/* Problem 9--Number of Numbers
   Although this was the easiest problem to solve, there was a
   deliberate vagueness in the description that led people not to test
   for negative numbers at all, thus slowing down the competition for
   everyone. */

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

int main (int argc, char **argv);
int mod (int a, int b);

FILE *in, *out;

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

  int first, last, diff, ct, cs=0;

 in = fopen ("prob9.in","r");
 out = fopen ("prob9.out","w");
 while (fscanf (in,"%d %d %d",&first,&last,&diff),first<=last) {
  if (first==last) ct = 1; /* Certain easy special cases */
  else if (diff==0) ct = 2;
  else {
   diff = abs(diff);
   ct = (last-first)/diff+1; /* How many numbers */
   if (mod (last,diff)!=mod(first,diff)) ct*=2; /*Doubled if necessary*/
  }
  if (ct!=1)
   fprintf (out,"Case %d: Set contains %d integers.\n\n",++cs,ct);
  else
   fprintf (out,"Case %d: Set contains %d integer.\n\n",++cs,ct);
 }
 fclose (in);
 fclose (out);
 return EXIT_SUCCESS;
}

/* Mod computes a normal modulo unless a is negative, in which case it
   is adjusted to ensure that the answer is nonnegative. */
int mod (int a, int b) {

 if (a>=0) return a%b;
 return ((b-(-a)%b)%b);
}