XX Open Cup Grand Prix of Gomel takes place on Sunday, February 16, 2020, at 11:00 AM Petrozavodsk time.
The links to the contest:
You need an Open Cup login to participate.
I'm the writer of all the problems. Let's discuss them here after the contest!
UPD: Thanks for your participation! Editorial is available.
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3690 |
| 2 | jiangly | 3647 |
| 3 | Benq | 3581 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | Radewoosh | 3509 |
| 8 | ecnerwala | 3486 |
| 9 | jqdai0815 | 3474 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 163 |
| 4 | TheScrasse | 159 |
| 5 | nor | 157 |
| 6 | maroonrk | 155 |
| 7 | -is-this-fft- | 152 |
| 8 | Petr | 146 |
| 8 | orz | 146 |
| 10 | BledDest | 145 |
When I login into the page I get the following message: "The virtual contest is in progress. You are not allowed to participate".
EDIT: Now both links given for Div 1. seem broken to me.
EDIT2: fixed, thanks :D
How to solve B and H?
How to solve E?
Let $$$P(X) = \sum p_i \cdot x^i$$$. We need to find several first coefficients of $$$Q(x) = P(x) ^n$$$.
We can write equation $$$Q'(x) = (P(X) ^ n)' = n \cdot (P(x) ^ {n-1})P'(x) = n \cdot Q(x) / P(x) *P'(x)$$$
So $$$P(x) \cdot Q'(x) = n \cdot Q(x) \cdot P'(x)$$$. If we write everything about $$$x^m$$$ it will give us a way to represent (m+1)-th of coefficient of Q(x) from deg(P) previous.
Only after reading editorial for A I noticed that it's guaranteed that n is divisible by k. :)
How can I obtain an opencup account? I've messaged snarknews on CF but got no response.
There is linear solution for F based on fact we can calculate $$$f(k) = \sum\limits_{t\geq n} bin(k, t)$$$ for all k in linear time
Instead of matching algorithms for G we can use much much simpler approach (especially when compared to blossom): take arbitrary unmatched vertex and match it to its random neighbour — if it that random neighbour was already matched before then unmatch it
What about the bipartite part? This method works for it too?
Yeah, works like a charm for both bipartite and non-bipartite case :)
We can use the symmetric chain decomposition of the boolean lattice to solve the bipartite case for G constructively. You can read pages 5 and 6 of this slide.
orz jqdai0815 teacher
Comments like this make me realize that I haven't even started CP xD
Will the mirror of this contest available anytime soon on Codeforces gym?