Think like a programmer. Episode 2. Real solution

gistfile1.txt

// I like to start with the one that has the narrowest set first. That's just personal preference.
// police department must have an even number. no need to loop to 7 which is odd.
for(int policeDepNum = 2; policeDepNum <= 6; policeDepNum += 2) {
  for(int fireDepNum = 1; fireDepNum <= 7; fireDepNum++) {
    // if these are equal there is no solution this loop so continue on.
    if(fireDepNum == policeDepNum) {
      continue;
    }
    // sum must equal 12. no need to loop for last number.
    sanitDepNum = 12 - fireDepNum - policeDepNum;
    // check that sanitDepNum meets all conditions and continue if not.
    if(sanitDepNum < 1 || sanitDepNum > 7 || sanitDepNum == fireDepNum || sanitDepNum == policeDepNum) {
      continue;
    }
    cout << "Police Department Number: " << policeDepNum << "\nFire Department Number: " << fireDepNum << "\nSanitation Department Number: " << sanitDepNum << "\n";
  }
}