In fact, you don't need dp in this problem.

Let's count the following thing (just using a loop over bitmasks): how many there are strings of length L than consist only of A and B and are good (not hated, i. e., don't contain any hated subword).

Now let's iterate over all possible positions of non-AB characters (again as a loop over bitmasks). Obviously, if there is any hated word, it will be between two two consecutive non-AB characters. So we can use the thing we calculated earlier to fill all characters between non-AB ones.