/* Problem 1--Double Trouble
   Trial and error would take too long for this problem.  What you
   needed to observe--mathematically--was that this problem REALLY
   involved repeating decimals.  The correct answer was a multiple of
   1/(2b-1).  The double trouble number in base 10 is, for example,
   1/19. */

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

FILE *in, *out;
int b;

int main (int argc, char **argv);
void Process (void);
void MakeString (char **n, int *l, int i);

int main (int argc, char **argv) {

 in = fopen ("prob1.in","r");
 out = fopen ("prob1.out","w"); /* Process case */
 while (fscanf (in,"%d",&b),b > 0) Process ();
 fclose (in);
 fclose (out);
 return EXIT_SUCCESS;
}

/* Process processes a single test case. */
void Process (void) {

  char *f, *n;
  int fl, nl, i;

 MakeString (&f,&fl,1);     /* generate repeating decimals */
 for (i = 2; i < b; i++) {  /* the one with the smallest length */
  MakeString (&n,&nl,i);    /* is the double trouble number */
  if (nl < fl) {
   free (f); f = n; fl = nl;
  } else {
   free (n);
  }
 } /* Print out the number */
 fprintf (out,"For base %d the double-trouble number is\n",b);
 for (i=0; i < fl;i++)
  fprintf (out,"%d ",(int)(unsigned char)f[i]);
 fprintf (out,"\n\n");
}

/* MakeString accepts a string variable and an integer i, and builds */
/* the repeating decimal string i/(2b-1) in base b.  It is put into */
/* string n, and the length is put into l */
void MakeString (char **n, int *l, int i) {

  int j = i;

 *l = 0;
 *n = malloc (2*b-1);
 do { /* get the next digit */
  (*n)[(*l)++] =  j*b/(2*b-1);
  j = j*b % (2*b-1); /* get the next remainder */
 } while (j!=i);
}