I don't have any official information. But seems like at least 4 teams will advance to finals.

What's the problem? One team on-site got 10 and one in opencup got 11, and all 3 difficult problems were kinda solvable :) You did good!

First, triple {0, 1, k} is a solution (but we don't output it because of 0).
If we have a solution, we can replace c with k(a + b) - c and get another solution (similarly for a and b).
This way we can run BFS over the triples until we have enough to output. Simple set is enough to check for duplicates.

Let's consider some good subset V, i.e. subset of vertices that can be covered by some matching, L and R are subset of vertices in V from left side and right side of graph respectively. Then it can be easily seen that L and R are also good subsets. So let's find all good subsets of left side and right side independently. neighbour(S) is the subset of vertices which are neighbours to some of vertex in S. Now just iterate over all 2ⁿ different subsets in left side and check that subset. Using Hall's theorem.

After finding two list of good subsets just find number of pairs with sum  ≥ t using two pointers or binary search. Overall complexity: O(n * 2ⁿ).

P.S: How fast is the solution with iterating over all subsets and then trying Kuhn?

In I thought about a bit easier solution: iterate over bitmask of whether each hist is LTR or RTL, reverse if necessary. Then precalculate for each pair of hints transition cost, and then run your solution for one-sided hints in O(2ⁿ·n). So overallcomplexity is something like O(2ⁿ·(n²·digits + n·2ⁿ))

Will it also work or I am missing something?

I could not believe my eyes when saw D marked as medium! Also sad we didn't solve B, because I misread it, but I think that was also not hard :(

In the meantime, guys, I think that was one of the best rounds this season! Thanks so much for the great problems!

Editorial. K on page 5, H on page 49.