/**
 * Problems:
 * <pre>
 * 1) Print all the positive integers less than 100
 * that are divisible by 3.
 *
 * 1B) Then print those numbers only 5 per line.
 * </pre>
 *
 * Uncomment each solution to test it.
 *
 * @author Zach Tomaszewski
 * @version 10 Oct 2008
 */
public class DivisibleByThree {

  public static void main(String[] args) {

    //
    // 1) Print all the positive integers less than 100
    // that are divisible by 3.
    //
/*
    //
    // 1a) Go through the numbers 1 to 100,
    // printing each one only if it's divisible by 3
    //
    for (int i = 1; i < 100; i++) {
      if (i % 3 == 0) {
        System.out.println(i);
      }
    }
*/
/*
    //
    // 1b) Go through only the numbers we know we'll have to print.
    // That is, start at 3 and print every 3rd number from there
    //
    for (int i = 3; i < 100; i += 3) {
      System.out.println(i);
    }
*/
/*
    //
    // 1c) Who said we need to start at the bottom?
    // This is as 1a, but prints in reverse.
    //
    for (int i = 100; i > 0; i--) {
      if (i % 3 == 0) {
        System.out.println(i);
      }
    }
*/
/*
    //
    //1d) An interesting student-suggested approach,
    // this for loop effectively goes from 1 to 33 (looping through
    // the factors that we are multiplying by 3)
    //
    for (int i = 1; (i * 3) < 100; i++) {
      System.out.println(i * 3);
    }
*/


    //
    // 1B) As above, but print 5 numbers per line.
    //
/*
    //
    // 1Ba) One way to do this is just print a newline (println()) after
    // each number that is divisible by 5 (or by 15).  (Of course,
    // doing this depends on somewhat on the nature of our data--the fact
    // that only every 5th number that we print will be divisible by 5.)
    //
    for (int i = 1; i <= 100; i++) {
      if (i % 3 == 0) {
        System.out.print("\t" + i);
        if (i % 5 == 0) {
          System.out.println();
        }
      }
    }
    //Bug: What happens if we don't nest the ifs, but have them one after
    // the other?  Why does this change produce the output that is does?
*/
/*
    //
    // 1Bb) Here's a solution that doesn't depend on the nature of
    // the numbers/data that are being printed: we'll separately track the
    // number of time's we've printed a number on each line.
    //
    int printedThisLine = 0;  //how many numbers I've printed on this line
    for (int i = 1; i <= 100; i++) {
      if (i % 3 == 0) {
        System.out.print(i + "\t");
        printedThisLine++;
      }
      if (printedThisLine == 5) {
        //end the line and reset the counter
        System.out.println();
        printedThisLine = 0;
      }
    }
    System.out.println();
*/

  }
}