AndreyPavlov's blog

By AndreyPavlov, history, 23 months ago, translation, In English

Hello, Codeforces!

IzhtskiyTimofey, qualdoom and I are glad to invite everyone to participate in Codeforces Round 846 (Div. 2), which will take place in Jan/25/2023 17:35 (Moscow time).

This round will be rated for the participants with rating lower than 0x834 (i.e. 2100). Participants with a higher rating can take part in the round unofficially.

You will have 7 tasks and 2 hours to solve them.

One of the problems will be interactive. Make sure to read this blog and familiarize yourself with these types of problems before the round!

I want to sincerely thank everyone who provided invaluable help in preparing the round and made it many times better:

This is our first official round on Codeforces. We sincerely hope to your participation. We hope that you will like the proposed tasks!

The score will be announced closer to the start of the round.

We wish you good luck and have a good time! See you in the round!

UPD: Scoring distribution: $$$500-1000-1250-1500-1750-2000-2500$$$

UPD: Round is unrated. We're sorry — it's our fault.

UPD: Tutorial and comment about task C Once again, we apologize for the inconvenience caused.

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23 months ago, # |
  Vote: I like it +3 Vote: I do not like it

It clashes with codechef starters 75 https://www.codechef.com/START75?itm_medium=hpbanner_1&itm_campaign=START75. Is it possible to change the time ? Thanks

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    23 months ago, # ^ |
      Vote: I like it +174 Vote: I do not like it

    You should be knowing that Codeforces >>>>>> Codechef.

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      23 months ago, # ^ |
        Vote: I like it +63 Vote: I do not like it

      Codechef tasks can be pretty good too, at least the ones at the end. Well, clashes are inevitable so just upsolve the contest you decide to skip.

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        23 months ago, # ^ |
        Rev. 3   Vote: I like it +10 Vote: I do not like it

        their plagiarism checker is extremely bad... their community support is really bad...

        I was plagiarised for one contest, in which I solved zero problem, and tried to solve one problem 16 times,,, and still I was plagiarised...

        How can you plagiarise someone, who solved zero problems and tried to solve the problem 16 times !!!!!!...

        if you copy code from someone, why wouldn't u get it right ???

        https://discuss.codechef.com/t/successful-plagiarism/104943/2

        UPDATE : I received response from codechef moderator regarding the plagiarism. According to them, I had solved 2 problems in contest and got plagiarised on 3rd problem.

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      23 months ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Yup, it always have been Sir. Looking forward to it. will upsolve last 3 questions of codechef (they are worth it though).

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      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Oh no that is an outdated statement. These days codechef problems are really really good.

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it +3 Vote: I do not like it

    Well, Codechef postponed it, but now they shouldn't have.

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23 months ago, # |
  Vote: I like it +20 Vote: I do not like it

As a tester, the tasks are quite interesting and the statements are clear.

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    23 months ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    What a joke is this round ? Unrated....

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      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Why C cant be solved in given constraints? What's the problem??

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        23 months ago, # ^ |
          Vote: I like it +9 Vote: I do not like it

        Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains

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    23 months ago, # ^ |
      Vote: I like it +118 Vote: I do not like it

    Sorry about my shit testing(((

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      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Git gud LOL, no harm intended!

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      23 months ago, # ^ |
      Rev. 2   Vote: I like it +5 Vote: I do not like it

      It feels weird how this issue wasn't caught by so many people during the setting and testing phase.

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    23 months ago, # ^ |
      Vote: I like it -23 Vote: I do not like it

    i am happy because if this round was rated i would get minus :D

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      23 months ago, # ^ |
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      same, i barely made it to 1300 last round

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        23 months ago, # ^ |
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        how many questions you solved on this round for now

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          23 months ago, # ^ |
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          one completely, but i got stuck on some edge case on second one, not even trying now.

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    23 months ago, # ^ |
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    if this contest rated,@huanghaoxiang will get 1400+ rating!

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    23 months ago, # ^ |
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    As a clown**

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23 months ago, # |
  Vote: I like it -36 Vote: I do not like it

Tester is me

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23 months ago, # |
  Vote: I like it +31 Vote: I do not like it

As a tester I can say that I am a tester

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

clashing with codechef, it would be great if timing is changed

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23 months ago, # |
  Vote: I like it +6 Vote: I do not like it

I Think it's Bitmask Round , I hope it is a misconception

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    23 months ago, # ^ |
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    I hope it is not (I like those type of problems)

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23 months ago, # |
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almost OrangeForces lol

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23 months ago, # |
  Vote: I like it +23 Vote: I do not like it

Codeforces round is not clashing with codechef round. Codechef is clashing with codeforces.

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23 months ago, # |
  Vote: I like it +47 Vote: I do not like it
Meanwhile Codeforces Lovers
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23 months ago, # |
Rev. 2   Vote: I like it +1 Vote: I do not like it

I take my words back ;(
DISAPPOINTED

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

omg orange round

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23 months ago, # |
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Hoping to solve till D in this round.

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23 months ago, # |
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wish every contester good luck and happy rating++ !

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23 months ago, # |
  Vote: I like it +9 Vote: I do not like it

"One of the problems will be interactive."I think it will be "D".

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Masters' Round!

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23 months ago, # |
Rev. 2   Vote: I like it -7 Vote: I do not like it

Will the rating update of this round before educational #142?

Update: Now the rating has been updated.

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What should I do To become Specialist

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23 months ago, # |
  Vote: I like it +17 Vote: I do not like it

hope i can solve problem C,so that i can change a color 。 i dont like green

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    23 months ago, # ^ |
      Vote: I like it +73 Vote: I do not like it

    this didn't age well

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it +5 Vote: I do not like it

    Fortunately, I solved C in the last thirty minutes, but there was some wrong with C。IS i solve C?(cry)

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23 months ago, # |
  Vote: I like it +7 Vote: I do not like it

IS THAT A JOJO REFERENCE??????!!!!!!!!!!11!1!1!1!1!1!1!

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23 months ago, # |
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Unrated?

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why will this round be unrated?

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The solution of the Problem C which was solved by the author is wrong.

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23 months ago, # |
  Vote: I like it +3 Vote: I do not like it

I was off to a great start, and then they make the round unrated :)))))

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    so C is unsolvable or what?

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      23 months ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      Idk man, I just used a simple greedy which did pass the protests, but greedy algorithms are hard to prove

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        23 months ago, # ^ |
          Vote: I like it +26 Vote: I do not like it

        Greedy is wrong.

        Consider following example: 102 people like dish 1, 104 people like dish 2.

        Tables are: 51, 51, 26, 26, 26, 26

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          23 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Pretests must be weak as shit lmao

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            23 months ago, # ^ |
              Vote: I like it +1 Vote: I do not like it

            I think even setters must have thought greedy was right

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            23 months ago, # ^ |
              Vote: I like it +4 Vote: I do not like it

            It's not that the pretests are weak (the round wouldn't have been made unrated if that was the reason), but the problem setters probably didn't realize that the greedy solution is actually incorrect.

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          23 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          is the answer not 179?

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          23 months ago, # ^ |
            Vote: I like it +1 Vote: I do not like it

          The answer should be 206, right? I think that's what my solution would give?

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        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        If your strategy is to assign biggest group of people to biggest table then this approach fails on this test:

        1 9 3 1 1 1 1 1 2 2 2 2 4 3 2

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          23 months ago, # ^ |
          Rev. 2   Vote: I like it 0 Vote: I do not like it

          answer is 8? edit. got it, 9

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            23 months ago, # ^ |
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            It is 9.

            Assign 1's to the tables of size 3 and 2, 2's to table of size 4.

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          23 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          My strategy was assigning the smallest that fits the table. That would still work?

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            23 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            1

            11 3

            1 1 1 1 1 1 2 2 2 2 2

            5 3 3

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              23 months ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              11

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                23 months ago, # ^ |
                  Vote: I like it 0 Vote: I do not like it

                Explain your algorithm more deeply since I have misunderstood smth.

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                  23 months ago, # ^ |
                  Rev. 2   Vote: I like it 0 Vote: I do not like it

                  Maintain a (multi)set with the group sizes. For each table (sorted in descending order), find the smallest group that has at least that size, and put on the table, and adjust the group size in the set. If none exist, put the largest among the set.

                  EDIT: I didn't prove correctness, but I tried to anticipate the problem with the direct greedy approach

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                  23 months ago, # ^ |
                    Vote: I like it +1 Vote: I do not like it

                  your solution fails on this test case: 1 10 4 1 1 1 1 1 1 2 2 2 2 3 3 2 2

                  answer should be 10

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                  23 months ago, # ^ |
                    Vote: I like it 0 Vote: I do not like it

                  1

                  11 4

                  1 1 1 1 1 1 2 2 2 2 2

                  5 3 2 1

                  Ans: 11

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                  23 months ago, # ^ |
                    Vote: I like it 0 Vote: I do not like it

                  Yes, you're right; I get 9

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                  23 months ago, # ^ |
                  Rev. 2   Vote: I like it 0 Vote: I do not like it

                  My code worked too. It gave me result 11

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        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Hello sir, can u tell me ur greedy algorithm?

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Same

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23 months ago, # |
Rev. 2   Vote: I like it -20 Vote: I do not like it

Unrated??.. First time solved 3 questions in 40mins

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

"Unranted"

what an absolute bruh moment

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23 months ago, # |
  Vote: I like it +23 Vote: I do not like it

Solved A+B+C in 26mins , thought I would finally become a specialist :")

But turns out the round is unrated. Sad :(

Anyways, Nice problems , thanks to the authors <3 !

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Hello sir can u tell me ur solution for C? I am curious and ur help will be greatly appreciated. Thank you.

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      23 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      It's greedy approach. But will fail on certain testcases. so yeah..... No solution exits in given constraints.

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23 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Sad that round will be unrated. Anyway, I enjoyed solving problems, especially D

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23 months ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

I was getting some positive delta (110+) after a long time. and now it's unrated. was it really necessary to make it unrated?

Edit:- ohh c is not solvable that's why it became unrated!

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    23 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Yes, since problem C is unsolvable. The problem setters thought that a greedy approach would solve C but it turns out that greedy doesn't always work. During the contest they realized that C is actually unsolvable within the constraints.

    And no, it wouldn't be enough to just not count C towards the ratings, because different people spent different amounts of time on C and it just wouldn't be fair.

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23 months ago, # |
  Vote: I like it +2 Vote: I do not like it

What is "can't be solved under given constraints"?? Last I saw, 2752+ correct submissions are there on C

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    23 months ago, # ^ |
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    same question?!

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    23 months ago, # ^ |
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    Yeah I'm kinda confused since I thought C was kind of easy. Maybe the test cases are weak?

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      23 months ago, # ^ |
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      Hello sir, can u pls tell me your solution for C? It will be greatly appreciated. Thank you.

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        23 months ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        Though the round is unrated it would be great if this is discussed after contest, ig?

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      23 months ago, # ^ |
        Vote: I like it +11 Vote: I do not like it

      did you use greedy approach ??

      can you solve for this,,, lets say..

      25 people wants to eat dish 1. 15 people wants to eat dish 2.

      tables are 15 , 13 , 12 .

      greedy wont work here... I was stuck here... also, I got stuck in B somehow... got 3 wrong subs..

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        23 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        is the answer not 38? table 15 dish 1 -> 15 satisfied table 13 dish 2 -> 13 satisfied table 12 dish 1 -> 10 satisfied

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          23 months ago, # ^ |
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          Answer is 40.

          Everyone who wants to eat dish 2 sits at table 1.

          The rest of the people (people who like dish 1) split themselves between table 2 and table 3. Therefore, there are no dissatisfied customers.

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        23 months ago, # ^ |
          Vote: I like it -8 Vote: I do not like it

        why the greedy wont work here? isn't the answer 15 + 13 + 10 = 38?

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          23 months ago, # ^ |
            Vote: I like it +17 Vote: I do not like it

          Answer is 40 ...

          we will make 25 dish-1 people sit on 12 + 13 table...

          and 15 people from dish-2 on table 15 ...

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    23 months ago, # ^ |
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    idk man, maybe pretests are well below the constraints.

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    23 months ago, # ^ |
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    Maybe they mean that the mistake cannot be solved within the time constraints of this competition, as the mistake would need be corrected in just a few minutes.

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    23 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    I suppose they mean the actual testcases, not the pretests (which are not comprehensive).

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23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

I skipped C, solved D, and after 5 minutes round became unrated. Not cool.

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23 months ago, # |
  Vote: I like it +12 Vote: I do not like it

The way it was going was almost sure of becoming CM today and it became unrated

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23 months ago, # |
  Vote: I like it +2 Vote: I do not like it

almost 3k people solved a unsolvable problem :/ how? misread? :/

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    23 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Maybe they used greedy approach, but it was actually wrong.

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23 months ago, # |
  Vote: I like it +70 Vote: I do not like it

the round is unranted, not unrated guys.

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23 months ago, # |
Rev. 2   Vote: I like it +19 Vote: I do not like it

nothing here

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    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The problem maker have their faults, but you're not expected to be so rude.

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23 months ago, # |
  Vote: I like it +2 Vote: I do not like it

how is problem C not solvable

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    23 months ago, # ^ |
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    because it cant be solved using a greedy algorithm. If you have used greedy algo, try this : 7 people want dish "1", and 5 people want dish "2" and we are given 3 tables with accommodation 5, 4 and 3

    SPOILER : the solution is 12 guests can be made happy and not 10

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23 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Garam krke thanda kr diya -_- .

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Here We Go, After Solving ABC Under 30 mins, the round is unrated. WoW.

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I feel you, this was my best performance in a long time and the round becomes unrated

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    23 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    The only reason u solved c is because the problem is wrong

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    23 months ago, # ^ |
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    Can u tell me solution for C? It will be greatly appreciated. Thank you.

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      23 months ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it
      Testcase
      Greedy Answer(Probably Yours Too)
      Something Better
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        23 months ago, # ^ |
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        I wanted to know what the greedy solution actually was, cos the greedy solution that I came up with was something I already knew didn't work. So i was wondering what greedy solution others came up with and thought worked but actually didn't

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          23 months ago, # ^ |
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          maintain a priority queue and sort b in reverse, and then greedily pop the maximum element from the pq and and assign them to the maximum table currently available, if not all of them fit in that table then add num_of_people — size_of_table to the priority queue and repeat the process. But this fails on so many test cases so yeah

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            23 months ago, # ^ |
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            Oh right. That was actually the first greedy solution that I thought of as well. Kinda weird that so many ppl just assumed that it would work when it doesn't.

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23 months ago, # |
Rev. 2   Vote: I like it +2 Vote: I do not like it

How is problem C unsolvable ?

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    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains

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      23 months ago, # ^ |
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      Oh! I understand now. My code (greedy) is giving output for this as 12.

      But, ideally, it should be 13.

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Can anyone explain why Problem C can not be solved?

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    23 months ago, # ^ |
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    I think the intended solution was a greedy algo, but it appears, that there are some tests, where it doesn't work

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    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Maybe the intended solution of C is not fully correct and maybe exist some counter case of this solution which makes problem more complicated than supposed to.

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23 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Unrated. Thank you so much.

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    23 months ago, # ^ |
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    Why even this single comment can receive downvote I can't understand

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23 months ago, # |
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Why unrated? Sad. I think constraints are ok

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23 months ago, # |
Rev. 2   Vote: I like it +2 Vote: I do not like it

Codechef round was postponed for an unrated round.

Who would have thought the sequel would be as good as the original https://mirror.codeforces.com/blog/entry/103170 https://mirror.codeforces.com/blog/entry/103108

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23 months ago, # |
Rev. 3   Vote: I like it +8 Vote: I do not like it

18 coders could not find this before and what were the testers doing. what a waste of time:(

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23 months ago, # |
Rev. 2   Vote: I like it -19 Vote: I do not like it
  1. Is this Rated !!!!! ::: >>>
BIG SPOILER !!!!!!
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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I was on the verge of becoming a specialist after so long and now the round in unrated..sigh :-)

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23 months ago, # |
  Vote: I like it +9 Vote: I do not like it

Is it ranted?

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23 months ago, # |
  Vote: I like it +70 Vote: I do not like it

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    23 months ago, # ^ |
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    *unranted

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    23 months ago, # ^ |
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    I first time passed pretests in D problem (hopefully AC) in contest , At least there is something to be happy.

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23 months ago, # |
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What does "unranted" mean?Unrated?

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

How did so many people get AC in C?

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Only pretests are run during the contest and the solutions probably would've failed when run against the proper testcases.

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    23 months ago, # ^ |
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    The judge's code probably used the same greedy algorithm everyone else used. They didn't realize that it is actually incorrect before the contest.

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The test data is too weak and the testers used the wrong method to produce the test data.

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23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

can we make it the first contest then where editorial is updated before the contest ends.

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23 months ago, # |
Rev. 2   Vote: I like it +3 Vote: I do not like it

+165 Delta and it's all gone. Thanks for the great round !

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23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

It say's unranted, what does that mean?!

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No rate updates for the official participants. You can find your rate history in the graph in your profile.

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone explain what they mean by that problem C is unsolvable in given constraints?

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23 months ago, # |
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my rank would've increased this round :(

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23 months ago, # |
  Vote: I like it +3 Vote: I do not like it

This is just sad.

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23 months ago, # |
Rev. 4   Vote: I like it +4 Vote: I do not like it

My best performance so far.

Bye Bye +125. Top 500 performance.

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23 months ago, # |
  Vote: I like it +17 Vote: I do not like it

muda muda muda muda muda muda muda muda

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23 months ago, # |
  Vote: I like it -16 Vote: I do not like it

bye bye +100 (real)

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23 months ago, # |
  Vote: I like it +15 Vote: I do not like it

I was wondering that how come C be 1250 worth of points but seem impossible to me. I am dumb but not that that dumb(hopefully).

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    23 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    It had an easy-to-think-of greed solution, but now it seems to be a wrong one. The correct solution may be the Knapsack problem, but it cannot achieve the required time complexity.

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23 months ago, # |
  Vote: I like it +79 Vote: I do not like it

How did so many people falsely solve C? I stared at it for like 20mins and had no idea, but seeing that many people solving it so fast, I started to doubt myself. I tried some stupid greedy ideas but all failed on paper.

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    23 months ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    Easy hacking greedy passed the pretests.

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    23 months ago, # ^ |
      Vote: I like it +34 Vote: I do not like it

    Given it was only 1250 points, proof by AC is easier to try than a real proof or looking for a counter example :P

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      23 months ago, # ^ |
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      Can you please explain what is the issue with the constraints?

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      23 months ago, # ^ |
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      exactly what I thought lol

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    23 months ago, # ^ |
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    Speedcoding just does that to you

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    23 months ago, # ^ |
      Vote: I like it +124 Vote: I do not like it

    I guess this really says something about how many people "solve" problems by simply guessing a reasonable-looking greedy.

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      23 months ago, # ^ |
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      B was guessable too

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        23 months ago, # ^ |
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        I am still not sure, how to solve problem B optimally...

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          23 months ago, # ^ |
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          It's ideal to split the array into 2 subarrays.

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        23 months ago, # ^ |
        Rev. 2   Vote: I like it +11 Vote: I do not like it

        I mean maybe, but the point is that C is literally unsolvable — so anyone who actually proves their solutions wouldn't have solved it.

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          23 months ago, # ^ |
            Vote: I like it +5 Vote: I do not like it

          I think the problem is that everybody writing was trying to get AC — and the greedy prrof is somewhat easy to think up really fast. None of the testers really struggled on the task, except me.

          The issue should have been caught by me — when we were "testing", I did not manage to solve C (it was B then) in contest, and I submitted like 7 wrong (all greedy) solutions to it. Then, when I was asking the author about the solution, I was told that it is "a simple greedy". Then, I decided to believe him and did not upsolve that task. I should have caught it.

          So I think that the CF system of less points with more time will always incentivize this sort of "half-done" proofs.

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        23 months ago, # ^ |
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        I proved B before solving it. But when I came to C I just guessed some unproved greedy approach and got WA for silly mistake then I started doubting this approach and couldn't come with another one.

        My problem solving skill is slowly goes from random guesses to prove-first approach or even partially-proved approach after watching many streams from tourist, um_nik, and many other legends.

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      23 months ago, # ^ |
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      simply guessing a reasonable-looking greedy.

      Is this a common thing among highly experienced users when it comes to simpler problems (say div 1 A-C)?

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If you go to the "status" page and look at all C AC's, the fact that most of them are 15ms should point at a sub-quadratic solution

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23 months ago, # |
  Vote: I like it +11 Vote: I do not like it

I see many people solved C !!
why c is unsolvable!?
why the round unrated?◑﹏◐

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    23 months ago, # ^ |
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    Seems that greedy solution of C is not correct

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    23 months ago, # ^ |
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    Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

+114 ...and then its unrated

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23 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Testcase for problem c that dosen't work whit the greedy metohd 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5

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23 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

don't do it unrated pls

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23 months ago, # |
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Not Happy with the contest making today :(

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23 months ago, # |
  Vote: I like it +15 Vote: I do not like it

This contest is a disgrace to the Joestar Bloodline!

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23 months ago, # |
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What was the problem with C?

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    23 months ago, # ^ |
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    did you use greedy approach ??

    can you solve for this,,, lets say..

    25 people wants to eat dish 1. 15 people wants to eat dish 2.

    tables are 15 , 13 , 12 .

    greedy approch will fail,, for 25 dish guests we can pick 12 + 13 = 25 , and 15 for rest.

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      23 months ago, # ^ |
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      I see. How did I not see that lol? Speedcoding I guess. What would be a good dp formulation for this assuming bounds are low enough?

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        23 months ago, # ^ |
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        after solving A , I tried solving C... couldn't solve that...

        it is basically knapsack problem with 'K' sacks given to us...

        where 'K' is number of distinct elements given in array.

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    23 months ago, # ^ |
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    Testcase for problem c that dosent work whit the greedy 1 13 4 1 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 5 And there is no solution that will solve this type of testcases that can fit in the constrains

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      23 months ago, # ^ |
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      Is the answer 12 ? I mean its working fine on my local environment. can someone hack my solution please I am feeling to much smart for getting my solution accepted.

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        23 months ago, # ^ |
        Rev. 2   Vote: I like it +1 Vote: I do not like it

        u put the 9 ones at the three 3 ppl tables and the 4 guys at the table of 5 and the answer is 13

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23 months ago, # |
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What the fuck?

The solution is $$$8$$$, not $$$9$$$?

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    23 months ago, # ^ |
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    It is 9, author's solution is wrong.

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    23 months ago, # ^ |
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    Maybe the standard code also used the wrong greedy method and fails on this.

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23 months ago, # |
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C looked so hard to me with given constraints. Looked like a multiple knapsack problem. The knapsacks are your guests that like dish $$$i$$$ and the items are the tables. In this version you can keep feeding a full knapsack but gain no score. I tried greedy strategy on papers, they all had edge cases. Couldn't find a dp. Best algo I found was like $$$O(m^{\sqrt{n}})$$$. I really wonder what happened there :)

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    23 months ago, # ^ |
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    the dp i came up with was something like dp[i][j]->max satisfied customers considering till ith type and till j seats. so dp[i+1][j]=max(dp[i+1][j],dp[i][k=0 to j] + min(count[i+1],summation till k)) ps:i did sort the tables though

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      23 months ago, # ^ |
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      I'm not sure what you mean with "count[i+1]" and "summation till k". Have you tried your solution with the counterexemples to many greedy approaches that were given in the comment ?

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23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Only if the contest makers had a stand for stopping time like ZA WARUDO

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23 months ago, # |
Rev. 4   Vote: I like it +20 Vote: I do not like it

Unfortunately,the round will be unranted.We apologize,we made a mistake in problem C and it cannot be solved with in the given constraints.

Notice that it is unranted instead of unrated. Does it mean that this round still needs to be rated?

UPD:Now this round has become unrated. It is really a frustrating round.

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23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

In an interactive problem, TLE means i am taking more operation than the available operations ?

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    23 months ago, # ^ |
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    Read the interaction instructions carefully: "If your program performs more than 30 operations for one test case, subtracts a number x greater than n, or makes an incorrect request, then response to the request will be -1, after receiving such response, your program must exit immediately to receive the Wrong Answer verdict. Otherwise, you can get any other verdict."

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why this round is unrated??

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23 months ago, # |
Rev. 3   Vote: I like it +19 Vote: I do not like it

sorry to hear that it's unrated..

and a wrong example for many solution including mine:

1
7 3
1 1 1 1 2 2 2
3 2 2

the answer is 7 instead of 6.

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    23 months ago, # ^ |
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    What is that extension bro?

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      23 months ago, # ^ |
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      Do you mean why the answer should be 7?

      let the room be $$$c_1, c_2, c_3$$$ , then we can match $$$c_1 \rightarrow 2*3$$$, $$$c_2 \rightarrow 1*2$$$ and $$$c_3$$$ the same.

      If you use a greedy like me, the match would be $$$c_1 \rightarrow 1*3$$$, which is obviously wrong ;_;

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      23 months ago, # ^ |
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      Carrot

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      23 months ago, # ^ |
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      sorry, i misunderstood...

      it's Carrot, also CF predictor is another good choice.

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23 months ago, # |
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I'm curious how these testers test this round?

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23 months ago, # |
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Bye bye my hopes and chances to become pupil...

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23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

Funny that a large portion of the top 100 participants didn't even attempt C because they knew it to be impossible

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23 months ago, # |
  Vote: I like it -36 Vote: I do not like it

Trash Russian Round.

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23 months ago, # |
Rev. 2   Vote: I like it +10 Vote: I do not like it

There has been 2 times in the past year when CF rounds and CC starters rounds are planned to clash with each other. One of it was postponed both times, and in both occasions, the round that is not postponed became unrated [Lol]

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23 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Why the announcement says

Unfortunately, the round will be unranted. We apologize, we made a mistake in problem C and it cannot be solved within the given constraints.

instead of "unrated"?

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Well, Unra'n'ted.

So, does the std solve this problem with greedy algorithm? lmao.

1
7 3
1 1 1 1 2 2 2
3 2 2

answer: 7

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23 months ago, # |
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Could have attended the Codechef contest instead. Whatever ...

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23 months ago, # |
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How to prove B?

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    23 months ago, # ^ |
    Rev. 4   Vote: I like it +1 Vote: I do not like it

    If g divides a and b, it divides a + b. That implies [there was typo mistake] gcd(a + b, c) >= gcd(a, b, c), which means that if you have some partition you wouldn't get worse solution if you delete some intervals.

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    23 months ago, # ^ |
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    Let's say your optimal answer has k partitions, whose gcd is 'x'. We can merge the first k-1 partitions and the gcd will either increase or stay the same.

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    23 months ago, # ^ |
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    Suppose that k >= 3 in optimal answer. Let d be the answer for testcase. Then you can combine some two neighboring segments in one segment and get a solution no worse than the previous one. It's because if d | a and d | b then d | (a+b) and you got answer for k — 1. So in optimal answer k = 2 and you get your solution

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    23 months ago, # ^ |
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    you only need to split array into 2, If you get some gcd 'x' by splitting more than once, then you can club all untill there are 2 subarrays because each subarray is multiple of 'x' and sum of them will also be multiple of 'x'.

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23 months ago, # |
  Vote: I like it -8 Vote: I do not like it

Why not just remove problem C and extend the round by 15-30 mins rather than declare it unrated?

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    23 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    because some people have spent time on this problem, and some people haven't

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23 months ago, # |
Rev. 2   Vote: I like it -20 Vote: I do not like it

Guys go easy on the Downvote Button
Mistakes were made

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23 months ago, # |
  Vote: I like it +11 Vote: I do not like it

To all the people that solved C, How did you fakesolved it? I'm interested in the "expected solution".

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    23 months ago, # ^ |
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    Greedy. Didn't even realize it was wrong lol

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    23 months ago, # ^ |
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    As a contestant it is highly expected to come up fast with a wrong solution by intuition especially with greedy. But as a round tester a plenty of time is there to test the problems thoroughly. Such kind of mistakes wastes time of others.

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23 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Such a frustrating moment solved 3 problems in 30 mins and then what??UNRATED. Good Bye +170

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    23 months ago, # ^ |
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    considering solution of B is cringe and everyone typed it without any proof i guess its fair

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      23 months ago, # ^ |
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      The proof is simple though. If $$$gcd(a,b,c) = k$$$ then $$$gcd(a + b, c) \geq k$$$ since $$$ k \mid (a + b)$$$.

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23 months ago, # |
  Vote: I like it +44 Vote: I do not like it

Hi @AndreyPavlov & fellow setters and testers

Questions are great, mistakes happen! I really enjoyed the questions and logic used. Doesnt matter if round is unrated but I really enjoyed the questions. Thanks for the contest. Much love <3

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    23 months ago, # ^ |
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    Indeed. The problems(except for C) are very interesting.

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23 months ago, # |
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For C, 1 7 3 1 1 1 2 2 2 2 2 2 3 Actual answer is 7 with optimal selection type 1 table 3 type 2 table 1, 2 But Greedy gives 6 type 2 table 3 type 1 table 1, 2 one person with type 2 is not satisfied

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23 months ago, # |
  Vote: I like it +14 Vote: I do not like it

Volveré y seré millones

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23 months ago, # |
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respect. GOA T

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23 months ago, # |
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Those who solved C what was your approach?

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23 months ago, # |
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I wonder why coordinator and many testers didn't even realize this problem.

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23 months ago, # |
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Imagine Masters not recognizing NP-Hard problem when they see one.

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    23 months ago, # ^ |
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    when masters see NP-Hard problems they say: no problem.

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23 months ago, # |
Rev. 2   Vote: I like it -13 Vote: I do not like it

Is C unsolvable? We could just add stronger tests and re-judge all submissions. It is solvable surely because the first AC solutions were by GMs. Would like to know more about the reason behind taking such a big decision, skipping other alternatives like re-judging solutions.

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it +100 Vote: I do not like it

    It can be proved to be unsolvable in polynomial time by reduction to 3-partition. The grandmasters who solved it were wrong.

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      23 months ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      topg for a reason

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        23 months ago, # ^ |
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        Can you please elaborate a little?

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          23 months ago, # ^ |
          Rev. 4   Vote: I like it +10 Vote: I do not like it

          to prove that a problem is NP-hard you can consider another problem known to be NP-hard, 3-partition in the comment by Everule, and then show a relation such that if this problem is solvable then 3-partition is solving.

          this means that this problems is atleast as hard as 3-partition which we do not believe to be solvable in polynomial time.

          everule is topG because he never guesses and anyone who never guesses is topG

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            23 months ago, # ^ |
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            I see thank you. :)

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            23 months ago, # ^ |
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            In C, sum of all numbers is bounded so you can't really make a reduction.

            3-partition is only NP-hard in number of elements, not when the sum if bounded

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              23 months ago, # ^ |
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              Prolly, I didn’t look into the reduction, was just explaining what the process is

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      23 months ago, # ^ |
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      You mean the other way around, reducing 3-partition to this problem?

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        23 months ago, # ^ |
          Vote: I like it +12 Vote: I do not like it

        And even then it doesn't imply unsolvability in polynomial time (probably)

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          23 months ago, # ^ |
            Vote: I like it +8 Vote: I do not like it

          I may have messed up the direction of reduction, because I don't have experience in theoretical CS, but solving this problem in polynomial time breaks the fact that 3-partition is strongly NP-complete. i.e., That even if the integers are bounded by polynomial in $$$n$$$, which it is, the problem is still not solvable in polynomial time.

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            23 months ago, # ^ |
            Rev. 2   Vote: I like it +22 Vote: I do not like it
            • The original version was incorrect

            Let us pick a set $$$S$$$ of with sum and size divisible by $$$3$$$ to 3-partition. We bound every element in $$$S$$$ to $$$sum(S) \times \frac{3}{4n}$$$ and $$$sum(S) \times \frac{3}{2n}$$$. This is also strongly NP-complete. Now we make $$$n/3$$$ groups of people, each consisting of $$$sum(S) \times \frac{3}{n}$$$ people, and tables consisting of the sizes of $$$S$$$. The optimal solution to this is a 3-partition if it exists. Notice that any solution with all must be a 3-partition due to the bounds on set sizes of $$$S$$$.

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              23 months ago, # ^ |
                Vote: I like it +3 Vote: I do not like it

              I just say that if P = NP then you can solve NP-complete problems in polynomial time

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              23 months ago, # ^ |
                Vote: I like it +10 Vote: I do not like it

              That's not polynomial solution, that's pseudo polynomial solution, as the total sum, T, is bounded.

              I believe 3-partition can be solved in $$$O(NT^3)$$$ (which is pseudo polynomial) with a 3D DP, similar to double knapsack.

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                23 months ago, # ^ |
                Rev. 2   Vote: I like it +10 Vote: I do not like it

                3-partition is strongly np-hard, so probably you can't even solve it in poly-time when the input is given in unary (base 1).

                it looks like you are confusing dividing a set into 3 subsets with dividing an array into triplets

                edit: oops, looks like it was everule who confused them. you're right that the reduction everule posted is solvable in pseudopolynomial time.

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                  23 months ago, # ^ |
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                  Yeah I was wrong, splitting an array into 3 sets with the same sum isn't the same as splitting it into triplets...

                  I wonder if it's a first proven NP-hard problem on Codeforces

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                  23 months ago, # ^ |
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                  Yeah, I've fixed the reduction, I messed up while looking for it.

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                23 months ago, # ^ |
                Rev. 2   Vote: I like it +21 Vote: I do not like it

                3-partition can't actually be solved in pseudopolynomial time (assuming P != NP). And it's probably a different problem then you're thinking.

                The problem is: given an integer B, and a set of $$$3n$$$ integers, partition the set into n groups, such that sum in each group is equal to $$$B$$$.

                This problem is strongly NP-complete, so even if the numerical values in the input are encoded in unary (so a value of $$$A$$$, takes $$$A$$$ bits to encode), it is still NP-complete. The reduction from 3-partition to problem C is as follows:

                Given arbitrary instance of 3-partition, where numbers are encoded in unary. Then make $$$n \times B$$$ guests, divided into n groups of equal $$$a_i$$$, each of size $$$B$$$. The set of $$$3n$$$ integers S, just make that into $$$c_i$$$'s. Now you can show that answer to problem C for this input is $$$n \times B$$$ iff the instance of 3-partition was solvable, Almost... One problem we face is that inside one bucket there can be more or less than exactly 3 elements. To fix this, we can add a sufficiently big enough number to each number in the set S, and add three times that number to B. This enforces that not too many numbers can end up in one bucket.

                And it's a polynomial time reduction, because $$$n \times B$$$ is polynomial in the input size, (and even if you increase by that sufficiently big number, you can still choose that as polynomially as big as the input).

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                  23 months ago, # ^ |
                    Vote: I like it +5 Vote: I do not like it

                  Jeroen there is another cool result, that makes the constraint on triplets quite easy in the wikipedia article

                  The 3-partition problem remains strongly NP-complete under the restriction that every integer in S is strictly between T/4 and T/2.

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23 months ago, # |
  Vote: I like it +43 Vote: I do not like it

I find it so funny to see many people complaining about the round going unrated saying: "I solved three problems" when they literally fakesolved one of them.

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it +3 Vote: I do not like it

    Some of them solved A, B, D. So their complain is genuine.

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23 months ago, # |
  Vote: I like it +4 Vote: I do not like it

"Rant" means "roar".

Then "Unrant" means "No roar"

Then writers means that "The contest is without roar."

Well...

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For problem D,

Spoiler

Is this wrong? I think it's probably wrong, but looking at the #of submissions is making me think that smth like this might work. Can anyone tell me if this is the intended sol? Please no other spoilers about any intended solution btw.

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    23 months ago, # ^ |
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    try with 35

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      23 months ago, # ^ |
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      Thanks, I see the problem now. I'll try smth else and pray it works lol

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    23 months ago, # ^ |
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    good idea, but i think there is an arbitrarily large number with weird binary sequence that can counter that, because the number could've just decided to troll you lol (A.K.A the input is always small, in which the queries would exceed 30). better way is to subtract it in a really interesting way to the point that its final form would be 2^input-1.

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      23 months ago, # ^ |
      Rev. 3   Vote: I like it 0 Vote: I do not like it

      Yeah, I found a solution that will work. I'm such an idiot lol. Will code it up after school. Edit: Got AC

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    23 months ago, # ^ |
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    I tried the same.

    It does not work. Offline I tried for 10**9 and it needed more queries than the limit.

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23 months ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

I would've become red today, if only it was rated :( /s

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23 months ago, # |
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author, tester and contestant all guessed solutions

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23 months ago, # |
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Contest would have been fine if C was just removed during testing or something

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    23 months ago, # ^ |
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    People wasted time on it , that's why this is not fair

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      23 months ago, # ^ |
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      "If C was removed during testing"

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        23 months ago, # ^ |
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        Yes !

        some people got pretests passed fast and others took time

        this time affected other problems

        so it is not fair(although I was really good in the contest)

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          23 months ago, # ^ |
          Rev. 2   Vote: I like it +3 Vote: I do not like it

          No, you don't understand. By testing I mean (and I suppose the original commenter also meant) the phase where testers try the problems before the contest.

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    23 months ago, # ^ |
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    Nah , some people who knew that greedy solution is wrong may have had wasted time on C which they could have used to solve other problems.

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23 months ago, # |
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ABCD in 45 minutes and really near of reaching CM

If the author did a mistake it's ok ,but what were the testers doing !!!

I think the main testers job is to find things like this

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23 months ago, # |
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There is no need to be harsh on tester guys! Everyone makes mistakes . I hope fellow testers learn from this and try not to do such mistakes in future!

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23 months ago, # |
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Does anyone figure out the correct answer to Problem C even if it may get a TLE or MLE?

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    23 months ago, # ^ |
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    You can just brute force all the possible combinations. There is probably also some dp solution, but it has non-polynomial time complexity.

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23 months ago, # |
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I've found an easier version of problem C, how to solve it?

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    23 months ago, # ^ |
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    Try to break your limits and make a solution

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    23 months ago, # ^ |
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    If I'm not mistaken, this task is also NP-hard and you just need to write some heuristic to solve it.

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23 months ago, # |
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I would get about + 120 and the round was unrated ToT. Anw still a good contest

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    23 months ago, # ^ |
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    Is the rating change just a guess or is there anywhere to calculate this estimate?

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      23 months ago, # ^ |
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      There are extensions like CF Predictor or Carrot that let you see the rating change you are going to get while the contest is still running.

      These update live during the contest depenging on how you're doing compared to others but they are not always exactly correct (typically the actual rating change is within +/- 5 points compared to the predictor).

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23 months ago, # |
  Vote: I like it +8 Vote: I do not like it

F

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23 months ago, # |
Rev. 2   Vote: I like it +68 Vote: I do not like it

Actually today's situation remind me of a contest held many years ago, whose div1.b was also a greedy hacked by a talented coder Um_nik

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    23 months ago, # ^ |
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    Funny you should mention me. Screencast, timestamp on when I send the clarification. Another good moment

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      23 months ago, # ^ |
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      I think the testers/setters should be required to submit the bruteforce solution (for small but as large as possible cases) along with the intended solution.

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23 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Maybe C was a blunder, but after all last 3 problems are quite a challenge, so the setters are not as bad, as you think

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23 months ago, # |
Rev. 2   Vote: I like it +70 Vote: I do not like it

I'm curious. You didn't even prepare a brute force solution to stress-test problem C? I thought stress-testing is a "must" when you prepare a problem?

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23 months ago, # |
  Vote: I like it +41 Vote: I do not like it

Most normal contest in Ohio 💀

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23 months ago, # |
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Thanks for the problems. D was a good problem btw but it seems more suited as a replacement for C.

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23 months ago, # |
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gcd round :(

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23 months ago, # |
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And this round

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23 months ago, # |
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its still nice problemset mens, shit happens, but i like this round anyway

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23 months ago, # |
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Use me as a dislike button

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23 months ago, # |
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so sad it got unrated, even though D is a fun question to do (as a math nerd and interactive problem fans lol :p). also, how do you dp problem C?

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    23 months ago, # ^ |
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    Problem C is NP-complete meaning that there is no known solution that can solve it fast.

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      23 months ago, # ^ |
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      i thought about using dp but with the priority guest group being different for each dp (A.K.A dp brute force because better than brute force^2 i guess :p), but idk how to do that. Guess my concept is kinda similar to that.

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    23 months ago, # ^ |
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    You can solve it as a greedy

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23 months ago, # |
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I want divide something, is it OK?

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23 months ago, # |
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Me after the announcement:

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23 months ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

for the sake of interest, it's still worth doing strong system tests against the greedy approach. my solution has been undergoing stress testing for more than half an hour. sorry for poor english

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23 months ago, # |
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Problem F seems intresting. How to solve it?

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    23 months ago, # ^ |
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    Sort the numbers, find the divisors for each number and then use mobius function with prefix sums and some tricks.

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23 months ago, # |
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Regardless of problem C and the round being unrated, I really enjoyed the problems and it could have been my master round.
Thanks for making an amazing contest :D

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    23 months ago, # ^ |
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    Agree if C was removed with the problems starting from D being shifted, it could've been one the best rounds made recently :)

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23 months ago, # |
Rev. 2   Vote: I like it +244 Vote: I do not like it

Yes, the author's solution in C was incorrect. We haven't noticed it in 9 months. We stressed this solution (as it turned out, we stressed it terribly), I even wrote an analysis with a proof of this greed. Now an exercise for the reader, find the error in this tutorial:

Tutorial
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    23 months ago, # ^ |
    Rev. 2   Vote: I like it -48 Vote: I do not like it

    1
    15 3
    1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
    7 4 4

    Courtsey — NemanjaSo2005

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      23 months ago, # ^ |
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      I think he means to find which part of the tutorial is wrong not to find a stress test.

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    23 months ago, # ^ |
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    In the second case, if $$$b_i$$$ are seated, then there are $$$0$$$ empty seats, but there are also $$$b_i-c_k$$$ extra people, compared to $$$c_k-b_j$$$ empty seats if $$$b_j$$$ are seated. It's not obvious that seating $$$b_i$$$ is better.

    In the third case, it's not necessarily true that having a larger maximum is better.

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    23 months ago, # ^ |
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    It happens :) I am happy you did not let me lose my sanity over 1 hours while trying to solve C :P

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    23 months ago, # ^ |
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    It's kinda hard to find a particular error in a text that doesn't make any sense. For some reason you are minimizing empty seats, how is it connected to the actual problem — I don't know. Then there are just random phrases like "it is better for us to have a maximum as large as possible". Why is it better?

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      23 months ago, # ^ |
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      This is how I understand it: for each table they choose people from the same group — these are going to be the "satisfied" people for that table. The number of free seats is then some constant minus total number of satisfied people, so we try to minimize it. What they do in the "proof" is they prove that the solution is greedy, i.e. at each step we are indeed minimizing the number of free seats for the current table. The error is that they never prove that greedy actually works.

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    23 months ago, # ^ |
    Rev. 2   Vote: I like it +12 Vote: I do not like it

    "stressed this decision" -- solution, not decision

    It's incorrect translation from Russian, I guess

    Upd: fixed

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    23 months ago, # ^ |
      Vote: I like it +97 Vote: I do not like it

    The main problem in the proof is that you never took into account that your choice for the current table might affect the other tables. If you were assigning people to only one table, the greedy is perfectly valid. Additionally, if all tables had the same capacity, your greedy is completely valid. Bullet point 3 is the one about which I am talking. You said that it is better for us to have maximum as large as possible, without providing proof for why. As it turns, that statement is incorrect as can be seen in the example above. I know that the proof looks quite logical and it seems to intuitively make sense. However, intuition can be really misleading sometimes.

    I honestly feel quite bad for you and the rest of the authors as this was your first contest and it went the way it did. I cannot imagine how bad you all are feeling, especially with all the hateful comments from some individuals in codeforces community. But I do not suggest you quit either CP or problemsetting because of that. (Your choice, I am just giving a recommendation) Everyone makes mistakes and it can be a great opportunity to learn and improve.

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    23 months ago, # ^ |
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    Local best solution doesn't equal to global best solution I think. For example, let's seat 4*1 and 3*2 into 3 2 2, and surely placing 1 or 2 into 3 can fill this single table, but the cases after that become very different. Just like why we can't use DP sometimes.

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    23 months ago, # ^ |
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    I think one problem is that you are assuming that "at the next iterations of the algorithm, it is better for us to have a maximum as large as possible", which is incorrect. If we have c = {3, 2}, b = {3, 2} gives a score 5 while b = {4, 1} gives only 4 although it has a larger maximum.

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    23 months ago, # ^ |
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    what you are trying to convince in the last section of 2nd point is that for a fixed c array the lexicographically bigger b1 is always better compared to a smaller b2 (both having same sum and b1,b2 are sorted),now,what if c is exactly same as b2?

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23 months ago, # |
Rev. 2   Vote: I like it -67 Vote: I do not like it

Could you please update rating only for those, who has positive delta?

UPD: Nevermind, just found out about C a little bit late

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    23 months ago, # ^ |
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    Could you please update rating only for those, who have 0 delta?

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23 months ago, # |
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I know the complex progress of contest preparation. It needs a lot of effort and attention, but there are many testers I am shocked by this annoying situation. The reason for my participation in Codeforces contests is not ranking. I am joining contests for having fun and learning some new things. But I thought my solution will get the wrong answer verdict while submitting but when it got accepted I got shocked. I hope testers and problem setters of this round will be more careful in the future.

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23 months ago, # |
Rev. 2   Vote: I like it +3 Vote: I do not like it

why "locale=ru" in the tutorial link
Oh it's fixed now.

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23 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

A: If there's no odd numbers, or n==3 and there's 2 odd numbers, no solution. Otherwise we can find 3 odd numbers or 1 odd number and 2 even numbers.

B: Note that common divisor of (a,b,c) would also be common divisor of (a,b+c), it's always optimal to divide the array into 2 segments.

D: First let q=1 and query(q). If the bitcount is reduced, the last bit of n is 1 and you've just cancelled it, then you need to try for q=q<<1 (which means, left-shift q to the next unknown bit). Else the last bit of n is 0, and by count the difference of bitcount before and after the query, you'll know how many zero bits was on the tail of n, and you've just fliped them to 1 (and a 1-bit to 0). Then you can cancel these 1-bits in one query, and left-shift q to the next unknown bit. In the first situation, you've got the information of the least significant bit by one query, and in the second situation, you've got the information of at least 2 bits by 2 queries. Thus by repeating this process we can get the answer.

E: Sqrt decomposition. Consider for gcd(u,v) in [1,ceil(sqrt(L))-1], [ceil(sqrt(L)),L-1] and [L,R/2] respectively.

PS: It's regrettable that the contest was ruined by an incorrect problem, and many great contestants lost their positive delta. It's a pity that problem D,E are pretty well-designed.

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23 months ago, # |
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So what happens to C about it being it problemset. I think C must be removed from problemset too.

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23 months ago, # |
Rev. 3   Vote: I like it +48 Vote: I do not like it

Brief solutions(hints) besides G,I'm weak in string problems.

A
B
C
D
E
F
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    23 months ago, # ^ |
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    Unfortunately, G is just a suffix automaton exercise :(

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23 months ago, # |
  Vote: I like it +120 Vote: I do not like it

Solve count for C during contest shows why cf is just guessforces nowadays

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23 months ago, # |
Rev. 2   Vote: I like it +5 Vote: I do not like it

rip my >100 positive delta ;-;
Anyways it’s still a fun round with other problems being interesting

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    23 months ago, # ^ |
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    How did you calc your delta?

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      23 months ago, # ^ |
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      Carrot extension

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        23 months ago, # ^ |
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        Is it possible without extensions?

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          23 months ago, # ^ |
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          Yes, but you have to do a few thousands of calculations

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            23 months ago, # ^ |
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            hahaha so funny mr obvious

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              23 months ago, # ^ |
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              what answer did you expect?) lol

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                23 months ago, # ^ |
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                I dont know what answer should i expect...thats why i am asking

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23 months ago, # |
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First time I solve 5 problems in div2. 140 positive delta. I would've been div1. Big bruh moment.

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23 months ago, # |
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first time got approach of D but unrated , its OK :)

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23 months ago, # |
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E could have been written as "Constructing a graph with vertices from $$$l$$$ to $$$r$$$" instead of "deleting all other vertices and all other edges", would be much easier to read and understand lol

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    23 months ago, # ^ |
      Vote: I like it +90 Vote: I do not like it

    E could have been written without any graphs at all. You just assign weights and right after that throw them in the set to count the number of unique values. There is nothing graph-y about this problem.

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23 months ago, # |
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this is my first time that I solve the last problem and rk110- in div.2,although it is unrated,I think it still be a good contest

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23 months ago, # |
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meme

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23 months ago, # |
  Vote: I like it +57 Vote: I do not like it

AndreyPavlov Please remove problem C from codeforces as it can be harmful for people who try to solve it later during practice or something.

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23 months ago, # |
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What ever happen, I still love codeforces

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23 months ago, # |
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what's wrong with problem c? this is the quickest i solved a,b,c and this round gonna be unrated fek

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23 months ago, # |
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It's so terrible to do system testing C with the wrong test case.

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23 months ago, # |
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I am kind of curious: how were there so many AC on C, even tho it could not be solved? Did people just submit a greedy solution?

Also curious: what would be the best possible runtime for a correct solution?

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23 months ago, # |
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can problem B solved using binary search on answer?

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23 months ago, # |
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Can someone give a test case where this approach gives a wrong answer to C?

  1. Calculate how many meals of each type and how many tables of each size there are.

  2. Go through the meals, if there is a table with the same size as the count of that meal, then put those people to that table.

  3. Put remaining meal and table counts to two priority queues. While there are still some empty tables or not all people has a spot: take the top of meals and put them at the biggest empty table, if some people who like that meal are still left, check if there is empty table with that amount of spots, if there is put them there, otherwise put that number back to priority queue.

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    1

    15 4

    1 1 1 1 1 1 1 1 2 2 2 2 2 2 2

    5 4 4 2

    The answer is 15, but your approach would give 14.

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      23 months ago, # ^ |
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      Thank you sir

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      23 months ago, # ^ |
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      hello sir, my approach works for some of these type of test cases, can you tell me where my approach can go wrong? Sort the tables array in descending order, and we keep the frequency of elements in a multi set, and for every table from largest to smallest we find the lower bound of the table size in the multiset, and we assign the table size to that many people of same type, and erase the people that are sitting in the table in the multiset and the extra people left will be put inside the multiset again, we do this if the lower bound exists, and if the lower bound does not exist, we just assign the maximum element of multiset to the corresponding table. In test case 16: the array was 1 1 1 1 1 1 2 2 2 2 2, and the table given was 3 3 5 And the answer to this test case according to my solution is 11, but the given answer is 10 which is wrong.

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        23 months ago, # ^ |
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        Your approach assigns people to the table without caring about what kind of effect it might have on other tables. That's the main reason why no greedy solution works.

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23 months ago, # |
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    23 months ago, # ^ |
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    Nice one. I didn't solve C during the contest and was wondering how 1000 people solved it. Shows how people don't understand their solutions and can't prove it.

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23 months ago, # |
  Vote: I like it -34 Vote: I do not like it

I hope Codeforces will hold a good competition next time, instead of being as rubbish as this one

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23 months ago, # |
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D is really an interesting problem

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23 months ago, # |
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i didn't participant in the contest yesterday, and only see people are talking about the unsolvable problem C. but now actually i even can't see what exactly is the problem statement, where can i find it? thanks

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    23 months ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    The problem was something like this:

    You are inviting people to a party. There are $$$n$$$ people, and the $$$i$$$ th person likes the dish $$$a_i$$$ $$$(1\leq i\leq n)$$$. You have $$$m$$$ tables, and the $$$j$$$ th table has a capacity of seating $$$c_j$$$ people $$$(1\leq j\leq m)$$$. Now, for every $$$1\leq j\leq m$$$ you have to select a dish $$$x_j$$$ to serve at the $$$j$$$ th table, and the people seated at that table who like that dish will become happy. You have to maximise the total number of happy people by selecting what dish to serve at each table.

    Input: Arrays $$$a$$$ and $$$c$$$ were given in input

    Output: Just print a single integer, i.e. the maximum number of happy people possible

    Constraints: $$$1\leq n\leq 2\cdot10^3$$$, $$$1\leq m\leq 2\cdot10^3$$$, $$$1\leq a_i\leq n$$$, $$$1\leq c_j\leq 2\cdot10^3$$$ and $$$\sum_{j}c_j \geq n$$$ (i.e. total number of seats is more than number of people, so every person can be seated)

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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi guys! If you wanna know why problem C was wrong, and the editorials for A,B,C,D then you can check it here — https://www.youtube.com/@GrindCoding

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23 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Does this solution for C work?

Take the array of guests and put 1 extra element in it, let's call it breaker. Now go trough each permutation of that array. To process a permutation, you are going to only assign people before breaker to the correct table and you can ignore the rest. (ignore means to assign them to any table) How do you assign them? Consider each consecutive subarray [l,r] of equal elements such that a[l-1]!=a[l] and a[r+1]!=a[r]. (a is the permutation) You will add it's length (r-l+1) to the array b. Obviously, you are ignoring the elements after the breaker. Now that you have array b, you are going to assign each b[i] (apart from the last one) to exactly one table, such that the sum of capacities of tables that you assigned is minimal. That problem can be solved with greedy. Now, fill the remaining tables with the last element. If it is possible to do so, such assignment is valid. Now you just select the best of all valid arrangements and output it as a solution.

Why does greedy work? So, let's say that we have 2 arrays x and y who will describe any optimal arrangement of people. On table i, I will put x[i] people who like dish y[j]. Now, if we rearrange x and y in such a way that none y[i]=y[i+1] other than on some suffix of array y, we get a permutation. First x[1] elements are y[1], then x[2] elements are y[2] and so on. After that we can put the breaker and unused elements. Now, it is clear that all b[i] apart from the last one will be assigned to exactly 1 table. The optimal assignment of b[i] is so that capacity of unoccupied tables is maximized, so that last b[i] has enough tables to arrange the people from it.

Complexity of the solution is O((N+1)! * (N+M) * log (N+M)). Meaning that for N=10 and M=10, the solution would use around 4 Billion operations. Note that this is probably not the optimal solution and that it is likely that there exists a faster one.

Also, if you noticed any mistakes in my solution or proof, please let me know. Example:

15 4

1 1 1 1 1 1 1 1 2 2 2 2 2 2 2

5 4 4 2

One of the optimal permutations is:

2 2 2 2 2 1 1 1 1 2 2 1 1 1 1 B

It will result in an array b like this: 5 4 2 4. We will assign 5 4 and 2 to the tables, leaving one table with capacity 4. We can assign the last element of array b (4) to it.

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    23 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Obviously you can do dp on subsets in $$$O(3^n m)$$$. Store the mask of people you didn't already sat somewhere, iterate over all tables, choose a mask of people to sit at this table. That's basically memoization of your bruteforce.

    But then we only care about the multiset of sizes of groups with the same taste, and the sum of these numbers is at most $$$n$$$. So we can write dp with memoization on reachable states, there will be at most $$$O(m \sum_{i=0}^{n} p(i))$$$ states where $$$p(x)$$$ is the [partition number](https://en.wikipedia.org/wiki/Partition_(number_theory)). This should work reasonably fast for $$$n \le 50$$$ or something.

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23 months ago, # |
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Problem F says "Kira is not very strong in computer science, so he asks you to count the number of ways to invide friends."

It should say "invite" not "invide" I think.