Codeforces celebrates 10 years! We are pleased to announce the crowdfunding-campaign. Congratulate us by the link https://codeforces.com/10years. ×

### selfcompiler's blog

By selfcompiler, 5 years ago, ,

Here is Problem Link i read its editorial but unable to understand it would anyone like to make me understand this problem ?

• +9

 » 5 years ago, # |   -18 Did you mean "understand its editorial"?
 » 5 years ago, # |   +5 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