vovuh's blog

By vovuh, history, 3 years ago, In English

Hello! Codeforces Round 786 (Div. 3) will start at May/02/2022 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participants of the third division, you must:

  • take part in at least two rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Alexander fcspartakm Frolov, Ivan BledDest Androsov and Mikhail awoo Piklyaev. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Also huge thanks to ashmelev, Vladosiya, mesanu and I.AM.THE.WILL for testing the round and valuable feedback on the problems!

Good luck!

UPD: Editorial is published!

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3 years ago, # |
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Good luck everyone!

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3 years ago, # |
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omg vovuh div3s are back!!! :D

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Yes, I hope no smurfs and good luck to everyone!

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3 years ago, # |
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Best way to celebrate my birthday!! Vovuh's Div3s are lit!

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3 years ago, # |
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I have been participating in Div3 for quite a long but this contest never went unrated for me. I wish it doesn't happen again after this contest.BTW, Best of luck everyone.

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3 years ago, # |
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Best Wishes everyone for the contest and Happy Eid to all of you.

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3 years ago, # |
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vovuh, welcome back

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3 years ago, # |
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I have a question why some of the announcements don't give us exact numbers of problems?

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    3 years ago, # ^ |
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    I think because they don't have a final problemset yet, maybe because they have 8 or more problems made but they are not sure which problem to put on 1-2 places, and they will decide later

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3 years ago, # |
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hopefully i get specialist yoooo

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3 years ago, # |
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Eid Mubarak today

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3 years ago, # |
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Did someone notice, its Codeforces Round #786(the holy number) on the eve of Eid-al-Fitr!!! Such coincidence Much wow!

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3 years ago, # |
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vovuh welcome back in div 3s

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3 years ago, # |
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welcome back in div 3s, vovuh

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3 years ago, # |
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You will be offered 6 or 7 problems (or 8)

or 9

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3 years ago, # |
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Eyeing specialist from a distance 0_0

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3 years ago, # |
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Hope I will be specialist after this div3 contest!!!

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3 years ago, # |
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I want be specialist after this contest.

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3 years ago, # |
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vovuh welcome back

please make the time of the contest 2 : 15

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3 years ago, # |
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Long time these authors have not done div 3. All good solutions!

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3 years ago, # |
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I think this will be a div 3 round.

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3 years ago, # |
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Very Bad contest

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Marinush

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3 years ago, # |
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Why there is not any example or description note for test cases in problem F... can't really understand the test cases :(

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Spent 30 minutes to understand F :(

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3 years ago, # |
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I has closed div 3. Very good problems! Upd: hacked E :(

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I think D is the hardest problem :))

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    3 years ago, # ^ |
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    what was your approach? I tried like this but failed — the pairwise mins in array A must have an increasing order. for odd n the pairing excludes the first element, but it has a constraint to be the smallest integer.

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      3 years ago, # ^ |
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      sort each pair of (N, N — 1), (N — 2, N — 3).... and check if you can get sorted array. your solution does not work for 2 4 3 5.

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    3 years ago, # ^ |
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    Orz

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3 years ago, # |
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G is a simple but educational problem. Hope there will be more problems like this in the following div2/3 contests.

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    3 years ago, # ^ |
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    Can you explain it please?I think it's the longest path in the graph after deleting some edges but there are some cases that I choose edges that I shouldn't

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      3 years ago, # ^ |
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      It's a simple dp problem where you first topsort the vertices, and then for each vertex $$$u$$$ with in-degree greater than 1, you iterate over the parent nodes $$$v$$$ ($$$v \rightarrow u$$$), and if the parent node has out-degree greater than 1, you update the dp value as $$$dp_u = dp_v + 1$$$.

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        3 years ago, # ^ |
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        Since the graph is a DAG, you can use memoized search instead of topsort

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3 years ago, # |
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E — Breaking the Wall, Wrong answer on test 32. What was that?

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3 years ago, # |
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can u unfreeze the standings + let us submit

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Interesting tasks and good round! Thanks :)

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I did'n noticed that problem G statement:'a valid directed acyclic graph', I tried tarjan to shrink strongly connected components into points (poor english).Although it was unnecessary, I'm just wondering if this algorithm would be correct without such a condition.I would be grateful if someone could help me.

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Why results are frozen?

I can't submit((

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Spoiler

How it should work? It just reloads the page

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I spent 40 minutes on D and got TLE because I'm using native linked lists, there is no way you wanted me to implement an optimized liked list for this right?

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    3 years ago, # ^ |
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    Nope, note that you can only change the relative order between two elements after sorting ($$$a_{n} \leftrightarrow a_{n - 1}, a_{n - 2} \leftrightarrow a_{n - 3}, ...$$$)

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    3 years ago, # ^ |
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    you don't really need to do the operation.

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      3 years ago, # ^ |
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      can you explain how to calculate it then please

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        3 years ago, # ^ |
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        do some simulation and you will find you only can pairwisely revert elements from the right side.

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        3 years ago, # ^ |
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        Problem D
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vovuh rocks!!

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anyone explain F pls

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    3 years ago, # ^ |
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    remember the total number N of current icons on the screen, and the number of icons in the first N cells. The result is the difference between them. Use fenwick tree to find out the second value.

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      3 years ago, # ^ |
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      Actually there is no need for a Fenwick tree. You can simply iterate through the columns for each query and sum the amount of operations needed

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        3 years ago, # ^ |
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        Looks like with this time complexity it will work only for c++

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    3 years ago, # ^ |
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    You can consider unraveling the matrix resulting in a vector (i.e. taking the first column, then second, then third, and stick them to make a new vector). Then, the apps need to be in a prefix of this vector. when adding a image, you extend the prefix, otherwise you need to shrink the prefix (where you need to move the icons). Then, to find how many images you need to move is to calculate the number of images you have in the prefix you currently are in. If this number is X, and you have Y images on the desktop, then $$$Y - X$$$ is the answer. To calculate X, you can do some if's when erasing/inserting images (to see if they spawn/despawn directly from your prefix)

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    3 years ago, # ^ |
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    You had to count gaps in the icon arrangement using segment/fenwick trees.

    Video Solutions: A-F

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      3 years ago, # ^ |
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      No need of segment tree or fenwick tree it can be done without any extra log factor just have a two variables currow curcol which will denote the position for last icon after rearrangement. And whenever a new * is added somewhere just change the curpos to next postion by curcol+=(currow+1)/n currow=(currow+1)%n; and calculate total stars and number of stars inside the consecutive region. Submission- https://codeforces.net/contest/1674/submission/155735790

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3 years ago, # |
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Aight, guys. This comment may seem a bit toxic or something, but I still want to express how do I feel about this situation. Firstly, I want to say that the problem F is mine (I'm an author of the idea and I prepared this problem).

I get when people are confused about palindrome bracket sequence (as it was in some recent Educational Round), this thing can be really confusing when you see it. But I can't get how people are not familiar with the order of icons appearance on the desktop. Like, if you code from your phone or from fedora or some other OS without GUI, the desktop itself for you can be really confusing, but I don't think there are a lot of such people among Div.3 participants. So, I can't believe nobody saw that when you install some application and mark the field ''add the icon to the desktop'', this icon appears in the first column that is not full, and then in the first position from the top inside this column. Also, when you sort your icons, they're rearranging exactly as in this problem. I thought I created a very life-based problem and people will understand it easily, but I was wrong.

Moreover, the statement wasn't perfect, but was clear enough. There was the following sentence in the second paragraph: ''In other words, some amount of first columns will be filled with icons and, possibly, some amount of first cells of the next (after the last full column) column will be also filled with icons''. I agree that I could add some pictures or maybe highlight some words in the second paragraph, but I thought this is very-very life-based thing that anyone will understand without problems.

At last, I agree that I'm not completely right in this situation, because our task is to make statements as clear as possible for anyone. But, as I said before, this is just my personal opinion, so you can discuss it with me here or maybe just downvote this comment if you disagree.

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    3 years ago, # ^ |
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    The statement is a bit hard to understand, but anyway it's still OK to me.

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    3 years ago, # ^ |
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    Hey, thanks for the problem, and for the years of service!

    I think the statement was okay, but when reading it for the first time, I felt like an image with the statement would be perfect. Or maybe, explanation of one test case.

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    3 years ago, # ^ |
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    As a tester, I agree with vovuh. I also thought everyone encountered desktop arranging at least once and would understand the problem perfectly. To me, it seems like a very nice life-based problem so I didn't suggest any changes to the statement. Will definitely keep such things in mind from now on though.

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    3 years ago, # ^ |
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    For problem F, why did the question allow a time limit of 3 seconds? It'd be reasonable to make the participants to use a time limit of 1 second instead.

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      3 years ago, # ^ |
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      Yeah, that's probably true. I was just scared that some python/java/(any other non-CP language) users couldn't solve this problem because there is too many IO. But I was completely wrong, so there are also some $$$O((n+m)q)$$$ solutions in C++, and this is a bit disappointing. I also discussed this with Ivan, and we decided that 3 seconds is okay, but turned out it wasn't.

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    3 years ago, # ^ |
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    "But I can't get how people are not familiar with the order of icons appearance on the desktop."

    On mac, by default it is right to left:

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3 years ago, # |
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Problem E's pretests are so weak!!!!!!! I have already hacked myself!!!!

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    3 years ago, # ^ |
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    Educational rounds and div3s are meant to have weak pretest. Why do you think there is a 12 hour hacking period?

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      3 years ago, # ^ |
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      But it is the responsibility of questions setters to have strong pretests that contains most situation of the problem,rather than called the competitors to successfully hack more than one thousand accepted answer.

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    3 years ago, # ^ |
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    Dear contestant...

    Please don't worry if you had to hack yourself in order to see if your solution is good. In a programming competition, the tests used can be useless, you should have assured your code works. Sorry if this does not fit in your view.

    Author of xem sdod

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      3 years ago, # ^ |
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      Imposter?

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        3 years ago, # ^ |
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        Dear non-contestant,

        Please don't worry if someone else is a fan of EGOI and swapped two letters from your name. In a programming community thoughts as such are completly irrelevant. Apologies if this does not fit your view.

        So, can we please get back to programming ?

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Thank you vovuh and the team for organizing this Div 3 contest. I love the questions and how they were being framed.

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Amazed to see some red+ coders getting hacked on E. What is the hack case ?

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thank you AHMADUL for solving A!!!

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in D I thought that the operation from b to c was the same as from a to b. F :(

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    3 years ago, # ^ |
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    I spent about 30 minutes thinking of a solution for this problem before I realize It was not the real problem

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Testing data for problem E is far too weak.

We can hack a lot of solutions using this data.

2
1 5
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3 years ago, # |
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idea for d? and whats on pretest 32 in e?

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so many hacks to E...I have rised from 600+ to 400+.

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;-; Got WA on C just because I wrote (1ll<<a.size()) as pow(2,a.length())

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Can someone explain to me, what's the point of the removing condition in G? Doesn't it allow us to remove any edge?

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    3 years ago, # ^ |
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    The condition is that, after removing any number of edges, the final graph MUST have in-degree and out-degree of EVERY vertex to be strictly less than the original graph (except if the in/out degree was 0 for any vertex in original graph).

    For example, if there is a graph 1->2->3->4, then we can't just remove 1->2, because in the resulting graph (2->3->4), the in-degree and out-degree of 3 are the same as that in the original graph.

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thanx for good and strong E test's!

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any hint for E?

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    3 years ago, # ^ |
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    I'm trying to explain my solution while my solution is hacked...so I deleted all my notes...

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      3 years ago, # ^ |
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      One minute I was hacking my junior, the next I was being hacked by myself...

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I didn't noticed acyclic graph in G...

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Wow tons of hacks on E. Accept number decreases in every second!

I forgot a corner case (1 10000 3), the answer is 2(shoot middle, then right). Weak sample and tests.

Upd: till now, about a half solutions hacked!

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E's pretests are a joke

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These hacks are crazy, I have gone from 170th to 89th in 45 minutes after the contest. Maybe I will be next...

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3 years ago, # |
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Cursed spoiler:

open this spoiler
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3 years ago, # |
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Only for Newbies
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What does E's pretest mean...I wrote this chmin(ans, (mx + mx + 1) / 2); Then it's got AC????? Of course it's hacked...

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I am a simple man, I solved 5, my rank was initially around 1800, then everyone started getting hacked and my rank became around 1200 :) , and then ( you guessed it right ) I myself got hacked! now my rank is 2200+

Please CF don't troll me XD

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Problem E: My solution

Whats wrong with this?

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2000 participants: "I solved E!" Joury: "No You didn't)"

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    3 years ago, # ^ |
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    There are more than 50 pages of successful hacking to E. Never imagined to this...

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The pretests are weak — I solved E, then got hacked. I forgot about the case a[i] > a[i + 1] * 2(and vice versa) (then it's not (a[i] + a[i + 1] + 2) / 3 but (a[i] + 1) / 2)). Such a simple one, but pretests didn't cover that case..

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    3 years ago, # ^ |
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    There are no pretests in div.3, div.4 or educational rounds. It's full feedback meaning submissions are judged on all tests. Although, the tests on E were weak no doubt. The solves went from ~3000 to ~550 after system testing was over.

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      3 years ago, # ^ |
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      Thank you, didn`t know that

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interesting in hacking problem E the real game

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Enjoyed the few minutes when I thought I solved E

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I have been hacked :(

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Too weak E pretests!!

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proof by AC isn't working with problem E

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This round is amazing ,Because some of my friends get higher ranking than one hour ago。

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3 years ago, # |
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Meme
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3 years ago, # |
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If everyone is hacked, it means everyone is not hacked.

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I might be making a rude comment but the number of hacks makes me wonder that is the Jury's solution completely correct? Is it not possible for some really correct solutions to get hacked just because the jury's solution happens to be not perfect.

Just a thought.

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    3 years ago, # ^ |
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    My dear friend, you are absolutely wrong jury always have a proof to defend their solutions

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    3 years ago, # ^ |
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    The hacks are tested against jury's solution, that means the jury solution gives different output from the hacked solutions. Which means it's correct.

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      3 years ago, # ^ |
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      Moreover, there is not just ONE jury solution, usually there are at least 5 to 10 different valid solutions (written in different languages)

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"You should always prove your solutions before submitting" — Tourist

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Alright, for those that had wrong answer on pretest 32 in problem E like me, maybe this one will help you:
2
3 6
Answer is 3. I had 4 because I was trying to reduce by 3 both numbers while possible (applying two operations, subtracting 1 2 and 2 1) but this is not optimal, as you can just shoot number 6 three times.

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3 years ago, # |
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Is it May's fool? :')

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3 years ago, # |
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Problem E: failing just like the question name : Breaking the Wall (of pretests)

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If you prefer command line, checkout CF Doctor for more flexibility and faster feedback.

If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints.

If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).

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3 years ago, # |
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A B C D E-Hacked

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3 years ago, # |
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I have just refreshed the page and 620 accepted solutions for E became 570(during 2min). I am sure till tomorrow there will not be correct solutions

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3 years ago, # |
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What does "the prefix of the next column" in F means? Does it mean that I need to put all remaining icons to the upper side?

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    3 years ago, # ^ |
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    Yes

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    3 years ago, # ^ |
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    Yes. Effectively it means that if you took all the icons out, you would fill the columns greedily left to right, and within each column greedily top to bottom.

    Correct configuration:

    ***..
    ***..
    **...
    **...
    

    Incorrect configuration:

    ***..
    **...
    ***..
    **...
    
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3 years ago, # |
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Can anyone help me know why i am not included in common standing. I do not have a rating of 1900 ever. Am I not trusted participant ?

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3 years ago, # |
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do successful hacks give extra points for this round?

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    3 years ago, # ^ |
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    nope!

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      3 years ago, # ^ |
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      really? Damn! I hacked 10 submissions. I guess it was a waste of my time... So why do problem setters provide 12 hours of hacking phase for participants? Do they want us to make strong pretests for them?

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        3 years ago, # ^ |
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        It was a waste of your time if you didn't learn anything from it. Otherwise it wasn't.

        Div 3 (and 4) rounds are run according to Educational Round rules. The purpose of the 12 hour hacking phase is to encourage the studying of others' code and learning how to stress test, debug, and spot holes in algorithms. Often the tests deliberately leave out certain cases to make hacks more likely (though rarely as extreme as problem E today).

        If you only did it for points then yes I suppose you did waste your time; otherwise you did exactly what the round is designed for.

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          3 years ago, # ^ |
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          I meant to say that doing 10 hacks was a waste of my time since nearly all of them failed on same test case, and obviously I learned why those submissions failed

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        3 years ago, # ^ |
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        it's not a waste of time if you hacked people with better ranking than yours

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        3 years ago, # ^ |
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        Anyways, You don't get points for hacking in Div 3

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3 years ago, # |
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hacks ;-;

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3 years ago, # |
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3 years ago, # |
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Was hacked in E because of the case for (i,i+1) we have equations

2*k1+k2>=a[i] 2*k2+k1>=a[i+1]

when k1<0 or k2<0 my solution was not working. This case should be in pretests!

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3 years ago, # |
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E: WA on pretest 14 and 32 -> could not even enjoy the temporary happiness

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3 years ago, # |
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and if you had no participation and zero rating, then the rating will not rise?

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3 years ago, # |
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Could someone please try and hack 155731727

I want to see if I have written the correct solution. Thanks,

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3 years ago, # |
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I have a question when this contest is rated.

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3 years ago, # |
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Here are video Solutions to all problems in case people are interested.

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3 years ago, # |
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Why I haven't get any rating after participating in the contest and solving 3 questions there?

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3 years ago, # |
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Can you explain me why I haven't got any rating after participating in the contest and solving 3 problems there?

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3 years ago, # |
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This F problem is not as difficult as F problem. Original two-dimensional problem can be very very easy converted to one-dimensional.

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3 years ago, # |
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Hello MikeMirzayanov,BledDest,vovuh,adedalic.System

I find out some plagiarism activity in Codeforces Round 786 (Div. 3).

Please check out submissions of problem 1674B - Dictionary,submissions are 155621247 and 155678427 They both have same function with same variable but just comment down it and wrote it again with some variable change. Please look out at it.

Proofs given below and also you can check submissions...