MrPaul_TUser's blog

Codeforces Round #748 (Div. 3)
By MrPaul_TUser, history, 3 years ago, In English

Hello, Codeforces!

I'm glad to invite you to amazing (we've tried to make it such) Codeforces Round 748 (Div. 3) which will start on Oct/13/2021 17:35 (Moscow time). This round is made by me (MrPaul_TUser), a significant contribution to the round was also made by MikeMirzayanov and BledDest.

The round contains 6-8 problems. The difficulties of the problems are expected to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round, it will be a 12-hour phase of open hacks. We tried to make really strong tests — just like you will be surprised if many solutions fail after the contest is over.

You will be given 6-8 problems and 2 hours 15 minutes to solve them.

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

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

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

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

Thanks to _c_k_r_, Vladosiya, Ahmed_Salama, BitHashTech, powergee101, ncduy0303, ashmelev, God_Of_Code, OlegZubkov, mahade31, arjunsanjeev7, and Killever for testing the round and improving tasks.

Good luck and have fun!

UPD Editorial

  • Vote: I like it
  • +193
  • Vote: I do not like it

Comments Comments (252)
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3 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Looking forward to an interesting round!

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

Hoping for a good contest :)

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3 years ago, # |
← Rev. 3   Vote: I like it +41 Vote: I do not like it

Looking forward to change my color :)

Spoiler

Plot twist:

Spoiler

Plot twist again:

Spoiler
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3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

My first unrated round.

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

Nevermind, but can someone remove cheaters in last round

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +12 Vote: I do not like it

    +1

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      3 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      I know bro... I feel sad a bit and very unfair

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      3 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Now, I am feeling it . :(

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

      That's not true, in the round before the last one, I got +24 after the contest, getting to 1599, and then, cheaters were caught and I hit 1600! Additionally, seeing this comment, we can conclude that cheaters have been/are being caught for the last round as well. Codeforces staff puts in a lot of effort, please don't belittle it with baseless accusations/claims.

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

    Could anyone explain me about what happened in the last round?

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      3 years ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      apocalypse and destruction

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

      I became eligible to participate in this round. :P

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3 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

-1

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    3 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    VERY UNPAIR

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    It would be nice to recalculate the rating for honest users.

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      3 years ago, # ^ |
      ← Rev. 2   Vote: I like it -20 Vote: I do not like it

      +1

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        3 years ago, # ^ |
          Vote: I like it +14 Vote: I do not like it

        Did you just say "Shame!!!"? What in the world do you think gives you the right to say that? Even you know that it's not a right thing to say, else you wouldn't be cowering behind an alt.
        Please don't ruin the comments section of every blog you see. If there are some issues, create a blog or comment on blogs that already exist for that issue.

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          3 years ago, # ^ |
          ← Rev. 3   Vote: I like it -33 Vote: I do not like it

          -1

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            3 years ago, # ^ |
            ← Rev. 2   Vote: I like it +33 Vote: I do not like it

            So you think that it is fair not to remove the cheaters?

            Where did you see him say it is fair not to remove cheaters ? Aren't you making up the truth? Do you think it's fun to deliberately destroy the environment of the comment section? Aren't you ashamed of yourself? And you just put the blame on Codeforces?

            It don't even take 1 hour to remove cheaters & update the ratings but don't know why they are not doing this from last 2 weeks.

            Don't you know that admins have to do a lot of things every day? And how can you be so confident? Less than an hour? Don't you know how much manpower and material resources it takes to run a website? Do not know how cumbersome it is to process all the competition information of more than 20000 contestants and then calculate the rating? Take your brain before you talk, and don't talk nonsense if you don't understand these things, okay?

            Also, can you stop yelling "Remove cheaters! Remove cheaters!" in the comment section of each official blog? You think admins doesn't know these things? Don't know how many contests recently? Don't you know how much admins workload? Please, be considerate of others. You're happy to make others hate you, aren't you?

            I didn't want to say anything at first, but I can't stand seeing the comment section like this. I hope this is the last time I get angry about such kind of shitposts.

            Sorry for my bad English.

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

              As an ex-admin of several huge online projects, I would like to agree with EasonCF: you can't imagine what enormous amount of work such kind of projects require.

              And the subject of the problem is insanely ridiculous: cheaters? Does it really bother you??? Rating? Are you serious?

              You've got a lot of great CP problems WEEKLY! Great tests, fantastic database of solutions and editorials, archive, and what are you talking about: cheaters and rating??? Be ashamed!

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                3 years ago, # ^ |
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                Big fan of all the positivity you just radiated :)

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    3 years ago, # ^ |
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    Can you create a blog for this problem??

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3 years ago, # |
  Vote: I like it +20 Vote: I do not like it

Despite my earlier criticism , I have come to appreciate naman1601's light hearted memes under every contest announcement. I would request him to bless us with another light hearted meme.

P.S. He just became purple again. Everyone please upvote him as a congratulations.

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3 years ago, # |
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Looking forward to increase my rating.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

In div-3 do all questions have same score? Like A and F have same value upon solving?

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    3 years ago, # ^ |
      Vote: I like it -9 Vote: I do not like it

    no, rating distribution will be updated later most likely.

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      3 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      no, in Div 3 and Education rounds the scores for all tasks are the same.

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

It really hurts when rating is exactly 1600 :(

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Hoping to be a blue coder to change my profile picture xD

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3 years ago, # |
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ah yes, all that's left is to convince my parents to let me stay up till 1am.

Or...
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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it
Spoiler
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3 years ago, # |
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Best of luck to everyone on this round!

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

But what if you're a beginner like me who hasn't taken part in two rated contests yet?

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    3 years ago, # ^ |
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    it says it'll be rated anyway. we're just being excluded from the official standings, i don't think it'll affect us in any way tho

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I hope everyone can raise rating

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3 years ago, # |
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Question
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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Allfather bless me so that my color changes (to the better side) this time.

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

Hi, Im New, but what languages will be accepted, like python c++?

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

    both, my dear friend! There are still more language available for you :)

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3 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Today is my birthday, and I will participate in this round.... Yoooo

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

Hope my color changes after this round ! )

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

It's cool that the penalty for a wrong attempt will be 10 minutes

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3 years ago, # |
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I hope to stay newbie after contest :)

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

    You are an inspiration truly! You are trying hard and didn't give up! [user:Khedr_]Best of luck!

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3 years ago, # |
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Good luck.

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3 years ago, # |
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Hoping for change my color =))

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

As a tester, I tested…
GL&HF

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I hope that the penalty of all contests will be 10 minutes instead of 20, not only Div3

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3 years ago, # |
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I'm not reading Announcement comments anymore.. Man I feel too old

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

    You're old in this platform.

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

I hope my rating will reach 1900 after this contest Good luck !!!!

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Any hints on F?

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    3 years ago, # ^ |
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    I think it can be solved using digit dp.

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3 years ago, # |
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solved G faster than rainboy pog

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    3 years ago, # ^ |
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    btw G felt too easy for its position...

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

      can you give me idea how to solve G, i got stuck on it used dfs and take max depth of node , give me bad output.
      edit my bad i mean problem E sry

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        3 years ago, # ^ |
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        dfs? how?

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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it +12 Vote: I do not like it

        my $$$O(n+q)$$$ solution for pG:

        note that we can change between $$$($$$ and $$$)$$$ (similarly, $$$[$$$ and $$$]$$$) at no cost, let's transform $$$($$$, $$$)$$$ into $$$a$$$ and $$$[$$$, $$$]$$$ into $$$b$$$

        then we can subdivide each even length substring into blocks of length 2. there are 4 possible blocks: $$$aa, bb, ab, ba$$$.

        A block of $$$aa$$$ or $$$bb$$$ cost 0, while $$$ab$$$, $$$ba$$$ cost 1. $$$aa, bb$$$ cancels themselves, and $$$ab, ba$$$ cancels each other. (both $$$abba$$$ and $$$baab$$$ cost 0) so the problem reduced to find how many $$$ab$$$ and $$$ba$$$ blocks are there in a range, and cancel them while possible.

        the answer is the difference between the number of $$$ab$$$ and the number of $$$ba$$$ in a given range, using prefix sum can solve each query in $$$O(1)$$$ time with $$$O(n)$$$ preprocessing.

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          3 years ago, # ^ |
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          Thanks bro!I never imagined G could be that easy so never gave it an honest try T_T

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        3 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        in pE do DFS $$$k$$$ times result in a worst-case $$$O(nk)$$$ time complexity, it occurs when the tree is also a "chain" (edges are $$$(1, 2), (2, 3), ..., (n-1, n)$$$)

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

Is D2 about iterating over the numbers which have more than half of the bits set in $$$[1,2^{N}]$$$ ?

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    you should just get the difference of between every 2 elements in the array and try all of their divisors

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

      Yeah dumb of me to not notice that, btw please accept my apologies I hacked your E :(

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        3 years ago, # ^ |
          Vote: I like it +21 Vote: I do not like it

        That's the evilest thing I have ever seen

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        3 years ago, # ^ |
        ← Rev. 3   Vote: I like it +1 Vote: I do not like it

        How is it TLE when I'm moving from the leaves directly to the next operation leaves

        Edit OMG I forgot to add break after i edited my code the last time it was correct

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    3 years ago, # ^ |
    ← Rev. 4   Vote: I like it +16 Vote: I do not like it

    My method:

    If any element appears >= n/2 times, answer is -1.

    Otherwise: for each choice of reference value a(i) from the array, consider the set of differences abs(a(j) — a(i)) for all j. Where the difference is 0, increment a counter z. Otherwise find the factors of abs(a(j) — a(i)), which can be done in O(sqrt(max(a))) time. Keep a count of how many times each factor occurs. Any factor such that count(factor) + z >= n/2 is a possible answer, and the max of these candidates is the final answer. Remember to reset the factor counter for each choice of reference value a(i). Total time complexity O(N^2 * sqrt(max(a)) )

    Solution: link

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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

hoooolly shit that was intense i accidentally solved D1 like how it should be solved in D2

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

:( ? Why D1. n<=40 but sum (n) <= 100 ?

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3 years ago, # |
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Any idea how to do D1? I tried doing it by taking the GCD of all the differences of pairs but I was getting WA on test case 3

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

    first find the minimum element, then take GCD on the difference between the minimum value and each element (except the ones with minimum value).

    output -1 if all elements are same.

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    3 years ago, # ^ |
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    I guess you forgot to sort array

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

      Sorting array is unnecessary. Any member of the array can be used as the reference value.

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

    You are also required to check if all items are equal. In such case, answer is -1(Infinite)

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

    I checked your solution here https://codeforces.net/contest/1593/submission/131802472

    and you have an integer overflow issue as you use vll a(n); which expands to vector<ll> of which ll expands to unsigned long long int.

    When you perform a[i] - a[j] and it is unsigned, you must take extra precaution.

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

Quite interesting problemset! Thanks for this round!

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3 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Really nice D2

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

What the heck is this second test for E?

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

can someone help me by telling what i am doing wrong in Problem-E

solution link : https://codeforces.net/contest/1593/submission/131852543

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

    Draw the following case on paper.

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

    consider this example.

    n=6 k=2

    tree

    Correct answer:

    • step 1 -> 1 4 6 removed
    • step 2 -> 2 5 removed

    Answer is 1

    Your answer

    • step 1 -> 1 4 6 removed
    • step 2 -> 2 5 3 removed

    Answer is 0

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3 years ago, # |
  Vote: I like it +87 Vote: I do not like it

I liked the problems, but this is really irresponsible behaviour. I understand sometimes issues may escape the testing phase too, but quietly changing the statement without announcements trying to not address the that there was an issue is not acceptable. I spent a lot of time trying to debug my initial submission, and finally when I couldn't believe that this was incorrect, I tried refreshing the page.

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +9 Vote: I do not like it

    Following up on this, problem E did not have limit of sum of K over test cases, neither had tests which had this large sum. I don't understand why authors didn't put max tests? Weak tests in general on many problems, as you can see from the number of hacks and FSTs.

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3 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Isn't this solution correct for problem E

Number of nodes left = total nodes — number of nodes which have distance less than equal to k to the closest leaf node.

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    3 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it 
    1
6 2
1 2
2 5
5 6
2 3
3 4

    Your answer should be 1, but using the above logic, you will get 0

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

How to solve C?

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

    the cat moves exactly $$$n-1$$$ times, so the total times you can move a mouse is also $$$n-1$$$.

    To save as many mice as possible, always move the mouse nearest to the hole first. You can sort the array of the distance of each mouse to the hole, then try to save the mice from the nearest until the total distance exceeds $$$n-1$$$.

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

Looking at the hacks, I foresee many TLEs on E, especially for those that BFS from leaves and keep track of degree of the nodes

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

    If done smartly this approach is fine. Only perform updates when a node is trimmed, and the total number of updates is equal to the total number of edges, which is N-1. Why should that TLE?

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      3 years ago, # ^ |
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      Ok looking closer at the ones that got hacked and unsuccessful hacks by galen_colin (don't wanna tag for no reason), during one cycle of BFS they push the next elements into a separate queue/DS, and then replace the queue/DS. Maybe overhead from replacement is causing the TLE.

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

        Yes I see. I just used a priority queue and did it Djikstra-style to avoid anything like that. That's not to say I won't be hacked though — who knows :)

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

        I think the problem is that sum of k is not bounded and many people (including three of my div 1 Ukrainian friends and myself) just had a while(k--) loop or something similar.

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3 years ago, # |
  Vote: I like it -9 Vote: I do not like it

The problems were good but the samples were really weak

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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    samples are not supposed to be strong lol... what's next? all tests are opened?

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      3 years ago, # ^ |
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      They could've added a test case for -1 in problem D1

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        3 years ago, # ^ |
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        imo, finding of cornercases must be a part of a solution

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          3 years ago, # ^ |
            Vote: I like it -12 Vote: I do not like it

          Imo there is no need for pretests at all let hackers do the deed.

          Seriously this round is a joke.

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        3 years ago, # ^ |
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        They did, I forgot that, got WA on pretest-6.

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

I know you are impatiently waiting for the editorial, so I've decided to share with you my easy-to-understand solution for problem A.

int t; cin >> t;
while (t--) {
    int a, b, c; cin >> a >> b >> c;
    cout << max(max(b, c) + 1 - a, 0) << ' ';
    cout << max(max(c, a) + 1 - b, 0) << ' ';
    cout << max(max(a, b) + 1 - c, 0) << '\n';
}
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    3 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Wow, very simple and efficient code. But I write all possible way with if else statements that what newbies do.

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

Can someone help me figure out why am I getting a different output (which is one more than the required output) when on codeforces judge, but when I run it locally on my compiler it gives the correct output for the same test case.

Here is my submission

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

    Because you are including death[0] in your answer. death[0] is not set anywhere, so its value on your local machine may differ to that on the server.

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      3 years ago, # ^ |
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      Nice catch, but after fixing that, I have a similar problem but this time a 3 instead of 1, I tested it on my compiler and it gives the correct result,

      Submission

      PS: I have realised my solution is incorrect but I still want to know the reason for the inconsistency in the outputs for future reference.

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        3 years ago, # ^ |
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        It's very difficult to say for sure but I have (recent) experience of similar things and my finding has always been that I've been accessing something that I didn't mean to, or that I didn't think was there, by mistake.

        A hypothesis might be that your adj is not resetting itself in the same way online (since you are not explicitly clearing the elements). But I confess I cannot say for certain.

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        3 years ago, # ^ |
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        You're setting some death[i] to be 1, but I think you might be assuming that the others would be 0, which is not correct.

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3 years ago, # |
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Why multisource bfs isn't working in E?

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    3 years ago, # ^ |
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    Reading your code, your method appears to be "ans(node) = min distance from node to any leaf". This method is not correct. Consider why this is incorrect with the below case, to which the answer is 1 but to which your code gives 0. During the 2nd iteration, node 1 remains.

    1

    6 2

    1 2

    1 3

    3 4

    1 5

    5 6

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3 years ago, # |
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Can Anyone Help Me Out to find why am i getting wa for problem E ,My Submission : 131857557

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

These types of Solutions are getting hacked with TLE, but it doesn't seem to exceed TL. This is O(KlogK) solution. Perhaps this is because of Java?

Correct me if I am wrong. Thanks.

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    3 years ago, # ^ |
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    I don't use java much but Arrays.sort(a) in java has worst-case time complexity of order n^2. Therefore the worst-case time complexity of your program is n^2 which gave tle.

    You can read more here

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      3 years ago, # ^ |
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      Oh okay, thanks for the information. So there are more chances of java solutions getting hacked whenever sorting is needed in the solution, right?

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

        Yes, you are correct. I have seen tons of solutions in java getting hacked on codeforces due to Array.sort(a). You should switch to Collections.sort(a) or suffle array before using Array.sort(a)

        Array.sort(a) uses quick sort only on primitive data types. So if there is no primitive data type Array.sort() is ok.

        An example of Collections.sort() can be seen here in the SecondThread solution

        Notice that he uses collections.sort and fastio for speeding up the code. Collections.sort() wont get hacked as the worst case of it is nlogn

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3 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Anyone else with the following on Test case 2 for E

wrong answer 37th numbers differ - expected: '1', found: '0' ?

What is this test case possibly? I did multi-source bfs starting with all the leaves of the tree.

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    3 years ago, # ^ |
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    I have this WA too, but i did dfs from a random leave

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    3 years ago, # ^ |
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    1
1 0

    Yours gives 0. Ans should be 1

    Edit: My bad k >= 1

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    3 years ago, # ^ |
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    1 6 2 1 2 2 5 5 6 2 3 3 4 The ans should be 1. Multisource bfs gives 0

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    3 years ago, # ^ |
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    try only pushing those nodes which are connected to the current node being processes(u) after removing u from the tree in the multisource bfs. If you didn't do it this way!

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3 years ago, # |
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Video Editorial of E (Gardener and Tree)

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3 years ago, # |
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It seems that D2 has many suceesful hack. Is pretest weak or some conercase ?

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3 years ago, # |
← Rev. 3   Vote: I like it -9 Vote: I do not like it

when I run this code on some other compilers like gfg ide, visual studio I get correct O/P for sample test case 1st but on codeforces I am getting W/A on test case 1st, IDK how this happens ??

Code:

https://ide.geeksforgeeks.org/zN7ERcyWaB

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    3 years ago, # ^ |
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    Try with language GNU G++17 9.2.0 (64 bit, msys 2) .

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3 years ago, # |
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Anyone solved E in O(n) by finding 2nd largest length from every node to leaves?

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    3 years ago, # ^ |
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    Nice idea :)

    I think your algorithm is actually O(K + N log N) though. You use set (insertion log N). I think this explains your long run time — set is slow and you perform a lot of insert and erase operations.

    If you wanted this to be truly O(N) you could store, for each node, just the two greatest distances you currently have, and update them in place without need for actually sorting (either through use of set or otherwise).

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      3 years ago, # ^ |
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      I brute forced like everyone.

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    3 years ago, # ^ |
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    Yup! I did just now

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3 years ago, # |
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Why there is so much hacking in E? Which test case it is failing???

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    3 years ago, # ^ |
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    The majority of hacks seem to be time limit related, where the user has, K times over, created a whole new queue which could have size of up to N.

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      3 years ago, # ^ |
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      How though? I believe only the new leaves would be added to the queue and that should be of the order of the nodes since any leaf would be added at most once.

      here's my code

      Sorry for the poor variable naming. Link to submission

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3 years ago, # |
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Nice problems, but weak testcases. Thanx.

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3 years ago, # |
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Were the weak tests on D1 and D2 on purpose? MrPaul_TUser

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3 years ago, # |
← Rev. 2   Vote: I like it +67 Vote: I do not like it

really strong tests

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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Some people hacked a lot of solutions, it's hard to read a lot of solutions (at least for me) and find which ones can be hacked and which ones are not.

I think there is a way to test a lot of solutions automatically. Is that true ? and is it allowed ? and do you have any idea how to do that ? it would be great if you can help.

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3 years ago, # |
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I got wrong on TC2 on C (my submission).... Please help!

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    3 years ago, # ^ |
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    If you would like a testcase that fails:

    10 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999 1000000000 1 999999999

    It has to do with not checking for your vector being out of bounds when you incremented i.

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3 years ago, # |
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Here's my Solution to E 131874293. Do you guys have a better approach or this solution is fine.

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    3 years ago, # ^ |
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    I think the algorithm itself is fine, and the implementation has room for improvement.

    deg is not necessary, you can use ar[i].size() instead.

    leafs and new_leafs can actually be vector rather than set

    And consider output '\n' for new line instead of endl since endl flush the output buffer each time, which may make the running time longer

    You can refer to my submission, which has basically the same algorithm as yours and has 3x faster in running time.

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      3 years ago, # ^ |
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      Thanks for your time, also isn't set better for removal of leaf nodes.?

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        3 years ago, # ^ |
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        But in leafs and new_leafs, all removals are removing all elements, which is actually clear the set.

        If you need to remove individual elements, using set is better (so we use set not vector for the adjacency list in this problem).

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3 years ago, # |
  Vote: I like it -32 Vote: I do not like it

I have a serious concern about the problem A.

Is it fair to allow such submission getting AC?

To compare, this one gets TL. Of course, the only difference is the line 3: #pragma GCC optimize ("O0")

I guess that checker must prohibit the loop elimination explicitly when compiling the solution, because that's what GCC does in this case.

Would be nice to listen to the contest authors' opinion. cc MikeMirzayanov MrPaul_TUser

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    3 years ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    Well, I'm kinda confused with people's reaction. Could somebody from those who downvoted to comment on that?

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

      Why in the world would getting AC by this submission not be fair? Modern compilers nowadays apply an overwhelming number of optimizations, you cannot know all of them for sure. Which of the optimizations must be considered "unfair"?

      Loop elimination is one of the simplest optimizations a compiler can do. If G++ is able to do it while others are not, well, one has to blame the other compilers for that.

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        3 years ago, # ^ |
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        Which of the optimizations must be considered "unfair"?

        I didn't tell the optimization itself was unfair. And I know well what compiler optimization is.

        Also, I didn't participate the contest and I don't ask for some other people's solutions to be retested, etc.

        I'm just pointing out that the solution which presumes 10^13 operations to be performed gets AC verdict.

        Moreover, it does perform those when being compiled with -O0 or -O1. I guess, there's also some sort of flag to disable such optimization while having -O2 or -O3 enabled too.

        So my intention was to raise a question about potential elimination of those optimizations while testing a solution.

        Why? Because the compiler must not influence the correctness of the solution. And the solution above is wrong indeed, it must not be accepted as a valid one, IMHO.

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          3 years ago, # ^ |
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          The solution is not wrong. It gives correct answers for all the tests. If it is compiled without optimizations, then it will get TLE, but it does not mean that it's wrong.

          But the question is: where is the margin between optimizations which influence correctness and optimizations which do not?

          If you really want to be sure that your code will not be affected by obstinate optimizations, then write code in the assembly language. The processor will unquestioningly execute all the instructions you would write.

          After all, this is not a theoretical olympiad in informatics where participants are supposed to describe the algorithm and prove its complexity. If we code in high-level programming languages, then it's essential for a compiler to speed up the code wherever it's possible.

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            3 years ago, # ^ |
              Vote: I like it -10 Vote: I do not like it

            The solution is not wrong. It gives correct answers for all the tests.

            The fact that it gets AC for all test doesn't mean it's not wrong.

            But the question is: where is the margin between optimizations which influence correctness and optimizations which do not?

            Of course we can't answer this for sure, but at least loop elimination prohibiting could be a good place to start.

            I don't insist on it, I was just curious of the authors' opinion.

            Anyway, what motivated me to start this conversation was getting AC by the wrong solution, and I'll keep claiming it is wrong just because it's not completely correct (whether or not you agree with such interpretation).

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              3 years ago, # ^ |
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              Once again: the solution is not wrong. It's slow. It's a slow correct solution. Not all correct solutions are supposed to fit into the time limit, as well as not all slow solutions are supposed to be rejected. Some of them may pass, though.

              at least loop elimination prohibiting could be a good place to start

              Sorry, but this is not a constructive way of discussion.

              Here is the discussion on turning on some optimization flags (-O3, -funroll-loops etc.) for G++ by default on Codeforces. You can't stop progress.

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3 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Has anyone done F using meet in the middle technique, instead of using normal dp. In which divide the string into 2 parts and checking each part individually ? I tried but got TLE in TC 7. Though I think it will pass if 2s is given as approx complexity is $$$O(2^{n/2}*{n/2})$$$

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

Can 131830101 be hacked? I didn't stop the while (k--) loop even after the queue (q in my code) is empty and my solution has extra $$$\log N$$$ factor because of std::set.

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    3 years ago, # ^ |
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    after q becomes empty the while (k--) does nothing (and won't consume a lot of time) because the for (int sz = LEN(q); sz > 0; --sz) inside your while (k--).

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3 years ago, # |
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Can somebody please explain where I am going wrong in problem E?
My Submission
This code is giving wrong answer on Test 3.

My Approach:
First, I found the centroid of the tree. Then I ran a DFS from this centroid and maintained an $$$ans$$$ vector where $$$ans_i$$$ will store the number of operations after which $$$i^{th}$$$ node will be removed.
1. ans[leaf node]=1.
 2. ans[node having only one child] = ans[child]+1.
 3. ans[any other node] = Second_Max(ans[all children])+1.

I am taking second maximum because for a node having $$$C$$$ children, atleast $$$C-1$$$ of its children must be removed for this node to become a leaf node(removable).

Final answer will be number of nodes having $$$ans[node]>k$$$.

Thanks in advance.

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    3 years ago, # ^ |
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    oooooh! Don't know where is the error but the logic is extraordinarily..

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    3 years ago, # ^ |
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    Can you prove that taking the centroid of the tree is always correct in your solution?

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      3 years ago, # ^ |
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      Well, it was kind of a intuition. But I think about it this way, since centroid is kind of a balance point because it divides the tree into forest of subtrees with size at most half of original, so centroid must be last one to be removed if we follow the procedure from the problem. Please correct me if I'm wrong.

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3 years ago, # |
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Hello everyone here, Yesterday I solved 3 problems and got an accepted verdict. Today about half an hour back it still showed an accepted verdict. But now 2 out of 3 problems are showing a queue verdict. May I know the reason for this?

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3 years ago, # |
← Rev. 4   Vote: I like it -20 Vote: I do not like it

We tried to make really strong tests — just like you will be surprised if many solutions fail after the contest is over.

You intentionally gave n <= 40 in d1 where O(n) soln is easily working just to mislead participants in directly using brute force without much thinking , however if u would have given n in range 1e5 then this thing wouldn't have happened , so it is really not expected from ur side , And also in problem C , u gave time limit 4 seconds and as a result of which some O(1e9 range) solutions got accepted and now they failed system tests.

Just because of your really strong tests i got +2000 my expected place since i solved d1 before b and c so it was really disheartening to see :(

Those who are downvoting will get to know how it feels when it happens to them :)

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    3 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    That's your problem. They all remind that passing pretests doesn't mean to pass system judge :))

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      3 years ago, # ^ |
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      When you will be at the receiving end you will understand. I hope you will remember what you preached at that moment.

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3 years ago, # |
← Rev. 15   Vote: I like it 0 Vote: I do not like it

Accepted.

TLE.

Just changed vector with array and now it is accepted. It feel very bad (T_T) During the contest my solution was accepted and after systest it gave TLE). First time I had got under 2000 rank. It was meant a lot for me. And now this happened because of vector vs array:|

Spoiler

PS: Do anyone know how to use vectors as fast as arrays?

PS: TLE code accepted on removing #define int long long.

PS: But, creating dynamic array example: int*m = new int[k + 1]; gives MLE. 131911303

PS: Global vector working fine.

By the way nice contest ^_^. IMO there should have such testcases during the contest so that we can correct this things during the contest.

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3 years ago, # |
  Vote: I like it -45 Vote: I do not like it

Dear MrPaul_TUser and MikeMirzayanov

Testing the D2 question cases seems very inappropriate !

See my submission , it gets wrong on test 56! While it received accept in pretests. It costs me a sharp drop in rankings so from near 1500 i become 3500 because of weak testcases! So i really please you to reduce the testcases for D2 and rejudge this question.

131848968

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    3 years ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    You mean your code fails on a certain testcase meaning that your code isn't correct within the given constraints and hence the testcases which your code can't handle should be removed?

    Nice

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3 years ago, # |
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Can anyone explain why my submission causes RTE, please? 131823111 Same code in Python3, I passed test

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    3 years ago, # ^ |
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    Technically, in C and C++ (and most other languages I guess), any integer other than 0 is considered true, so your comparator function that you passed into the sort function to sort your input in descending order is not working as intended, I replaced return a.first - b.first; with return a.first > b.first and the code worked perfectly.

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

I can't get over this that was my best performance since I started on this site I had a break command in my previous WA submission for E and i fixed everything but deleted the break by mistake it passed and got hacked :'( I could have reached expert without that stupid break command went from 250 to 670 someone give me 10 bags of tissues :'( :'( :'(

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    3 years ago, # ^ |
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    Don't worry man. Just do your best next time.

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    3 years ago, # ^ |
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    dont worry bro! soon u ll become expert

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3 years ago, # |
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When will the rating changed?

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3 years ago, # |
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When will the rating changed?

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3 years ago, # |
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Why do I see this contest in my unrated contest. My rating is < 1600

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    3 years ago, # ^ |
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    The rating hasn't updated yet. You need to keep waiting for a while until the rating get updated.

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    3 years ago, # ^ |
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    You may be not a trusted participant because you played only one rated round before this one.

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3 years ago, # |
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I often see it is written in the open hack or system testing phase that only trusted participants. What does this mean?

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    3 years ago, # ^ |
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    Read the announcement carefully and you'll find:

    Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as trusted participants of the third division, you must:
    take part in at least two rated rounds (and solve at least one problem in each of them),
    do not have a point of 1900 or higher in the rating.

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3 years ago, # |
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When will the editorial be released Waiting for the explanation of D2 and E problems

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3 years ago, # |
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Well, I have waited for 4 hours from the open hacking ended. However, I didn't see any signs about rating. I'm very nervous and I want to ask when will Codeforces rate? Thanks and I know a lot of people are waiting like me =))

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3 years ago, # |
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i am wondering(HELP) what is the significance of n is even in D1 problem

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    3 years ago, # ^ |
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    Nothing. This detail is more important in the harder version, D2.

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3 years ago, # |
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.

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3 years ago, # |
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Can we use meet in the middle for D2??

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    3 years ago, # ^ |
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    Possibly someone smarter than me will have a way but I can’t see it. For me this is brute force over all possible smallest values of your set where you try to find the largest factor which exists in N/2 of the differences with that element (itself included).

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      3 years ago, # ^ |
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      Now, I think that (meet in the middle one)approach would result in TLE since in that approach you will divide the array into 2 equal sizes and look for all possible subsequences that would result in TC(2^20) which is roughly 10^6 and if you try all possible values of k you will get TLE

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

In problem C: "The actions are performed until any mouse hasn't been hidden or isn't eaten."

That should be while not until. Every programmer should know it regardless of English proficiency.

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3 years ago, # |
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Congratulation @Viettran, 8528190072TuanHD

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3 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it

It's sad to see that top1 and top2 in div3 are cheaters (U need to use unofficial leaderboard to see them). They have multiple submissions to different problems within one minute using completely different templates xD MrPaul_TUser maybe something can be done about that, althought I doubit it.

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3 years ago, # |
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I have given all three rounds made by you MrPaul_TUser and personally found all this rounds very interesting now I am excited and waiting for your next rounds and thanks alot for writing such nice and interesting problems:)

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3 years ago, # |
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I have solved the problem E in the following way, however getting TLE #4. Here's the solution.

I'm maintaining a sorted set of pairs of vertexId and its adjacency list (in fact, set in the implementation). In each operation, removing pairs from the set where the adjacency list size is one. And also removing it from its the only neighbour. And this solution gets timelimit but I believe it shouldn't have very bad complexity.

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    3 years ago, # ^ |
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    Actually the constraints of n and k both are 10^5 . But the interesting thigh which we have to notice in question is that author only given that the sum of n over all test cases will not exceed 4×10^5.So if you solve this problem in O(n) then it will be O(n) only but if you will solve this question in O(k) so actually you are solving it in O(T*k). That's why your code is getting TLE. I hope that I cleared your doubt if not then let me know I will try to elaborate it or simply it.

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      3 years ago, # ^ |
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      Well, when the sorted set is empty, I break, it shouldn't go over n, even n/2 operations as in each operation there are at least 2 leaves/vertices erased. I think my solution is O(nlogn) due to the sorted set.

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

        Yes, Actually sorry I didn't read your code completely so I thought you are not breaking that for loop of k. If you will get the reason for which your code is getting TLE then please also let me know. It will be informative and helpful for me also :)

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

When will the editorial be released?

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

There is a brute force solution for D2 that passes https://codeforces.net/contest/1593/submission/131966435 . Did you intend for optimally written brute force to pass for D2 ?

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3 years ago, # |
← Rev. 6   Vote: I like it 0 Vote: I do not like it

ok

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

    Your comment was very funny, so I checked your profile out of curiosity. Useless code inserted to avoid plag checker (successfully in many contests). I didn't expect you were a cheater. What a plot twist.

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

According to here, the editorial should be posted no later than 1 day after a round. So why the editorial of this round hasn't been posted yet?

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3 years ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

just my opinion nothing else:

Spoiler

thnx for the contest authors! eagerly waiting for editorials :D

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

Can anyone help me why this 131912878 fails for problem E ?

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

Editorials???

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3 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it
meme
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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

No Editorial till now

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

2 days and still no editorial . WTF!!!

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

I hope to be blue before the editorial is published

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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

someone know why I get TLE for this java submit 132153643?

It's totally same solution with my c++ submit -> 132153310

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3 years ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it

Both this announcement and the editorial do not seem to be attached to the round. Thanks for the round!

Update: fixed already.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

why rating havn't change after remove cheaters? i think it should increased a bit

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

About my code similarity with user "grafOn". That is my second account that i use for checking if my code is right and right after that I submit that code on my profile. I didn't know that was breaking the rules. I am sorry for that. I don't know how to prove to you that it is my second profile. Please tell me if i broke the rules of CF.

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

    read the Terms of agreement

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

      I read it, but there was nothing about having two accounts at the same time. I just don't understand why i can't submit the same code on two accounts during the contest. After all this two accounts are mine and there is no second person that coding and submitting the same code as mine.

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        3 years ago, # ^ |
          Vote: I like it +6 Vote: I do not like it

        I just don't understand why i can't submit the same code on two accounts during the contest.

        I just don't understand why every LGM does not make 10 alt accounts and stay on top of the leaderboard themselves :)

        After all this two accounts are mine and there is no second person that coding and submitting the same code as mine

        Well you yourself are the second person. What you are doing is checking if your submission is correct so you don't get a penalty on your original and can stay on top of the leaderboard. You are registered as 2 different people in the contest.

        there was nothing about having two accounts at the same time

        Next time when you click register for a contest, try reading what the box above it says.

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

Can somebody help me find a bug in my solution of problem F pls:

Spoiler
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