Consider the sequence of cards that will be taken until the end of the game. For example, if n = 5, m = 4, k = 6 then it can look like:

           V these cards weren't revealed, so they could be chosen arbitrarly
bcaabacccaaXXXX
          ^ after this card Alice wins

Alice wins if the sequence consists of exactly n a's, no more than m b's and no more than k c's. Let's say there are s b's and c's totally. The rest of cards may be chosen with 3^{m + k - s} ways, and the revealed cards may be chosen with ways where f(s) is the number of strings consisting of s characters a and b in total from which no more than m are a's and no more than k are b's.

It's easy to see that

If we draw a Pascal Triangle, this is a section of rectangle (0, 0) — (m, k) with a diagonal line that looks like a segment consisting of binomial coefficients with fixed first argument. There is no closed form for such sum, but it's easy to see thath f(s + 1) is almost equal to 2f(s), you only need to carefully consider O(1) binomial coefficients at the border. So, the solution is: calculate using DP the function f(s + 1) = 2f(s) + [O(1) binomial, 2016 - 09 - 12 coefficients], and then output the answer . Binomial coefficients may be calculated in O(1) if we precompute all factorials. Overall complexity is O(n + m + k).

Can someone explain problem E also..