Блог пользователя Um_nik

Автор Um_nik, история, 5 лет назад, По-английски

Hello!

I'm glad to invite you all to Round 576 which will take place on Jul/30/2019 17:35 (Moscow time).

There will be 6 problem in both divisions.

Round is based on Team Olympiad in Computer Science Summer School. It is (yet another) summer school for schoolchildren organized by Higher School of Economics and "Strategy" Center in Lipetsk. Almost all the problems are authored and prepared by teachers and teaching assistants in CSSS: Um_nik, Burunduk1, fake123_loves_me, MakArtKar, Villen3tenmerth, Aphanasiy, Gadget. One of the problems is authored by Merkurev (just because we are friends :) ). One more problem for the round was added by KAN.

I would like to thank KAN for CF round coordination, I_love_Tanya_Romanova, Merkurev, Rox and 74TrAkToR for testing, and Codeforces and Polygon team for these beautiful platforms.

Scoring will be announced.

Upd: We added one more problem to div.1 contest, now both contests have 6 problems (4 in common). The round is not combined, if it were, I would write "combined" in the title.

Scoring distribution:
div2: 500-750-1250-1750-2500-3000
div1: 500-750-1250-1500-1750-2250

Congratulations to our winners!
div.1:
1. Radewoosh
2. tourist
3. mnbvmar
4. Benq
5. pashka
div.2:
1. ChthollyNotaSeniorious
2. Honour_34
3. idxcalcal
4. shogunator
5. yijan

Editorial won't be published.

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5 лет назад, # |
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Auto comment: topic has been updated by Um_nik (previous revision, new revision, compare).

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5 лет назад, # |
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Wow wide range of writers ( LGM to pupil )

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5 лет назад, # |
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Looking for a crazy contest :P

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5 лет назад, # |
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5 лет назад, # |
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Will there be stories about Um_nik's teal hair?

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5 лет назад, # |
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Why You forgot to thank specially to MikeMirzayanov for Codeforces and polygon platform? He can block your account chinese!

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5 лет назад, # |
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Oh~,long time without a div1+2 conbined contest.

I was excited as soon as I saw this contest appear in the list of upcoming contests.

Hope the problem ststements will be as short as the announcement!\XD

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5 лет назад, # |
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I wanna get high rating.

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5 лет назад, # |
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points or plus ???

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5 лет назад, # |
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Anybody knows the difference among div.1+div2 and 2 separate rounds?

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    5 лет назад, # ^ |
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    Div.1 + Div.2 Round is One contest in which Div.1 participants and Div,2 participants participate altogether and 2 seperate rounds are literally Two contests in which participants participate in his own Division.

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5 лет назад, # |
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Why results of div1 and div2 is n

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5 лет назад, # |
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I think Um_nik has not much experience to make a good contest, so I will not write this contest.

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5 лет назад, # |
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Usually when one puts + between "Div. 1" and "Div. 2" it means a combined round, and names for posts for regular separate rounds don't contain a substring Div at all. You explained the separateness of the round in the update and wrote explicitly that each division has 6 problems (4 in common), but anyway I suggest not to use the + sign to denote separateness next time(s), otherwise someone will be confused.

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    5 лет назад, # ^ |
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    Yes, I understand the confusion. I didn't know that there is a fixed convention, sorry.

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5 лет назад, # |
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How long the contests are?

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5 лет назад, # |
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"We added one more problem to div.1 contest" <3

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5 лет назад, # |
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why start at 17:35, why not at 17:30?

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    5 лет назад, # ^ |
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    It may be farfetched, but (because) it's usual start time of the contest. So, is there a reason why the competition must start at 17:30?

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    5 лет назад, # ^ |
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    It's good to think that the contest starts at 17:30, because of the registration.

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5 лет назад, # |
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Is it rated?

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5 лет назад, # |
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hello :D I have a doubt. What is a hack?

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    5 лет назад, # ^ |
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    When you are able to prove some submission wrong by providing a clever test case whose output is wrong according to tester code/verification of output by the system. Hope it helps:D

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    5 лет назад, # ^ |
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    At codeforces , you can see the code written by others. So after you see others code which have passed the pretest, if you think there may be some data that his or her code may give the wrong answer, you can submit a piece of data ,and if his or code give the wrong answer , that means you hacks successfully, and you will get point, otherwise, you will loss point.

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5 лет назад, # |
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gl & hf all

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5 лет назад, # |
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the number of registration in div.2 must be over than 10k before contest!

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5 лет назад, # |
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Разбалловка будет?

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5 лет назад, # |
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Scoring doesn't seem to add up for common problems. Interesting...

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5 лет назад, # |
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It would be a good contest excited to solve problems.

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5 лет назад, # |
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I can see there's nothing strange about the performance of HanwhaEagles in this contest.

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5 лет назад, # |
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After a long time you see so many full scores in Div 1 and hacks too, indicating "Author's mind is weak" Um_nik

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    5 лет назад, # ^ |
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    What is your problem with full scores?

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      5 лет назад, # ^ |
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      However hard I try, I can't get it right. My mind is weak. But seeing so many full scores makes me feel Author doesn't have hard/interesting questions.

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      5 лет назад, # ^ |
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      In general, it's not good to break ties between too many contestants at the top just based on time. Doesn't matter much in a CF round compared to World Finals of Serious Competition, though.

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5 лет назад, # |
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Where is tag "square round"? (576 = 24*24)

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5 лет назад, # |
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How to solve D

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5 лет назад, # |
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This contest made me sad

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5 лет назад, # |
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How to solve Div2D?

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    5 лет назад, # ^ |
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    Process queries in offline mode. Hint: we don't care what queries of type "2 x" there were before the last query "1 p x" for every p. We just need to take the maximum of the x in the last "1 p x" query for a certain p and all x's from queries of type "2 x" that occurred later. Suffix maximums may be calculated easily in O(n).

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5 лет назад, # |
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How to solve D?

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How to solve B :D

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    5 лет назад, # ^ |
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    l^2 — h^2 = 2hx. Find x. Because length of segment A-top of flower remains same.

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    5 лет назад, # ^ |
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    I think this is an isosceles triangle :3

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      5 лет назад, # ^ |
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      Which triangle is an isosceles?

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        5 лет назад, # ^ |
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        If you connect the line from two flowers like a drawing for you to create a triangle. Calling the three vertices of that triangle respectively A, B, C with B and C are respectively flowers. I noticed that triangle ABC is an isosceles triangle

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    5 лет назад, # ^ |
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    A^2 + L^2 = (H+A)^2

    Use binary search to find A. For me binsearch is simpler than separating A in that formular.

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5 лет назад, # |
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i think E can be solve by blossom algorithm,

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    5 лет назад, # ^ |
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    but i dont know how to implement in O(V sqrt E). Can you give me some link, please?

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      5 лет назад, # ^ |
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      There is no way to find the maximum matching in general graph in $$$O(V\sqrt{E})$$$. As far as I know, the complexity of the fastest algorithm for this task is $$$O(VE)$$$.

      This is my solution for E (I cannot implement it in time :( ):

      1. Find a maximal matching $$$M$$$ in the given graph (a matching such that no more edge can be added to $$$M$$$).
      2. If $$$|M| \geq n$$$, then print out $$$M$$$. If $$$M < n$$$, then create a set of vertices $$$S$$$ that has all vertices that are not in $$$M$$$.

      I can prove that $$$S$$$ is an independent set we need to find.

      Suppose that $$$S$$$ is not an independent set. That means there are two vertices $$$u, v \in S$$$ such that there is an edge $$$(u,v)$$$ in the given graph. However, that also means we can add $$$(u,v)$$$ to the matching $$$M$$$ (contradiction). Therefore, $$$S$$$ must be an independent set.

      It is easy to see that $$$|S| > 3n - 2|M| > 3n - 2n = n$$$, therefore $$$S$$$ is an independent set we need to find.

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5 лет назад, # |
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is div2 E is a coloring problem(vertice coloring and edge coloring)?

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5 лет назад, # |
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DP state for Div2 F?

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    5 лет назад, # ^ |
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    Yes. It's in $$$O\left(n^5\right)$$$. $$$f(x1,y1,x2,y2)$$$ mean the answer to the range $$$(x1,y1)$$$ to $$$(x2,y2)$$$.Use memorized search to maintain the answer.

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      5 лет назад, # ^ |
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      Can you please explain how O(n^5) is passing? If we directly calculate then it's 3*10^8 computations. The more complicated calculation would be — (50*(51)/2)^2 (states) * 2 * ~25 (dividing each rectangle into two rectangles, horizontally and vertically.

      This is easily crossing 10^7 computations and almost touching 10^8. Generally, 1 second means less than 10^7 computations, right? (Help me out with that please, could never get a proper idea)

      With reference to my submission — F

      On a side note, hot to make comments look pretty? I know there's bold and italics but for example, how to use superscript?

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5 лет назад, # |
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I was in despair until I found that $$$O(n^5)$$$ solution can pass 2F/1D.

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5 лет назад, # |
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5 лет назад, # |
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Div1C одна из лучших задач за последнее время, спасибо автору.

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5 лет назад, # |
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Will the random solution in Div1 F fst?

58029804

Hope it won't. :D

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5 лет назад, # |
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how to solve D, I wrote some stupid n^6 dp and thought it would pass, of course it did not pass

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    5 лет назад, # ^ |
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    Observation 1: An optimal solution exists by only painting squares

    Observation 2: An optimal solution exists such that no two painting regions intersect, otherwise we can merge them into a bigger square.

    Thus we can do a naive dp using submatrix as states, divide horizontally or vertically for transferring.

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      5 лет назад, # ^ |
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      Can you explain observation 1 a little bit please?

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        5 лет назад, # ^ |
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        If I paint a rectangle with size $$$w\times h(w>h)$$$, it costs $$$w$$$, if I paint $$$w\times w$$$ instead, it still costs $$$w$$$ and it won't be worse.

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        5 лет назад, # ^ |
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        If we paint a rectangle of size $$$n\times m$$$ where $$$n<m$$$, we can expand it to an $$$m\times m$$$ square which only makes the solution better.

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          5 лет назад, # ^ |
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          can you prove your observation using another argument like for a given construction can you provide a way to convert a rectangle submatrix to some subset of square submatrices while maintaining the same cost . i can't seem to prove both of your observations simultaneously using your argument , suppose i give you an optimal construction then i can choose larger side of a rectangle as a square and the total cost remains same and your observation1 is true but then we may have intersection between two regions and observation2 is not valid (I think we need to prove your observations in some other way ).

          Also you didn't use any of those observations in your code for div1D , main point was that we could divide horizantally or vertically a given submatrix , can you prove why this strategy gives optimal solution ?

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            5 лет назад, # ^ |
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            otherwise we can merge them into a bigger square.

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              5 лет назад, # ^ |
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              ohk i understand it now(skipped that part :p).

              can you prove the main part, why dividing horizantally/vertically while transferring is optimal.

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                5 лет назад, # ^ |
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                An optimal solution exists such that no two painting regions intersect Therefore it must be splittable.

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                  5 лет назад, # ^ |
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                  nice, i was trying to draw some counter example's but seem's i can't avoid both a horizantal cut & vertical cut simultaneously (except the whole rectangle ) inside a rectangle , such a strong statement.

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                  5 лет назад, # ^ |
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                  This conclusion is wrong, consider the following painting regions on a $$$9 \times 9$$$ board:

                  $$$\text{1 3 3 5}\\ \text{3 7 5 9}\\ \text{7 5 9 7}\\ \text{5 1 7 3}$$$

                  What you actually need is that if the optimal solution is not splittable, then the projections of the regions onto the longer side cover the longer side, so the cost is at least its length, which is easily achievable (just take the whole grid as a painting region).

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    5 лет назад, # ^ |
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    $$$f[x1][y1][x2][y2]$$$ denotes... Well, you know.

    When it is all white, it is $$$0$$$.

    $$$f[x1][y1][x2][y2]=\min(f[x1][y1][x2][i]+f[x1][i+1][x2][y2])$$$ (Split it vertically)

    (Split it horizonally too)

    And paint a square with length $$$\min(x2-x1+1,y2-y1+1)$$$, then split it to two smaller rectangles.

    It is $$$O(n^5)$$$ with small constant.

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      5 лет назад, # ^ |
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      What does "... Well, you know" mean?

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        5 лет назад, # ^ |
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        The minimum cost to paint the rectangle with left lower corner $$$(x1,y1)$$$ and right upper corner $$$(x2,y2)$$$ to all white.

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      5 лет назад, # ^ |
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      I think it should be max(x2-x1+1, y2-y1+1).

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        5 лет назад, # ^ |
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        Well, actually my solution is a little different to other solutions (including official solution). You mean using a square with length $$$\max(x2-x1+1,y2-y1+1)$$$ to paint all the rectangle white, while I mean using a square with length $$$\min(x2-x1+1,y2-y1+1)$$$ to paint only a part of the rectangle.

        The following picture describes my idea.

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    5 лет назад, # ^ |
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    got AC with n^6 dp (code). Can someone hack my solution?

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5 лет назад, # |
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how to solve div2 C?

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    5 лет назад, # ^ |
    Rev. 5   Проголосовать: нравится +4 Проголосовать: не нравится

    You can have at most 2^k distinct elements, where k is (8 * I / n).

    If there are m distinct elements in the sound file, you need to reduce d = m - 2^k elements to fit onto the disk.

    Let cnts be an array that:

    index: nth small element

    value: how many times it appears

    (We can use std::map to obtain such an array: counting with std::map, iterate over the map, collect its value part.)

    For i in [0..d], we change all the smallest i elements and the largest d - i elements, count how many element we changed, and the final answer is the smallest count. (That is, sum(cnts[1..i]) + sum(cnts[(m-(d-i)+1)..m])) (We can use prefix sum array to calculate the sum)

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5 лет назад, # |
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A is the hardest problem among ABCDE :(

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5 лет назад, # |
Rev. 2   Проголосовать: нравится -27 Проголосовать: не нравится

B is totally dump.

Maybe I will never write Um_nik contest anymore.

Now I am upset enough to give some bad words, but I just can't.

The last time: B is dump.

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5 лет назад, # |
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Wow, another implementationforces round, so cool (no)(2).

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5 лет назад, # |
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In Div1E if the grid was small enough, I believe it could be solved by bipartite matching(with the observation that it is always worth to choose an entire line or column). Is there a way to "normalize" the big bipartite graph and transform the task into flow problem or something similar?

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    5 лет назад, # ^ |
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    You can compress the coordinates to get a grid that is essentially 100 by 100. The observation is that only the boundaries of the rectangles matter.

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5 лет назад, # |
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Div1C and Div1E are so super standard, especially E

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5 лет назад, # |
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how to solve div2 c??

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    5 лет назад, # ^ |
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    First you can note that the position of numbers doesn't matter. So think of a set of numbers (multiset) instead of array. It's because when we select $$$[l, r]$$$ it will "influence" all numbers.

    So now we have to make all numbers in our set in range $$$[l, r]$$$. It means that when we select some $$$[l , r]$$$ all numbers that are outside this range will change.

    So our answer for selected $$$[l, r]$$$ is $$$ans = cnt[1] + cnt[2] + ... + cnt[l-1] + cnt[r+1] + cnt[r+2] ...$$$, where $$$cnt[i]$$$ count of number $$$i$$$

    Now let's think what we have to do in our problem :D

    $$$ourTotalMemoryUsed = n*ceil(log2(K))$$$, where $$$K$$$ — number of distinct elements

    $$$maxMemory = 8*I$$$

    $$$overallMemoryUsed \le maxMemory$$$ (overall memory should be less equal then max memory)

    $$$n*ceil(log2(needK)) \le 8*I$$$

    $$$ceil(log2(needK)) \le 8*I/n$$$

    $$$needK \le 2^{(8*I/n)}$$$

    That means that our result set should have less or equal then $$$2^{8*I/n}$$$ distinct elements.

    Now let's select some $$$l$$$. Now we have to select such maximum $$$r$$$ that number of distinct elements in that range will be less or equal then $$$needK$$$ and our answer for selected $$$[l, r]$$$ will be $$$n - (r - l + 1)$$$

    Why? Because elements outside this range will change.

    Now how can we calculate it? Iterate through all elements in our sorted array. It will be our $$$l$$$. Then using 2 pointers we can find max $$$r$$$, that number of distinct in range $$$[l,r]$$$ will be less or equal then $$$needK$$$

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      5 лет назад, # ^ |
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      Could you take a look into my solution, it is what you said but do not pass 10th test :(

      https://codeforces.net/contest/1199/submission/58049925

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        5 лет назад, # ^ |
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        Increase R before entering your for loop.

        Otherwise you add cnt[b[R]] twice. Once in while, next time in for loop.

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          5 лет назад, # ^ |
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          aijey can you please tell me why the position of element doesn't matter ?

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            5 лет назад, # ^ |
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            Any range you select, you should check all elements, whether they are in range, and change them if not.

            Example: 2 4 1 5 2 3 5. [3, 4]

            Checking a[1] = 2 -> changing...

            Checking a[2] = 4 -> in range(don't change)

            Checking a[3] = 1 -> changing...

            Checking a[4] = 5 -> changing...

            Checking a[5] = 2 -> changing...

            Checking a[6] = 3 -> in range

            Checking a[7] = 5 -> changing...

            So ans = 4. You can see, that any way you shuffle the array, the answer for [3,4] will be always 5

            UPD: a[5] isn't in range

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              5 лет назад, # ^ |
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              But can you please tell me why a[5] = 2 is in range . By the way thanks actually I misinterpreted the question .

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What was hacking test for Div2 C/Div1 A ?

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5 лет назад, # |
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When can I resubmit? I have almost solve. I want test my new code.

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5 лет назад, # |
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Messed up problem A itself. Can somebody explain the logic now?

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    5 лет назад, # ^ |
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    Iterate for whole array :

    Run two loops :

    for(int j=i-x;j<i && j>=0;j++)

    for(int j=i+1;j<=i+y && j<n;j++)

    if in any of them a[j]<a[i] than index i wont be answer else that index is answer and return that index.

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5 лет назад, # |
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How to solve Div2-C?

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    5 лет назад, # ^ |
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    From disk size and number of samples you calculate the available bits per sample. This determines the biggest difference from lowest to highest possible storagable value.

    Then you find among the input samples the biggest possible count of samples with a difference less than or equal to the above value.

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      5 лет назад, # ^ |
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      The problem statement is easy to misread. The number of bits needed is calculated based on the number of distinct values, not the range of the values. So you need to find the optimal range to clamp the values such that the number of distinct values after the clamp is small enough.

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5 лет назад, # |
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Can someone please explain an online solution for DIV2-D?

mango_lassi can u please share your approach?

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    5 лет назад, # ^ |
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    What I tried to do was sort all people on the basis of the last change of type 1, with that change.Start iteration from last 2nd type change, and maintain 2 pointers, 1 on people array(sorted by last change), and other on the array holding 2nd type of events, maintain a maximum variable and update the final value of in-range people with maximum of (current value of that person, largest of x till now) and decrease pointers, continue....

    Edit:: Sorry didn't read "online" in your comment.

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    5 лет назад, # ^ |
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    build a lazy propagation segment tree where the lazy values represent the value x in type 2 updates.

    A type 1 update would need to push down the lazy values from the root to the leaf, and then update the array value at the leaf. A type 2 update just sets the lazy value at the root. To push a lazy value of a node to its children, you update the lazy value of the children to the maximum of the node's lazy value and the child's lazy value, then set the node's lazy value to some garbage value which is at most 0.

    When querying for the value of one point, we push down the lazy values from the root to the leaf, and max the array value with the lazy value at the leaf.

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5 лет назад, # |
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How's that even possible that O(n(x+y)) solution does not pass A? 58001554

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    5 лет назад, # ^ |
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    It's (s — x) in the 2nd loop i think, buddy! :D

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    5 лет назад, # ^ |
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    You wrote in line

    $$$ for $$$ $$$($$$ $$$int$$$ $$$j$$$ $$$=$$$ $$$s$$$ $$$-$$$ $$$1$$$; $$$j$$$ $$$>=$$$ $$$j$$$ $$$-$$$ $$$x$$$ && $$$j$$$ $$$>=$$$ $$$0$$$ $$$;$$$ $$$--j)$$$ j >= j — x instead s — x

    I think its the error

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5 лет назад, # |
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Any ideas why it's TLE xd? 58001816

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For Div. 2 D, anybody has ideas about what is the 4th test case? Please share it :D

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5 лет назад, # |
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Btw, F can't be solved in polynomial time for big integers.

Hint
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5 лет назад, # |
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Претесты слишком слабые

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5 лет назад, # |
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If Um_nik in panel AND you use Java, Run Virtual Contest at same time.

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5 лет назад, # |
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Div. 2 C is a bad problem

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5 лет назад, # |
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@Um_nik: Out of curiosity, question about F: If there is no prime that divides at least n-2 numbers, is it true that if there exists a solution then solution trying out completely random assignments will find one?

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    5 лет назад, # ^ |
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    I don't know, tests were prepared by Burunduk1

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    5 лет назад, # ^ |
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    random assignments

    for me it's not obvious, what "random assignments" are

    i tried out "random shuffle + greedy assignments", on my tests it finds answer during first 100 tries.

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      5 лет назад, # ^ |
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      I meant simplest possible way, no greedy or whatever, just i-th number belongs to first group with prob 1/2 (all events independent)

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        5 лет назад, # ^ |
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        Wow =) i even could not imagine so simple solution.
        It gets WA 46 with cutting by TL=0.5 seconds

        WA 46

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          5 лет назад, # ^ |
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          But does this test satisfy the assumption that no prime divides at least n-2 numbers? Because my solution that got accepted in contest 1) does some weird mumbo-jumbo, 2) tries some random assignments satisfying some.constraints determined by first part. However if there is no prime dividing >=n-2 numbers then my first part does nothing and second part is trying simplest possible random assignments as I described (and it got accepted as I said). It seemed fine to me as 3-edges are far less limiting than 2-edges and I still can't have a lot of them, but I didn't have any formal proof and was curious whether I am right.

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            5 лет назад, # ^ |
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            does this test satisfy the assumption that no prime divides at least n-2 numbers?

            Hm... No. Moreover, it seems, none of tests, where random assignments fails, satisfies the assumption.

            You may investigate tests more precisely

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5 лет назад, # |
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How to apply Segment tree in problem D?

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слабые тесты в последней (div 2)!

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5 лет назад, # |
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hackforces!

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Div.2 C can be solved using binary search.

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Find the bug in code. Spoiler in edit history :(

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5 лет назад, # |
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Can someone explain me how to use segment tree in problem D(div 2) please

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Just out of curiosity, can I get a count of how many solutions failed system tests for Div2C/Div1A?

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5 лет назад, # |
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nobody:

pretests:

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Somehow I didn't think $$$O(n^5)$$$ could fit the time limit for D and tried to come up with an $$$O(n^4)$$$ one. My idea is basically to only consider rectangle $$$(x_1, y_1, x_2, y_2)$$$ where $$$x_1 = 0$$$ or $$$y_1 = 0$$$. I don't know how to prove it formally though. Code

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5 лет назад, # |
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can we solve C by Binary search ?

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    5 лет назад, # ^ |
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    Yep! I solved it using binary search, first applying BS on the answer, which would be b/w 1 to cnt (cnt being the number of distinct numbers) and then another binary search nested in it to find the min number of changes needed to obtain that particular distinct element count. Here's a link to my solution div2C

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5 лет назад, # |
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Correct me if I am wrong,

But my Div2 D's solution passed the system test, without being right. Test Case:

input

4

1 2 3 4

2

2 1000

2 1

Output 1 2 3 4

Expected 1000 1000 1000 1000

Link to Code https://codeforces.net/contest/1199/submission/58024929

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    5 лет назад, # ^ |
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    Yeah, it is incorrect, mine gives 1000 1000 1000 1000. And i was expecting the pretests to be strong, jokes on me :P.

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Auto comment: topic has been updated by Um_nik (previous revision, new revision, compare).

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    5 лет назад, # ^ |
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    One of the submissions was during the contest and the other was after the contest. The test case (Test 109) which the first submission failed on was added after the contest as part of uphacking, but is not used to judge submissions in the contest.

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5 лет назад, # |
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Div2C:

If there are exactly K distinct values in the array, then we need k=⌈log2K⌉ bits to store each value.

As they talking about bits, I thought the distinct values are continuous (that's how bits work right?) . Am I the only one who faced this confusion?

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    5 лет назад, # ^ |
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    Even the question talks about the range:

    We choose two integers l≤r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
    

    I thought every number from l to r (both inclusive) are represented by one of the bits. So the distinct values are contiguous. Did anyone else feel it ambiguous?

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      5 лет назад, # ^ |
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      Um_nik can you please look into this? I felt the question statement was confusing.
      As it says about K distinct values, but then it says about a range [l, r]. So I thought we can represent values [l, r] (contiguous) using the 'k' bits. But it turns out that we can use 'k' bits for non-contiguous values as well.
      In that case why did they talk about a range [l, r] and had confused the statement?

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        5 лет назад, # ^ |
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        The range is necessary for the problem statement. If you can select arbitrary values to modify then you would greedily retain the K distinct values with the most occurrences.

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      5 лет назад, # ^ |
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      I thought the same thing. I should really read questions more carefully next time.

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      5 лет назад, # ^ |
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      If there are exactly K distinct values in the array, then we need k=⌈log2K⌉ bits to store each value.

      I was confused too. But after looking at it a second time, I think this means to compress with a dictionary.

      For example, if there are 3 different values 'A', 'D', 'K', and message M ['K', 'A', 'D'].

      'A' => 00

      'D' => 01

      'K' => 02

      M => [02, 00, 01]

      With a dictionary, values don't have to be continuous.

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    5 лет назад, # ^ |
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    No it just means that if you had distinct values a1, a2, a3, .... an then you can describe each of them with ceil(log2n) bits.

    Example: if you know that you are given values 11, 12, 15, 257 then you don't need 9 bits to hold 257. Instead of it you can mapping 11 to (00), 12 to (01), 15 to (10) and 257 to (11).

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    5 лет назад, # ^ |
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    I didn't completely understand the statment too.

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Is there anyone like me who didn't understand the problem statement of C?

If you already got AC,then please share your idea.

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5 лет назад, # |
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2D/1B solution

Let request time be it's order number.

Let's keep two arrays: the first one will hold our values and the second one will hold the time for the last first type request (1 p x). Let's name them values and last_upd.

Iterate over the requests.

If it's first type request (1 p x), then values[p] = x; last_upd[p] = request_time;

If it's second type request (2 x), then save it as pair (request_time, x) in the array second_type_requests.

Right now there are the last values in the array values, which were changed by the first type requests. Let's find out which second type request can change this value. It goes without saying that the request_time of the second type request should be more than the request_time of the first type request. If such request is not alone, we should choose the one with the biggest value.

We've reduced our task to finding maximum on suffix of the array second_type_requests. We can find it fast using the suffix_max_array.

Time complexity is O(N + Q)

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5 лет назад, # |
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Hi. Can anyone point out any optimization to improve my solution? https://codeforces.net/problemset/submission/1199/58040359

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    5 лет назад, # ^ |
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    Fast I/O makes a huge difference. You can check my submission and see my comments marking the optimizations at the beginning. Include those lines at the beginning of all of your submissions (or use scanf/printf) and your programs should run considerably faster.

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Can someone explain how to solve div1E? I realized that I need to somehow reduce the problem to flow, it is clear that you need to compress coordinate, but I'm apparently stupid and can't understand what kind of graph here you need to build.

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    5 лет назад, # ^ |
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    Because the cost function is min(h, w), you can assume that all your actions are of two types only: paint whole row in white or paint whole column in white. Than we can say, that every black cell needs to be painted either from the corresponding row, either from the corresponding column. Let's create a bipartite graph. One half will be for the x coordinates, another one for the y coordinates. First lets assume that we don't compress the coordinates. We will add a single edge from ith row to jth column for each cell (i, j) which is black. Now you need to find the minimum vertex cover of this graph, because at least one of row/column needs to be painted. When you compress coordinates, every vertex will correspond to the range of columns/rows, and because of it will have a weight equal to the length of this range. Finding the minimum weighted vertex cover can be deduced to finding minimum cut of the network, where you add edges from source to each vertex of the first half and add edges from each vertex of the second half to the sink with with the capacities equal to their weights, and keep the original edges of the bipartite graph with the infinite capacities.

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    5 лет назад, # ^ |
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    A similar version of this problem is "Given a $$$N \times N$$$ grid containing white and black cells. In each operation one can color all black cells in a row or a column white. Determine the minimum number of operations required to turn all black cells into white".

    The solution is to construct a bipartite graph of $$$N$$$ vertices on both sides, and for each black cells $$$(r, c)$$$ we connect the $$$r$$$-th vertex on the left side to the $$$c$$$-th vertex on the right side. The minimum vertex cover of the resulting bipartite graph is the answer since each black cells is destroyed by the operation corresponding to the vertex which covers that edge.

    As for Div1 E, a similar bipartite graph $$$G = (X, Y)$$$ can be constructed on the compressed coordinates. So the problem is to find the "minimum weighted vertex cover" on the graph, which is equivalent to the min-cut on the following flow graph:

    1. For each $$$x \in X$$$, connect $$$S \rightarrow x$$$ with capacity $$$w_x$$$.
    2. For each $$$y \in Y$$$, connect $$$y \rightarrow T$$$ with capacity $$$w_y$$$.
    3. For each edge $$$(x, y) \in G$$$, connect $$$x \rightarrow y$$$ with capacity $$$\infty$$$.

    See this article for detailed explanation.

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5 лет назад, # |
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Solution for Div2 D without using segment tree.

Observation:

  1. a later higher payoff will cancel all previous lower payoffs.

  2. a citizen's final balance is determined by the last change and the highest payoff after that.

We go through events in reverse.

Record a visited array for all citizen and the highest payoff.

On event 1 p x:

If the p-th citizen has been visited, then its balance has been changed later, pass.

else, mark it as visited, update its balance to max(highest_payoff, x)

On event 2 x:

update highest_payoff to max(highest_payoff, x)

It may be easier to understand by reading the code: 58038489

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5 лет назад, # |
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Я, конечно, не специалист, но Div1 B и C какие-то безыдейные

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5 лет назад, # |
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My submission
58042687 on Div1 C
and
58041401 on Div1 F
passed the system test.

Of course, they failed during the contest, but after I tried to do some voodoo programming, I finally managed to write some randomized sols that (I think) should not be passed, but they finally got accepted (almost TLE/WA).

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5 лет назад, # |
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An interesting thing, cf predictor says that cnnfls_csy should get 583 pts, whilst he got only 296. Why is there such a difference?

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5 лет назад, # |
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someone please recommend some other problems like Div2 D where we can get answer without using segment tree?

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5 лет назад, # |
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Weak pretest, random algorithm bypassed, flood of fst.

Good contest.

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5 лет назад, # |
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Is there an optional solution to solve Rectangle Painting 2 if cost is max(h, w) ?

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5 лет назад, # |
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FST

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5 лет назад, # |
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How to solve Div2-D without using segment tree?? (I tried to understand through comments but can't)

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5 лет назад, # |
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Can someone please help me on DIV2 problem C question. Let me know if my logic is correct:

first find critical k <= min(n, 2^((8*I)/n)). Then sort the given array, and convert the array to its freq array i.e. if original array is 1, 2, 2, 3, 3, 4: change it to 1, 2, 2, 1 (each element in this new array represents the frequency of distinct elements in original array). Now original task is to minimize the number of changed elements, that is now equivalent to finding k-subarray with maximum sum in new array (using sliding window method in linear time), and then return new array size — max sum found.

Am I missing something here? It is failing on 10th test case. Link to my soln:

https://codeforces.net/contest/1199/submission/58049910

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    5 лет назад, # ^ |
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    Your logic is correct, but there was a flaw in your computing the window sum. In maintaining the maximum seen so far, your code takes the max of the current window and the next one instead of updating the window sum. This could mean that sometimes you include elements more than once, and sometimes consider a group of non-contiguous elements.

    I've made a slight modification (here), and marked the place where I changed code for convenience :)

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      5 лет назад, # ^ |
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      Oh yes! So silly of me. Thank you so much for taking your time & debugging, and even writing down the correct code :). Much appreciated. Finally, submitted the correct code. ✌

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5 лет назад, # |
  Проголосовать: нравится +4 Проголосовать: не нравится

I need editorial, please :(((((

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5 лет назад, # |
Rev. 2   Проголосовать: нравится -37 Проголосовать: не нравится

Fuck D, E's author. Losing free point for E just because of min/max. They didn't even bold it.

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    5 лет назад, # ^ |
    Rev. 2   Проголосовать: нравится +59 Проголосовать: не нравится

    Once again please. You thought that we gave two exact same problems, you can't open your eyes, and it is my fault somehow?

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    5 лет назад, # ^ |
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    Afaik, if the problems were the same but with different constraints, they would be named D1 and D2 instead of D and E. Moreover, the author always mentions it before the round whether there is a problem with 2 subtasks.

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    5 лет назад, # ^ |
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    Turn out it’s not free point when you still need to be able to read, right?

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    5 лет назад, # ^ |
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    I agree that it should have been bolded. It's a small thing that would prevent many people from misreading.

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    5 лет назад, # ^ |
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    These problems have very small statements, how is it possible not to notice the difference? There are different examples and descriptions for them. Why don't people read the whole text before solving the problem? Isn't it their problem?

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      5 лет назад, # ^ |
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      fhgfhgfdgfbecueiyukcbsycuwcbuaipcbsdivdsnuwdilcgwyiulncsauciowh8vioer;vbudsivnwioeprnvisdosdc bywduiveybukbyvlsdnuiweciwopbh8pec9jaspch8vr8v9dsov7sdv8ogw78veg7v8erovhe7ovshd8v9s

      fhgfhgfdgfbecueiyukcbsycuwcbuaipcbsdivdsnuwdilcgwyiulncsauciowh8vioer;vbudsivnwioeprnvisdosdc bywduiveybukbyvlsdnuiweciwopbh8pec9jaspchBvr8v9dsov7sdv8ogw78veg7v8erovhe7ovshd8v9s

      These two strings are definitely shorter than these problems statements, how is it possible not to notice the difference?

      Of course two very similarly looking statements could bring the impression that they are the same and we need to fight that feeling for like 1-2 minutes before actually finding the significant difference, but we could have been saved from that by just either bolding min/max or writing a short note "the main difference is different cost function".

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5 лет назад, # |
  Проголосовать: нравится +9 Проголосовать: не нравится

Some SegmentTree solution of Div2D shoule be hacked like this
https://codeforc.es/contest/1199/hacks/575652/test

why it passed system test?????

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    5 лет назад, # ^ |
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    Can you explain why it will break ST solutions

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      5 лет назад, # ^ |
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      I use SegmentTree by a method different from others.I maintain the interval minimum and the interval maximum and a tag which notes whether the interval has been modifyed.

      When it comes to Case 2 ,if the interval minimum >= x then return immediately , if the interval maximum < x then update tag[p] , else update its subintervals

      When it comes to my hack case it will update all the subintervals in a enquiry .This will make it TLE

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    5 лет назад, # ^ |
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    UPD: Hacking~

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    5 лет назад, # ^ |
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    I have no idea how others used ST solve this. I used a segment tree with lazy propagation. There is no case I can think of for which my solution will fail. Can you give a hack for my solution?

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5 лет назад, # |
  Проголосовать: нравится +3 Проголосовать: не нравится

Weak system tests. A lot of accepted Div2 C solutions give incorrect results for

9 1
1 2 2 2 2 2 2 2 2
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5 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Getting TLE. Can anyone check which test case is taking up time or can possibly give wrong answer?

58060352 Thanks!

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5 лет назад, # |
Rev. 2   Проголосовать: нравится -6 Проголосовать: не нравится

thanks to div1b's author. through this problem I found my segment tree is wrong, but it passed all the systests. what a bad contest

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5 лет назад, # |
Rev. 2   Проголосовать: нравится +8 Проголосовать: не нравится

Div 1F : Why the time limit have to be 0.5s ? I was scared that the naive $$$O(n * \sqrt(10 ^ 9))$$$ factorization can't pass so I implemented Polard's Rho and it's got TLE.

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    5 лет назад, # ^ |
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    It's probably set that low to disallow us from factorizing all the numbers. :P

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      5 лет назад, # ^ |
      Rev. 3   Проголосовать: нравится 0 Проголосовать: не нравится

      But if I find all $$$O(40000 / log(40000))$$$ primes in range [1, 4000] then I can factorize all the numbers. My code : 58053947

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      5 лет назад, # ^ |
      Rev. 4   Проголосовать: нравится +20 Проголосовать: не нравится

      SQUFOF still works: 58085119

      Edit: Some other part of the previous solution 58078132 has been hacked, but the fast factorization still stands.

      Edit2: Updated solution with better randomization.

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        5 лет назад, # ^ |
          Проголосовать: нравится +66 Проголосовать: не нравится

        Looking at the submissions above, I now think that they set the 0.5s time limit because [reasons], and they didn't think how fast the factorization can be. (Maybe they wrote a naive $$$O(\sqrt{n})$$$ factorization only and called it a day, who knows.)

        Personally, I really dislike tasks like F (some number theory stuff, you have to invent some construction, it's really hard to create a good test against any dumb heuristic, and some uninsightful 20-liners pass the systests). Future problem setters, please don't be afraid to drop/modify a task if you feel it's nearly impossible to make nice tests for it.

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        5 лет назад, # ^ |
          Проголосовать: нравится +38 Проголосовать: не нравится

        Yeah man, nice factorization, thank you; however, it's not enough to consider only 2-edges, sorry

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    5 лет назад, # ^ |
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    During the contest I implemented naive factorization and it was way too slow to pass (2.7s on my laptop which is significantly faster than cf). So I just ignored prime factors larger than 1000 xD. Seems like a completely foolish thing to do, but I actually had arguments why that should not change much performance of my code.

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5 лет назад, # |
Rev. 3   Проголосовать: нравится 0 Проголосовать: не нравится

My algorithm idea for Div1A/Div2C seems not working. Can someone give me counterexamples or i'm missing something in my implementation?

  1. Number of distinct elements in answer must be K = 2^(8*I/n);
  2. Calculate number of distinct elements in original array, let it be cnt;
  3. Calculate frequencies of elements and sort in ascending order;
  4. Answer = sum of (cnt — K) minimum frequencies.

Code implementation 58059706 — Python and 58039964 — C++ both fails with WA on test 11.

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5 лет назад, # |
  Проголосовать: нравится +1 Проголосовать: не нравится

can someone please explain how to solve DIV1-F?

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    5 лет назад, # ^ |
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    Yes, someone definitely can explain it. Not me, I haven't tried the problems yet.

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5 лет назад, # |
  Проголосовать: нравится +39 Проголосовать: не нравится

I didn't expect that a tester's or author's solution uses unordered_map...

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5 лет назад, # |
  Проголосовать: нравится +11 Проголосовать: не нравится

Why 400k numbers and 1s in Div1A?

Your solution is definitely O(n * log n) as well. Did you try to reject O(n * log^2 n), or what?

In my view, you only made it worse for those writing in Python/Java/Kotlin. My reasonable code in Kotlin with groupingBy() doesn't fit into 1s.

It would be really good if you thought about "Python etc." people when setting limits.

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    5 лет назад, # ^ |
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    I think that "Python etc." people don't deserve high TL. You can choose any language you want, Codeforces even supports C++. If you choose Python, you basically want to get TL.

    200k wasn't enough to cut out $$$O(n^{2})$$$ with pragmas in C++.

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      5 лет назад, # ^ |
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      Do you think it is still a bad idea to keep Python users in mind even when we are talking about beginner level problems (like div1A)?

      I also agree with "you can choose any language" part, but I think that discouraging beginners this way isn't beneficial for competitive programming community development and growth.

      I wonder if your point here is that we should care about it at all, or that div1A is already above the level at which we should care, or that it isn't actually beneficial for community (as everybody should realize the part about Python as early as possible)?

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        5 лет назад, # ^ |
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        I think that it is good when Python solutions are accepted for easier problems, but in no way it should be a decisive factor for choosing problems for the round

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5 лет назад, # |
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I submitted the following code for problem C(div 2) . The verdict provided was wrong answer on test case 2 while on my compiler it's returning the correct answer . Can anybody suggest anything? 58028994

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    5 лет назад, # ^ |
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    It seems like the issue was in your binary search. I modified it so the initial value for end was $$$n - 1 $$$ instead of $$$n$$$, and had it return a large value ($$$10^{9}$$$) if it hadn't returned anything during the search. You can see my changes marked by comments in the code.

    Without the modification to return $$$10^{9}$$$, it would not return anything, and without the modification of the initial value of 'end', it would access the vector distict out of bounds. I'm not sure what happened with your compiler, but your program may have just been lucky (or unlucky, in your case) and your compiler's C++ implementation allowed the function to return a value that led to the correct answer. Likely something with differing memory handling.

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5 лет назад, # |
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