/*
 *	\( (3) \) Let \(n\) and \( m\) be positive integers. An \( n \times  m \) rectangle is tiled 
 *  with unit squares. Let  \(r(n,m) \) denote the number of rectangles formed by the edge of these 
 *  unit squares. Thus, for example, \(r(2, 1) = 3\). Can you find \( r(11, 12) \)?.
 *
 *	In a row of length n,there are 
 *
 *	number		rectangle dimensions
 *	n			1x1 
 *	n-1			2x1 
 *	n-2			3x1
 *	n-3			4x1
 *	n-a			(a+1)x1
 *
 *	a+1 = n at most, a = n-1
 *	n-(n-1)	nx1
 *	1		nx1
 *
 *	Sum is from 1 to n which = n(n + 1)/2
 *	Because of the way I plan to do the rest of this, I'm going to exclude the 1x1 dimension rectangle. Sum is then from 0 to n-1, so:
 *	(n-1)((n-1) + 1)/2
 *	n(n-1)/2
 *	multiplied by m rows,
 *	There are,
 *	mn(n-1)/2 rectangles with a dimension of 1 in the whole rectangle LOOKING ONE WAYS. The rectangle can be flipped and the process repeated, so you must add:
 *	nm(m-1)/2
 *
 *	Now for 2x2 rectangles.
 *	They can extend for (n-1) lines across, and (m-1) down. So,
 *	(n-1)(m-1)
 *
 *	Since 2x2 are a square dimension, like 1x1 we only need to count it going one way.
 *	Sum becomes:
 *	mn + mn(n-1)/2 + nm(m-1)/2 + (n-1)(m-1)
 *
 *	2x3?
 *	(n-1)(m-2) and then also (m-1)(n-2)
 *
 *	Sum is then:
 *	mn + mn(n-1)/2 + nm(m-1)/2 + (n-1)(m-1) + (n-1)(m-2) + (m-1)(n-2)
 *
 *	I see a pattern, so I'm going to redo it.
 *
 *	(m - 0)(n - 0) for 1x1 rectangles
 *	(m - 1)(n - 0) for 2x1 rectangles
 *	(m - 2)(n - 0) for 3x1 rectangles
 *	...
 *	(m - (m-1))(n - 0) for mx1 rectangles
 *	(m - a)(n - b) for (a+1) by (b+1) rectangles
 *
 *	All of this is for going one way. If the rectangle is flipped, then it is done again IF a != b.
 *	I'm not good with sums, so I wrote a program to calculate this.
 *
 */
 
import java.lang.Math;
 
public class Main {
	public static void main (String[] args) { //no error checking on purpose, no need really
		int total = 0;
		int m = 11;
		int n = 12;
 
		for (int ai = 1; ai <= Math.min(m, n); ai++) { //loops through all the possible first dimensions
			for (int bi = 1; bi <= Math.max(m, n); bi++) { //loops through all possible second dimensions
				if (bi >= ai) { //so you don't count things twice
					total += numberOfaxbRectanglesInmxnRectangle(ai, bi, m, n);
				}//if
			}//for
		}//for
 
		System.out.println(total);
	}
 
	public static int numberOfaxbRectanglesInmxnRectangle(int a, int b, int m, int n) {
		int ret = 0;
		ret += (m - (a-1)) * (n - (b-1)); //see above explanation
		if (a != b && a <= n && b <= m) { //ensures you don't try and see how many 4x2s are in a 2x4
			ret += (n - (a-1)) * (m - (b-1)); //if it isn't a square, could have recursed but w/e
		}
		System.out.println(ret + " many " + a + " x " + b);
		return ret;
	}
}

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