| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3823 |
| 3 | Benq | 3738 |
| 4 | Radewoosh | 3633 |
| 5 | jqdai0815 | 3620 |
| 6 | orzdevinwang | 3529 |
| 7 | ecnerwala | 3446 |
| 8 | Um_nik | 3396 |
| 9 | ksun48 | 3390 |
| 10 | gamegame | 3386 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 165 |
| 2 | maomao90 | 163 |
| 2 | Um_nik | 163 |
| 4 | atcoder_official | 161 |
| 5 | adamant | 160 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | nor | 153 |
| 9 | Dominater069 | 153 |
Tomorrow Saturday November, 10 at 18:00 hours UTC. The Latin American Regional Contest(Open) will be held in Caribbean Online Judge coj.uci.cu. Everyone is invited.
The real contest will begin one hour before and the the caribbean standing will be show in live.uci.cu.
Is it the same contest ? http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=12&page=show_contest&contest=306
Apparently yes. But in the COJ starts an hour earlier.
I'm sure there will be a very insteresting problem set, and that open competition will be like the real one, 'cause will be the first time those problems will be publics. I bet everyone to participate and compare latter with the oficials results..!
lol, TopCoder Single Round Match 560 will be at 12:10 PM EST. Maybe a lot of coders here will select TopCoder. In my particular case, I will try to compete in Latin American Regional Contest but I will lose one hour.
I'm going to compete in the real contest. I thinks will be a good contest.
Sure, I will do it too in COJ. Good luck everybody from UCI in the real contest.
Nice contest, but 2 things:
a) Tests in A were very weak.
b) I hate something like "output a number with exactly 5 signs after decimal point". It means absolute precision when the answer is like "1.000005 — 1e-40".
Oh, you 're anonymous user :)
Congratulations, you solved 7 problems. First team in real contest only solved 5 problems.
Did you get AC on A with O(KN^2)?
I think the problemset was great but I didn't like a few things.
1) Problem E was not interesting at all. 2) The problem concerning string algorithms (Problem C) was in fact very easy. I was expecting something harder. 3) There were only ten problems, I would have liked 11. 4) There were four easy problems , three problems would have been enough.
Congratulations to the problemsetter of problem G and J, very interesting and nice problems.
I've seen problem like J before on some Codeforces gym.
How to solve G? Something like analyzing of 2-connected components?
Problem G is solve using bipartie matching.
Could you give a hint please ?
Try to use domino tiling.
I could see that. So you partition the board into the sets A, B (A is the set of possible states for one player, same for B) right ? What I can't see is the connection between this graph and the winning condition ?
Try to invent a winning strategy for a second player, if it exists (and you'll understand when it exists).
You have the fallowing problem: In a graph to players play a game, first player select a vertex and then they alternately move to and adjacent vertex of the current one (not visited yet). The one who can't play any more losses.
If there exist a perfect matching in the graph then player 2 has a winning strategy: No matter what vertex player 1 selects, player 2 can move to its corresponding vertex in the matching. If there is no perfect matching on the graph lets consider a maximum matching. Player 1 has a winning strategy which consist in selecting an unmatched vertex, player 2 has to move to a matched one, otherwise this edge could be added to the matching and it wont be maximum (notice that player 2 moves across edges that don't belong to the match and player 1 moves across edges that do belong). If at the end player 1 can't play it means that there is an augmenting path, but this is a contradiction because the matching is maximum so player 1 should be the last to make a move.
Does anyone knows the solution for problem B???
Problem B ( http://coj.uci.cu/24h/problem.xhtml?abb=2118 ) has to be really hard 'cause even here anybody has answer with a solution. I hope someone will read it and comment a solution. Thanks to that user in advance.!
My team solved it during the contest. It is really nice.
A first observation I made was that if there are at least 2 balls in the last box then the first player wins. So there must be 0 or 1 in the last box. But you could also get some balls from the previous box, and that box could get balls from a previous one.
So it becomes something like:
(xn + (xn-1 + (xn-2 + (...) / 2) / 2) / 2) / 2 == 0
Then you can calculate the winning positions for the second player with DP.