Here is Problem Link i read its editorial but unable to understand it would anyone like to make me understand this problem ?
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3843 |
2 | jiangly | 3705 |
3 | Benq | 3628 |
4 | orzdevinwang | 3571 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3530 |
8 | ecnerwala | 3499 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 171 |
2 | awoo | 164 |
3 | adamant | 162 |
4 | TheScrasse | 159 |
5 | nor | 154 |
6 | maroonrk | 153 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
9 | orz | 145 |
10 | pajenegod | 144 |
Here is Problem Link i read its editorial but unable to understand it would anyone like to make me understand this problem ?
Название |
---|
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