For the first subtask, it suffices to iterate over all possible 3-tuples or 4-tuples of lattice points in the grid, and to determine whether a rectangle can be formed from these points. Measures can be taken to avoid over-counting by dividing the final result by the number of times that a given rectangle is counted in this process.

Time complexity: ~\mathcal{O}((HV)^3)~ or ~\mathcal{O}((HV)^4)~

For the second subtask, we seek to eliminate the redundancy of counting congruent rectangles multiple times. To do this, we must find a method of enumerating all possible rectangles that can fit within the grid, and for each rectangle, calculate ~S~, the number of times that it can be translated in the grid. The answer is then the sum of all such ~S~.

One way of doing this is to iterate over all non-ordered pairs of points with one point on the top of the grid, and the other on the left of the grid. This fixes one side of the rectangle ~L_0~. Next, iterate over all ~L_1, L_2, \dots, L_N~, which are all translations of ~L_0~ in the direction orthogonal to ~L_0~ that fall on lattice points in the grid. This will enumerate over all rectangles such that the separation between the left-uppermost point and right-uppermost point is ~(x, y)~.

To find ~L_{i+1}~, apply the translation ~(dx,dy)~ to ~L_i~, where ~dx = \frac{y}{\gcd(x,y)}~ and ~dy = \frac{x}{\gcd(x,y)}~. It can be shown that applying this translation is the smallest translation such that ~L_{i+1}~ is different from ~L_i~ and is in the directional orthogonal to ~L_i~.

Pseudocode:

answer = 0
for x in [1, v)
    for y in [0, h)
        dx is y / gcd(x, y)
        dy is x / gcd(x, y)
        x_i is x + dx
        y_i is y + dy
        while x_i in bounds and y_i in bounds
            add (h - y_i) * (v - x_i) to answer
            add dx to x_i
            add dy to y_i

Diagram:

Time complexity: ~\mathcal{O}(HV \log^2(\min(H,V)))~