Did you mean "understand its editorial"?

I haven't understood the editorial, but I have another idea for you, and it may be shorter! :)

When you are at the i-th click, let L be the length of the previous consecutive "O"s block. If the i-th click is bad, your score increases 0, otherwise it increases (L+1)^2-L^2=2L+1.

We will canculate the expected score each click "gives". The answer to the problem is the sum of all expected score. For the i-th click, let X be the expected length of the previous consecutive "O"s block (after i-1 clicks). The expected score this click gives is p*(2*X+1), where p is the probability that the click will be correct. Then, the expected length X after this click changes to p*(X+1).

See my submission 7665345 for more information.

I hope you will understand it. Thanks for your attention :D