Hints for the 3rd midterm problem.

1. First write a function that prints out numbers with three
   digits or less. 
   Once you have it, break your number into parts, each no longer
   than three digits. Say, 23,570 breaks into  23  and  570,
                           23,070 breaks into  23  and  70  etc.

   These parts can be processed independently. After you process
   the first part, you only have to print "thousand." and 
   process the second part.

2. When processing a three digit number, you can print the number 
   of hundreds (if any) without looking at the last two digits.

3. When printing the number represented by the last two digits of
   a three digit number, you have to check if that number is above
   or below 20 (numbers below 20 have irregular names).

4. Test your program extensively.

For some tricks you might use, look at the following program that
counts the number of digits in a number, and then prints all of
the digits one by one. IMPORTANT: Do not try to use this code as
a model -- it solves a very different problem! It just shows you 
how you can count the digits in a number, etc. For best results,
step through this code, with pencil and paper, for  n = 2450 .

#include <iostream.h>
#include <string>

int main(){

  int n;
  cin >> n;

  // find out how many digits n contains, and print that number

  int count = 0;    // digit counter
  int m = n;        // make a working copy of n

  while( m != 0 ){  // let's divide m by 10 and see how many times
    count++;        // we can do it before we get a zero
    m = m/10;
  }
      
  cout << "The number " << n << " has " << count << " digits.\n";

  // now print these digits, one by one

  // let us first prepare an auxiliary variable
  int aux = 1;
  for( int i=0; i < count-1 ; i++) aux *= 10;
  cout << "aux = " << aux << endl;

  string digits[10] = { "zero", "one", "two", "three", "four", "five",
			"six", "seven", "eight", "nine" };

  int current_digit;
  m = n;            // again, make a working copy of our number

  for( int i=0; i < count; i++){
    current_digit = m/aux;  // the leading digit of m 

    cout << digits[ current_digit ] << endl;

    m = m % aux;    // prepare for the next iteration (i.e. next digit) 
    aux = aux/10;
  }

  return 0;
}