| # | User | Rating |
|---|---|---|
| 1 | Benq | 3857 |
| 2 | jiangly | 3810 |
| 3 | maroonrk | 3534 |
| 4 | tourist | 3528 |
| 5 | Kevin114514 | 3510 |
| 6 | turmax | 3411 |
| 7 | Um_nik | 3387 |
| 8 | Radewoosh | 3367 |
| 9 | heuristica | 3322 |
| 10 | strapple | 3317 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | maspy | 150 |
| 3 | Um_nik | 146 |
| 4 | Errichto | 139 |
| 5 | adamant | 136 |
| 6 | maroonrk | 134 |
| 7 | DNR | 133 |
| 8 | nik_exists | 131 |
| 8 | Dominater069 | 131 |
| 10 | Proof_by_QED | 130 |
|
+66
Div1C has the most confusing statement I've ever seen. Setters should really be careful when translating. |
|
+8
Very weird statement. Nice problem, but a better care with the statement would have made a great difference. |
|
+8
It would be ok since you can select the online judges you wish to be updated about. If you didn't care for hackerrank you could disable it. |
|
+3
Why the downvotes? Is it wrong? |
|
+13
Yeah, I'm sorry about saying this and not explaining. Actually I should give credit because yancouto first realized the fully dynamic approach was not necessary. Here's my code. Maybe the most natural approach for this problem is to try to solve the following recurrence (or something similar) where f(0) = 0 and d0 = 0: Well, this recurrence really requires a fully dynamic approach. We'll find one that doesn't. Instead of trying to solve the problem for each prefix, let's try to solve it for each suffix. We'll set g(n + 1) = 0, a0 = d0 = 0 and compute which can be written as now we notice that the function inside the max is actually a line with angular coefficient j and constant term aj + g(j) (which are constant on i) evaluated at - di. Apply convex trick there (the standart one) and we're done. Notifying possibly interested people after a long delay (sorry about that again): fofao_funk, Sgauwjwjj and -synx-. And sorry in advance for any mistake, the ideia for the solution is there. |
|
0
On 1D — Randomizer you said
I didn't understand the sentence. Can someone please explain what that means? Thanks :) |
|
0
You can undo your registration by visiting the page which lists the registered users and clicking x next to your name. Also, if you register but don't submit it counts as if you've never registered. |
|
+16
Which means it is an unusual statement. |
|
0
Not that good of an idea right? Tried that a couple times, never works out... |
|
0
Think of the cost function as a matrix C. The problem required is exactly finding diagonal blocks (squares in the matrix that are non intersecting and contain the diagonal on it's union, search for block-diagonal matrices and it may help to understand) of minimal sum this helps on devising a clear proof of the property. I wrote this on a rush, if it didn't help answer this and I may write a careful proof. Good luck. |
|
+88
Don't get sad :( We love you <3 |
|
+3
New link for Commando: |
|
0
GOODG can be solved with Type 1 EDIT: I explain that below. |
|
0
Just as a warning for future readers. You don't need the fully dynamic variant to solve this problem. You can solve it easily with a different approach. :) |
|
+18
Sugestion: Problem lists. If you're training with a university or a team, it's bad to make a big contest, it would be better to make shareable lists of problems for people to solve in infinite time. It's like a shareable personal tagging of problems, maybe. |
|
0
|
|
0
I guess that's why hacks are divided by rooms, right? |
|
0
Thanks a lot :) I was trying to do something similar but i tought the Binary search wasn't good enough for the time and the bitmasks got quite confusing. Your explanation was clarifying. Thanks. |
|
+4
marcoskwkm? Help? |
|
+12
Hey guys. Any tips on solving the I — Ominobox? I guess the trouble is to discover quickly the minimum height for each n-omino. |
| Name |
|---|


