998batrr's blog

IZhO 2021
By 998batrr, history, 4 years ago, In English

First: We apologize for any inconvenience caused. On the second day, we'll try to correct all today's mistakes.

We still hope that you enjoyed the contest.

You can upsolve here

Let's discuss the problems.

Upd: Day 1 unofficial results here

Upd: Day 1+2 unofficial results here

Upd: Day 2 now available to upsolving.

Upd: Editorials.

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4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Can you say how to solve A?

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    4 years ago, # ^ |
      Vote: I like it +63 Vote: I do not like it

    Let's assume that we are at point $$$p$$$ now. We have two options.

    1) Go to $$$t$$$ directly, paying extra $$$|p - t| * a_p$$$ money;

    2) Go to another "optimal" point. There are two choices for that. Either it is minimum $$$r$$$ such that $$$a_r < a_p$$$ and $$$r > p$$$, or maximum $$$l$$$ such that $$$a_l < a_p$$$ and $$$l < p$$$.

    So run Dijkstra, maintaining minimum distance. Update answer by the first option. Or go to the next optimal point paying $$$a_v * dist$$$ ($$$r - p$$$ or $$$p - l$$$) money. We are multiplying $$$a_v$$$ because each time we are going to the index with a lower value.

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      4 years ago, # ^ |
        Vote: I like it +46 Vote: I do not like it

      I also did djikstra, but there is no need for it. The graph you get is a DAG, so simple dp is enough.

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    4 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    If I'm not wrong, it can also be solved by the Convex hull trick. The naive solution will be using DP. Let's say $$$dp_i$$$ is the minimum time needed to reach the point $$$i$$$. Recurrent relation will look like $$$dp_i = dp_j + |i - j| \cdot a_j$$$. It is easy to notice that the next point's value will be always less than the current's (i.e $$$a_{cur} > a_{next}$$$). So the $$$a's$$$ will be decreasing.

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +22 Vote: I do not like it

      I also solved it with a convex hull trick. I used two convex hulls. One for queries for the left (i < j) decreasing "ladder" from the start, the other for the right (i > j).

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4 years ago, # |
  Vote: I like it +25 Vote: I do not like it

Can you say how to solve B?

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    4 years ago, # ^ |
    ← Rev. 5   Vote: I like it +10 Vote: I do not like it

    To solve in n^2logn, we can notice that actually if we reassing our element from L to R, their id to from 1 to R — L + 1, it is easy to notice that actually after each separation their indexes goes left shift. So having M different groups, we can answer the sum with the formula. Then we can notice that 1 to R — L + 1, does not matter and we can just add their initial ID. So with that we can do MO in Nsqrt(N)log(N), but it will also be TL. Then we can notice that it is enough to keep the last Position where such division starts, so just by fixing R we can answer L easily with that information. My realization was with Trie and Fenwich. Nlog^2

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      4 years ago, # ^ |
      ← Rev. 3   Vote: I like it +29 Vote: I do not like it

      I Don't understand what you mean by "such division starts". I had trie and Mo's algo but I still got wa-s(and also TLE :P). Here's what I did, it's probably the same thing. You are splitting the array into multiple subarrays based on prefixes in base 2. For 2 equal values you are trying to get to the point where the prefixes of the indexes differ. Another thing is that if A and B have a common prefix of some length, A — X and B — X will also have a common prefix of the same length. In my VALMAX trie I added indexes from the original array(thanks to the observation it still works) based on the value of respective index. I counted all the nodes that have at least 2 paths going through them, because that means you have to split up to that point(maybe even further down). I also had an unordered map which included all the points in which I have to split. From multiple values would've collided. It still got WA, so it's probably wrong.

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4 years ago, # |
  Vote: I like it +45 Vote: I do not like it

Can you say how to solve C?

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4 years ago, # |
  Vote: I like it +63 Vote: I do not like it

Can you say how to solve D?

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4 years ago, # |
← Rev. 2   Vote: I like it +52 Vote: I do not like it

Is the second problem reference to this post?

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    4 years ago, # ^ |
      Vote: I like it +54 Vote: I do not like it

    Yes. To meme and author

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      4 years ago, # ^ |
        Vote: I like it +22 Vote: I do not like it

      Wowowowow, it's extremely cool, thanks :D :D :D

      Is Aizhan a real person?

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        4 years ago, # ^ |
          Vote: I like it +25 Vote: I do not like it

        Is that meme a real life situation?

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          4 years ago, # ^ |
            Vote: I like it +22 Vote: I do not like it

          DavitMarg foresaw that he wasn't going to win a medal, so he decided to bring potatoes instead. He didn't bring potatoes either, though :D

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4 years ago, # |
← Rev. 2   Vote: I like it +69 Vote: I do not like it

Will be official results published today?

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4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Are all the participants official in the standings?

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4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Lol who was responsible for the 6th subtask of 3rd problem? It genuinely costed me 20 minutes :P

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    4 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    somehow, i had "ai = i; bi = i + 1" from the start of the contest, or i accidentally reloaded statements. im not sure

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4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

How to solve day2 A for n <= 30 and n <= 2000?

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4 years ago, # |
  Vote: I like it +66 Vote: I do not like it

Do you know the cutoff for the medalists?

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    4 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Or, at least, what it was (roughly) in the past years?

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      4 years ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      18 is gold

      43 is silver

      80 is bronze

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        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Okay but how many participants were there?

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          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Something around 290

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            4 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            There were 295 this year, yes. But how many were there in 2020, I heard somewhere that the no. of participants became quite bigger this year, right?

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Dies from cringe(

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4 years ago, # |
← Rev. 2   Vote: I like it -28 Vote: I do not like it

Why its not working in C on n <= 1e3?

Code
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    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Put your code in a spoiler, please.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Explain your idea

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      So. Let's for every vertex i and L border find max R, where in F(S[l, r]) no contains i. So clear, that with increasing L, the R monotonically increasing. So let's do this with 2 itterators, and check this. Condition is met, If there is a vertex in the subtree and depth LCA set <= d[i]. And i do this

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Или более человечески по-русски. Для каждой вершины найдем все моменты времени, когда она находится в этом "обтянутом дереве". Для этого для каждой левой границы найдем минимальную правую границу, для которой это выполняется (понятно, что тогда выполняется на всем суффиксе). И что правая граница не убывает. Тогда будем это делать двумя указателями, а проверкой будет следующее условие: в поддереве этой вершины кто-то есть, и LCA этого множества выше, то бишь его глубина меньше, что, как мне кажется, у меня написано

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        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        ты писал стресс? мб поможет

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          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Мне было лень( И я ручками 1.5 часа дебажил...

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            4 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            Имхо дебагать полтора часа +25, когда у тебя 0 и при этом по задаче с прочтения набирается 30 баллов — не очень хорошая идея :/

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              4 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              Ахаххахаха, я условие неправильно понял, тут было без шансов)

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                4 years ago, # ^ |
                  Vote: I like it 0 Vote: I do not like it

                I don't speak russian, but it with google translate it seems to me like you found the mistake yourself. Good job?

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                  4 years ago, # ^ |
                    Vote: I like it 0 Vote: I do not like it

                  Yes, im find mistake, wrong read statement)

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    4 years ago, # ^ |
      Vote: I like it +24 Vote: I do not like it

    Nice code, everyone in Novosibirsk writes so beautifully?

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4 years ago, # |
  Vote: I like it +25 Vote: I do not like it

Will there be editorial or author's solutions?

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4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Day 2 upsolve?

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Isn't there an easier solution for day 2 B (rooms)?

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