About problem C the idea to detect sides (as I said before) was to count corners. First a single square could be several corners at the same time

  12
1 OOO
2 .O.

In this sample square at position (1;1) represents two corners of the shape, Imagine that you want to check if a single cell could be a corner in one of the four directions: NW, NE, SW, SE (Directions are diagonal) The amazing idea of my teammate was the way to check that: (x;y) is corner in direction d(remember that d is a diagonal) <=> if (x;y) belongs to one shape(it isn't empty) and (x+dx[d];y+dy[d]) is empty and (x+dx[d];y) == (x;y+dy[d]) ie, you need to check if the cell in the diagonal is empty and the common neighbours cells between current cell and the one in the diagonal have the same state(are both empties or filled). For each corner you should add 2 to the total side, but then notice that you have counted each side twice (one per corner) so dividing the answer by 2 it turns out that the answer = total corners.

For problem H, the answer was: 2 * (C(o + e, o) * 5^(o + e) — C(o + e — 1, o — 1) * 5^(o + e — 1)) where C(a, b) = a! / (b! * (a — b)!) 2 times: because negative numbers. The substraction must be done only if o > 0, because you don't need to count numbers that start with zero(0), I remember that there was a tricky case in which we fail, it was something like 1 0, because number 0 should be counted, and the answer was 9.