mul_2exp(tmp1, numer, 1);

  mpz_add(accum, accum, tmp1);

  mpz_mul_ui(accum, accum, y2);

  mpz_mul_ui(numer, numer, k);

  mpz_mul_ui(denom, denom, y2);

}

static void eliminate_digit(unsigned int d)

{

  mpz_submul_ui(accum, denom, d);

  mpz_mul_ui(accum, accum, 10);

  mpz_mul_ui(numer, numer, 10);

}

static void pidigits(unsigned int n)

{

  int d;

  unsigned int i = 0, k = 0, m;

  mpz_init(tmp1);

  mpz_init(tmp2);

  mpz_init_set_ui(numer, 1);

  mpz_init_set_ui(accum, 0);

  mpz_init_set_ui(denom, 1);

  for(;;)

  {

    do {

      k++;

      next_term(k);

      d = extract_digit();

    } while(d == -1);

    putchar(d + '0');

    i++;

    m = i%10;

    if(m == 0)

      printf("\t:%d\n", i);

    if(i >= n)

      break;

    eliminate_digit(d);

  }

  if(m) {

    m = 10 - m;

    while(m--)

      putchar(' ');

    printf("\t:%d\n", n);

  }

}

int main(int argc, char **argv)

{

  pidigits(argc > 1 ? atoi(argv[1]) : 27);

  return 0;

}