jdurie's blog

By jdurie, 4 weeks ago, In English

Hi, Codeforces!

I welcome everyone to participate in Codeforces Round 941 (Div. 1) and Codeforces Round 941 (Div. 2), which will start on Apr/27/2024 17:35 (Moscow time).

Both divisions will be given 6 problems and 2 hours to solve them.

One of the problems was also used in the Yandex Cup finals, which was authored by tourist, so if you participated in the Yandex Cup finals or know the problems, please refrain from participating. All other problems were created and prepared by me.

I would like to thank:

Good luck to all the participants!

Score distribution:

Division 1: 500 — 1250 — 1500 — 2000 — 2500 — 3500

Division 2: 500 — 1000 — 1500 — 2000 — 2250 — 3000

Update: Editorial is up

Congratulations to the winners!

Div. 1:

  1. jiangly
  2. conqueror_of_tourist
  3. nocriz
  4. tatyam
  5. bruhopen

Div. 2:

  1. Chengxixi
  2. SlhShn
  3. SlavicC
  4. awooga
  5. zyh_helen
  • Vote: I like it
  • +254
  • Vote: I do not like it

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4 weeks ago, # |
Rev. 18   Vote: I like it -7 Vote: I do not like it

What if someone has seen this Yandex Cup problem and participates.

Will it be skipped?

UPD:Please do not downwote i just wanna learn what's gonna happen

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4 weeks ago, # |
Rev. 2   Vote: I like it +80 Vote: I do not like it

yandex cup qual? semifinal? or final? (Because if not final,I can participate.)

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

    Only final. Sorry for misunderstanding

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4 weeks ago, # |
Rev. 4   Vote: I like it +37 Vote: I do not like it
Finally my turn to say......
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    3 weeks ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    my first div1, too. But as you can see, I became Expert again. :(

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4 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

I hope we will be honest enough to refrain who have participated in Yandex Cup. Otherwise, it will adversely affect the rating distribution. Because you know, the problem authored by "tourist" will surely hold a huge score.

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

In which division one of the problems was also used in Yandex Cup, or both?

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

:O tourist problem

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4 weeks ago, # |
  Vote: I like it -75 Vote: I do not like it

As a participant, I would like to get upvotes as minus looks ugly in my profile!

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    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it -19 Vote: I do not like it

    Oh.. I feel so sadge. Bro got down vote to oblivion.

    Spoiler
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    3 weeks ago, # ^ |
      Vote: I like it -15 Vote: I do not like it

    I am really sorry, I clicked downvote by mistake.

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

tourist orz

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4 weeks ago, # |
Rev. 8   Vote: I like it -25 Vote: I do not like it

I HOPE I GET +1 DELTA IN THIS CONTEST

why soo many downvotes ?? just why?

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4 weeks ago, # |
  Vote: I like it +9 Vote: I do not like it
Meme
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4 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

jdurie

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4 weeks ago, # |
Rev. 6   Vote: I like it -42 Vote: I do not like it

I am Gona Become a Pupil in this one.

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4 weeks ago, # |
  Vote: I like it +151 Vote: I do not like it

I don't understand why risking the round with putting an already used problem usually when we find a known problem by mistakes it creates a lot of complaints from contestants, how about now when they know that the problem already exist , probably some people will try to find it now. I'm just curious about why did you go with this decision ?

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    3 weeks ago, # ^ |
      Vote: I like it +83 Vote: I do not like it

    the Yandex Cup participants were already informed to not leak the problems i believe. It is not any different situation than testers for a contest. If testers leak problems, there is nothing we can do, and similarly here

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

      I understand, that make sense, thanks for the clarification

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    3 weeks ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    "this is all history" ans "this is tourist"

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

O tourist problem. The problem gonna be nice.

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

congratulations on the first coordinating

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3 weeks ago, # |
  Vote: I like it -103 Vote: I do not like it

where can I find Yandex Club problems Please

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

My first div1 finally

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3 weeks ago, # |
  Vote: I like it +128 Vote: I do not like it

If only one problem is from the Yandex Cup Finals, then how about not using it and making a 5-problems round? I planned to participate in this round but now can't :(

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3 weeks ago, # |
  Vote: I like it +28 Vote: I do not like it

What is the point in including an already used problem? It excludes the people who have already seen it from participating, and the ones who haven't will have no guarantee that everyone else will be honest and not participate if they've seen it.

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    3 weeks ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    There are only the people that were at the Yandex Cup Finals and participated in algorithms that have seen the problem. I think making sure that $$$19$$$ people do not participate is not that hard to do, and no big concern. Although I do agree I would like to participate, but can't now.

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3 weeks ago, # |
Rev. 2   Vote: I like it -44 Vote: I do not like it

Could you tell us in which division the Yandex problem is included.

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3 weeks ago, # |
  Vote: I like it -46 Vote: I do not like it

Sorry to ask but is it rated? and if it is, would this problem from the Yandex Cup finals affect other participants if a leak happened?

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3 weeks ago, # |
  Vote: I like it -25 Vote: I do not like it

omg tourist round

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

Hope everyone will not get stuck in A, B and C

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

tourist question is a good question,If the original question exists, it will be an unfair competition.

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

What books gennady korotkevich(tourist) read to learn algorithms?

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3 weeks ago, # |
  Vote: I like it -11 Vote: I do not like it
Meme
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3 weeks ago, # |
  Vote: I like it -7 Vote: I do not like it

My first Div.1,ok I'm ready to go back to the Expert

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3 weeks ago, # |
  Vote: I like it -15 Vote: I do not like it

Is this contest rated?

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

How to solve D?

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

    i feel like it is related to binary

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

    I was able to do a construction like:

    Add powers of 2 starting from 1 to our ans array and keep a running sum of ans sum_ans until adding another power of 2 makes sum_ans >= k. In that case, add (k — sum_ans — 1) to ans so the ans array can have all subsequences from 1 to k — 1.

    And after that you are able to add 2 * k, 4 * k, 8 * k ... to the array until the sum of array exceeds or is equal to n. But we will have some gaps where subsequences don't sum to. We can add k + 1 and 3 * k to the array to fill these gaps.

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

      The tough part here is adding k + 1 and 3 * k. How to see that?

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

        I wrote it out in terms of k and it seems like the powers of 2 + k * powers of 2 that are >= 2 give subsequence intervals from:

        [1, k — 1], [2k, 3k — 1], [4k, 5k — 1], [6k, 7k — 1] ... and it seems like adding k + 1 will cover [3k + 1, 4k], [5k + 1, 6k], [7k + 1, 8k] ... and then adding 3k will ofc get 3k since we don't have it and it also gets 5k, 7k since they can just be 3k + 2k or 3k + 4k since we already have 2k, 4k ...

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

loved C , even though i couldn't solve it

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

how to do Div2 E?

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3 weeks ago, # |
  Vote: I like it +85 Vote: I do not like it

Loved the problems.
Thanks for the round!

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3 weeks ago, # |
Rev. 2   Vote: I like it +352 Vote: I do not like it

How I solved E: open Minecraft and

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    3 weeks ago, # ^ |
      Vote: I like it +44 Vote: I do not like it

    That's why I can't get to GM. They think different

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

      no they play Minecraft you don't

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    3 weeks ago, # ^ |
      Vote: I like it +84 Vote: I do not like it

    All images for the problem and editorial were created in Minecraft lol

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

How to do C? 😭 😭 😭

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

    What I tried is: iterate on blocks of 0 and 1, and maintain the minimum possible length of alternating 0/1 string behind us, say L. If we encounter an even length block, we can jump L blocks ahead by folding here. (for example, if L=2 and the suffix we are at is 1100011110101, we can jump to 0101 by folding). Feels correct but I cannot get ac

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

    on current element is one and its my turn then i cant do anything . but one i got control i can make sure for each element it will go first ( ex for x i will reduce x-1 and 1 will remain so opponent must pick 1) . to check who gets control find continous 1 2 3....

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

    Let's say you have an infinite string like ...010101010... You start somewhere in between 0 and 1. Then you can take every symbol in s and make step left or right depending on the current symbol si. Answer is your max position minus min position. I have no idea how to prove it, I just checked examples from the problem, noticed this pattern and submitted

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

      How do you notice random patterns like this? You must have some sort of intuition that leads you there?

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

        Thank you for the question.

        So the meta-learning rule for thinking process is to build simple idea, then build counter example to it, and try to understand why it doesn't work, and how it could be fixed.

        The most complicated example here is "110110110011" (answer is 3). I tried to solve it by hand and noticed that you can't fold in between 01 and 10, and all the remaining 00 and 11 were folded. So I was thinking what if I greedily just fold every 00 and 11, what could be a counter example? And then I realized that I have around 30 minutes left, I have no counter example, so why not submit it.

        Here is my submission: https://codeforces.net/contest/1966/submission/258461794

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3 weeks ago, # |
  Vote: I like it +22 Vote: I do not like it

Guessforces

Literally just guess greedy works in C and get +30 rating orz

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3 weeks ago, # |
  Vote: I like it -6 Vote: I do not like it

Good round. I enjoyed the problems.

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

How to solve D(div-2) B(div 1) ?

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3 weeks ago, # |
  Vote: I like it +72 Vote: I do not like it

Two boring implementation problems in 1D and 1E. Worst round of this year.

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

F! I submitted in the last 10 seconds, And I think. It did not get processed.

My code for Part 4 was (Will check if after system testing)

ts = int(input()) tc = 0 while (tc < ts): arr = [1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576] tc+=1 s = input().split(" ") n = int(s[0]) k = int(s[1]) for i in range(len(arr)): if arr[i] > k: break higher = arr[i] lower = arr[i-1] arr.remove(lower) if(k != higher — lower): arr.append(higher — k) arr.append(k+1) print(len(arr)) for i in range(len(arr)): print(arr[i], end=" ")

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3 weeks ago, # |
  Vote: I like it +96 Vote: I do not like it

I might be biased, but I find it impossible to appreciate Div1 D. I guess messy caseworks are not interesting to most participants, so problem setters might consider using fewer such problems.

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

i spent way too much time overthinking on A and B and wasted like an hour....

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3 weeks ago, # |
  Vote: I like it +36 Vote: I do not like it

I had a construction for 1E that used at most $$$4 \cdot 10^6$$$ operations instead of $$$4 \cdot 10^5$$$ :( though I guess that's where the difficulty of the problem is meant to come from.

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3 weeks ago, # |
  Vote: I like it +60 Vote: I do not like it

If you are failing to come up with good problems, try not to propose a contest next time

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    3 weeks ago, # ^ |
      Vote: I like it -27 Vote: I do not like it

    Nooooo, please don't say this. Jake loves the money and clout which he gets from contests.

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    3 weeks ago, # ^ |
      Vote: I like it -39 Vote: I do not like it

    Maybe should just quit CodeForces. Recent contests are like you win if and only if you code fast and neatly, and manage to avoid the case work(by luck sometimes, or usually), which sounds like a joke to me.

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      3 weeks ago, # ^ |
        Vote: I like it +73 Vote: I do not like it

      Skill issue

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      3 weeks ago, # ^ |
        Vote: I like it +37 Vote: I do not like it

      Totally agree. I enjoy the contest iff coordinator == antontrygubO_o

      Feeling that other coordinators don't know what a good contest should be like.

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      3 weeks ago, # ^ |
        Vote: I like it +112 Vote: I do not like it

      Just curious: how this complaint is related to todays contest? As it seems to me D is the only problem of that kind today. But A, B, C and (probably)E have pretty short implementation so D is compensated by this, no?

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      2 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I agree

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

    Bro lost 70 rating and be really mad

    Though tbh I would be pretty mad too if I lost 70 rating to random problems

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    3 weeks ago, # ^ |
      Vote: I like it -70 Vote: I do not like it

    Also, if you cannot think of a problem F and have to borrow it from some contest before, why would you not just simply discard one of the problems and propose a Div.2

    at least it makes the count of bad problems decrease

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      3 weeks ago, # ^ |
        Vote: I like it +27 Vote: I do not like it

      I don't think the Div 2 portion of the contest was particularly bad anyway.

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        3 weeks ago, # ^ |
          Vote: I like it -77 Vote: I do not like it

        yeah so giving up the idea on proposing a Div.1 would have made the contest better

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      3 weeks ago, # ^ |
        Vote: I like it +36 Vote: I do not like it

      We already have a div1 drought, why do you want to make it worse?

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    3 weeks ago, # ^ |
      Vote: I like it +81 Vote: I do not like it

    I thought each of ACD is quite nice (despite losing 15 rating) and B is ok Are you sure youre not biased due to losing rating?

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      3 weeks ago, # ^ |
      Rev. 4   Vote: I like it +19 Vote: I do not like it

      You are biased towards this particular problem style.

      (fwiw i found this at least median contest quality, no problem is too stupid, none is my favorite but it at least feels well thought.)

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      3 weeks ago, # ^ |
        Vote: I like it +48 Vote: I do not like it

      sorry, too mad last night, apologize to the authors and everyone else affected by the hate i sent

      really should reflect on my contest strategies and problem solving skills now

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

I think i have the logic for D1C/D2E, could anyone point out if there is anything wrong with it please.

Observations:

1) You can think of the binary string as segments of 1's and 0's. If the length of the segment is even reduce it to 2 and if it is odd reduce it to 1, it won't affect the final answer, can't proof this one arrived at this from casework.

2) After this you can greedily construct the answer from one end, since folding requires an even length palindrome and odd in some edge cases, just iterate on the string and fold it as soon as you see an opportunity to do so. This works since even if you were do some other folding operation on these elements later on, the final optimal answer would still end up being the same.

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

Bruh what is this problem B? Just guesswork honestly wtf.

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

    Just find a left-most position, right-most position, up-most position, and down-most position of 'W' and 'B', and if either all positions of 'B' or 'W' are on the edges of the grid, then we can make all the elements of the grid equal; otherwise it is impossible.

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

Though in the problem setter's defense

  • We have Russian probs (i.e. Guessforces)
  • We have a Chinese prob (Nim)
  • We have American probs (i.e. implementation)

So it's a pretty balanced round and it's just that everyone is used to Guessforces...

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    3 weeks ago, # ^ |
      Vote: I like it -7 Vote: I do not like it

    It would have been an icing on the cake if there were Indian probs (i.e. questions) as well. Xd !

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

    If you cannot solve problems, you need more practice. If you cannot afford to lose, don’t game.

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3 weeks ago, # |
  Vote: I like it +70 Vote: I do not like it

Div1 B / Div2 D seems quite similar to https://atcoder.jp/contests/DEGwer2023/tasks/1202Contest_j :)

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    3 weeks ago, # ^ |
      Vote: I like it -20 Vote: I do not like it

    And I was wondering why there was such a quick jump in number of solves…….

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      3 weeks ago, # ^ |
      Rev. 2   Vote: I like it +31 Vote: I do not like it

      I don't have a strong opinion on this and am just asking out of curiosity, but do you think a coincidence like this has a real impact on the standings? That problem had 67/104 submissions, and I feel like there are very few people whose performance on 1B increased because of that past problem.

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

    ;[ why

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

for Div2D/1B, is value of n really matter under constraint? n <= 1e6.

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

    for my Soultion it does 258446518 I used Bitset dp

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

    No, I did nothing with the value of $$$n$$$.

    We can satisfy the condition for all $$$i$$$ up to maximum possible $$$n$$$.

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

    I think it is because the checker needs ~ O(n) to check the correctness of the solution.

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

in div2C shouldn't answer for 1 2 5 6 and 1 2 5 6 7 different? ( d seems beautiful but couldn't solve it )

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    3 weeks ago, # ^ |
    Rev. 14   Vote: I like it +2 Vote: I do not like it

    both of them Alice can win

    • 1 2 5 6 Alice remove 1
    • 0 1 4 5 Bob remove 1
    • 0 0 3 4 Alice remove 2
    • 0 0 1 2 Bob remove 1
    • 0 0 0 1 Alice remove 1
    • then she win
    • =================================================
    • 2nd example
    • 1 2 5 6 7 Alice remove 1
    • 0 1 4 5 6 Bob remove 1
    • 0 0 3 4 5 Alice remove 3
    • 0 0 0 1 2 Bob remove 1
    • 0 0 0 0 1 Alice remove 1
    • then she win
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      3 weeks ago, # ^ |
      Rev. 2   Vote: I like it +1 Vote: I do not like it

      oh now i get it , first time if its one alice dont have any choice but if its other than one then she will pick x or x-1 .

      i am noob

      Thank you so much for clearing doubt.

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3 weeks ago, # |
Rev. 2   Vote: I like it +15 Vote: I do not like it

As a personal review by cyan-blue contestant participating Div2:

==

A is hard to find and prove the answer comparing with another D2A.

B also harder than before, but not as much as A and C.

C is awesome. It's very easy to get the wrong solution, and if then cannot get the answer after many challenges. Wrong solution using Sprague-Grundy get correct answer with all examples, and that's why I didn't even think about abandon wrong solution and do it again without anything.

D is also ad-hoc. I got answer with 1 and a half A4 paper with calculation.

Each problem was pretty satisfying, and I know that I shouldn't complain about the problem construction just because cannot getting a satisfactory result. But ABCD all are so ad-hoc.

I think one or two ad-hoc is enought for single Div2. I do not have any intent to complain. But at this point, I wonder why you brought too many ad-hoc problems...

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

    I feel almost the same

    A was unusually hard, given that most Div2A, 800 rated problems requires no more than counting, sorting, parity check..

    And 4 observation & ad-hoc problems in row is just bad.

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

    I genuinely want to know this. What exactly comes under an ad-hoc problem?

    If the problem is not directly asking some data structure/algorithm, does it come under ad-hoc? Because then I've seen people complaining that the problems are standard.

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

    Ad hoc problems are the purest forms of problem solving. An ideal contest would have 100% ad hoc problems. (Ad hoc literally means that this problem is unique from previously seen problems, so surely thats better right?)

    Im not saying adhoc => good rather good => ad hoc

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      3 weeks ago, # ^ |
        Vote: I like it -27 Vote: I do not like it

      Of course the adhoc simp is here. Didn't even need to check.

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

I wish everyone will do good in the contest :) ;

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

Great pretests for C.. Simply WOW

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

I had an interesting situation for div 1 problem A. In my first attempt I forgot to unique the elements and I got WA6. It looks weird for me, and I think this is made intentionally.

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

My submition to problem c did not entered the queue to be judged. it is still apears to me that it passed the pretest cases after the final standings. Does anyone know what should i do? 258453668

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3 weeks ago, # |
  Vote: I like it +50 Vote: I do not like it

Div1C is a much easier version of problem 1394E. I know that it is obviously an overkill to use the solution of that problem, but this obviously brought some unfairness to this round. Added that B was also originally an atcoder problem, this round just can't be called perfect.

I myself have done both the atcoder problem and 1394E beforehand, so I can quickly think of both solutions (though the one for C was the more complicated one). It was my fault to make a mistake when implementing and messing everything up, but I did feel that seeing these problems will indeed make solving them in the contest easier.

I'm not blaming the author or anything because the problems were not intentionally copied so that's OK. But maybe next time just devote more time in the contest? 1394E is right from codeforces and many high rated users have done it. If more testers were involved or if only the author could google a bit more about this folding process, situations like this could definitely be avoided.

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

My submission for problem C has not been judged 258443367. Please fix this soon.

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3 weeks ago, # |
  Vote: I like it +202 Vote: I do not like it

I see a lot of hate in the comments. I hope it doesn't discourage the author from making more contests. I personally found ALL (div1) the problems very interesting!

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3 weeks ago, # |
  Vote: I like it +33 Vote: I do not like it

literally one of the best contest statements easy to understand not annoying respect for problem setters

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3 weeks ago, # |
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todays C in div2 , many got wrong ans on test 14 , this gave me a chance for looking into the cheaters today as the solution which they copied most likely had wrong ans on test case 14 as many solutions are exact same and having wrong answer on test case 14 only . like for these: 258453115 258453415 258453136 258453124,258453112 258458444 i request to ban these cheaters once and for all and please once look for all those others who had wrong answer on test case 14 only . Those hacked ones would be having many more ones who tried cheating but failed due to WA . MikeMirzayanov Vladosiya

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

    i also request the problem setters to do similar in future too what they did this time . this infact do not effect honest members rank to get lost plus also result in many cheaters getting caught

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3 weeks ago, # |
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what would you say is the rating for Div2 E?

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    3 weeks ago, # ^ |
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    Around 1800? The tutorial for this problem was just too easy:( But very few submissions to it. I don't know whether it was because problem 2D or not. Anyway I don't think it can reach 2000.

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      3 weeks ago, # ^ |
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      I thought it has to be above 2000 given the number of submissions. I haven't seen the solution yet so don't know.

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3 weeks ago, # |
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It was too difficult to write the code of Div1 D.

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3 weeks ago, # |
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Congratulations to Jiangly!

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3 weeks ago, # |
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"Get ready to test your coding mettle in Codeforces Round 941! With an exciting lineup of problems across both divisions, this promises to be an exhilarating challenge. A special shoutout to the problem setters, testers, coordinators, and of course, MikeMirzayanov, for making it all possible. Wishing all participants the best of luck, may your algorithms be sharp and your solutions elegant! #Codeforces #ProgrammingContest #GoodLuck"

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3 weeks ago, # |
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Unknowingly violating the code similarity rule, I take full responsibility for this error. I'll be much more vigilant moving forward. In light of my unintentional mistake, I kindly request your understanding and hope to avoid penalties or an account block.

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3 weeks ago, # |
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I received a notification regarding a high degree of similarity between my solution (258432282) for problem 1966B and other submissions. I understand the platform's strict policy against plagiarism and unintentional leakage.

I want to assure you that the code similarity was purely coincidental. Given the nature of the problem involving matrix manipulation, it's possible for solutions to converge on similar structures and approaches, especially when following a common thought process.

I can confirm that I did not engage in any form of collaboration or share my code publicly (including using ideone.com with default settings).

I kindly request your understanding of this situation and hope to avoid any penalties or account blockage. I would be grateful if you would not skip my submissions in this contest. I value fair competition and would never intentionally violate the rules.

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3 weeks ago, # |
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Regarding the coincidence of my decision with another participant in the competition. I have no proof in the form of published code before the start of the competition. But I ask you to take into account that the solution to the problem that I came up with (and apparently not only me) is very simple and there are not many ways to implement it quickly and simply. I admit that I violated the rules of fair competition 1.5 years ago. I realized my mistakes and no longer violated the rules of fair competition. I hope you won't ban anyone for this coincidence.

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2 weeks ago, # |
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can anyone tell me why this is giving runtime error although it runs fine in local ide..259362903