/* This program demonstrates
 * 1. Nested Loops
 * 2. Working with loops
 * 3. Deriving loop conditions and formulae in general from examples.
 *
 * Nested loops
 * We can place loops inside loops. If we do so, the inner loop runs several times
 * for EVERY iteration of the outer loop.  We can also use nested loops in conjunction with selection statements in any
 * combination.
 *
 * Deriving loop conditions and formulae in general from examples.
 * Sometimes, it not obvious how many times a loop should run.
 * In this case, we can write out a few examples, and derive a formula for
 * the number of iterations from those examples.
 */

#include<iostream>
using namespace std;
int main()
{
    int num ;
    cout<<"Enter a number: ";
    cin>>num;

    /* Print the factors of a number
      * A number f is a factor of a number n if f divides n without reminders.
      * Naive way: check all of the numbers below n
      * Step 1: Read in n
      * Step 2: Loop from 1 until n.
      *       Step 2.1 If i divides n, print i
      * Step 3: Done
      * We do this for all 3 loops - to demonstrate how a problem can be solved by
      * choosing any of the 3 and onl making minor modifications
      */

    cout<<"Factors of "<< num << " (while): ";
    int f = 1;
    while( f<=num)
    {
        if(num % f == 0)
            cout<<f<<endl;
        f++;
    }

    cout<<"Factors of "<<num<<" (do=while): ";
    f = 1;
    do
    {
        if( num % f ==  0)
            cout<<f<<endl;
        f++;
    }while(f <= num);

    cout<<"Factors of "<<num<<" (for): ";
    for(f=1; f<=num; f++)
    {
        if(num % f == 0)
            cout<<f<<endl;
    }

    /* Printing ASCII ART letters.
      * We can assume the letters will be printed using a square grid of characters,
      * with "size" number of characters on each row and "size" number of rows.
      * We can assume "size" will be an odd number.
      * We mark any spot on the grid that should be a part of the letter with a "*"
      * All other spots should be filled with spaces.
      * We do one row at a time, top to bottom. In each line, we go left to right.
      * This is because, on the console, it is impossible to go back up to the previous
      * row or back left to the previous column. The console is like a typewriter.
      */
    int size;
    cout<<"enter the size: ";
    cin>>size;

    /* The letter I will look like this:
     *      *
     *      *
     *      *
     *      *
     *      *
     * Print one star per row, followed by newline
     */

    cout<<"Printing I\n";
    for( int r = 0; r <size; r++)
    {
        cout<<"*"<<endl;
    }

    /* Printing L
     * The letter L looks something like this:
     * 	    *
     * 	    *
     *	    *
     * 	    *
     * 	    *****
     * The bottom row is a row of "size" *. Every other row is 1 *.
     * Print a newline after each row.
     * How to print a row of *'s? Since we don't know "size" before the user
     * enters it, we need to print 1 star at a time. We need a loop for this
     * 
     * How do we find teh row number for the last row?
     * Size => 7, Rows (0-6). Last row => 6
     * Size => 9, Rows (0-8). Last row => 8
     * Size => 11, Rows (0-10). Last row => 10
     * Size => 13, Rows (0-11). Last row => 12
     * So, the last row is row number "size -1"
     */

    cout<<"\nPrinting L\n";
    for(int r=0; r<size; r++)
    {
        if ( r == size-1) // bottom row, always do the weirdest case first
        {
            for(int c=0; c<size; c++)
                cout<<"*";
        } else      // every other row
        {
            cout<<"*";
        }
        cout<<endl;
    }

    /* For the next example, we will print an ASCII art letter again.
     * This time, we will print the letter C.
     * The letter C looks something like this:
     * 	    *****
     * 	    *
     *	    *
     * 	    *
     * 	    *****
     * Here, we have "size" number of stars on the top and bottom row. Every
     * other row has only one star.
     * How do we know whoch row is the top row?
     * It is the "first", or in our case, the 0th row
     */
    cout<<"\nPrinting c"<<endl;
    for(int r=0; r<size; r++)
    {
        if( r==0 || r==size-1)      // top or bottom row
        {
            for(int c=0; c<size; c++)
                cout<<"*";
        }
        else cout<<"*";             // every other row
        cout<<endl;
    }
    return 0;
}