KAN's blog

By KAN, 2 weeks ago, translation, In English,

Hi!

Tomorrow, at Feb/07/2019 16:35 (Moscow time) we will host Codeforces Global Round 1.

It is the first round of a new series of Codeforces Global Rounds. The rounds are open for everybody, the rating will be updated for everybody.

The prizes for this round:

  • 30 best participants get a t-shirt.
  • 20 t-shirts are randomly distributed among those with ranks between 31 and 500, inclusive.

The prizes for the 6-round series in 2019:

  • In each round top-100 participants get points according to the table.
  • The final result for each participant is equal to the sum of points he gets in the four rounds he placed the highest.
  • The best 20 participants over all series get sweatshirts and place certificates.

The problems of this round were developed by a team of authors: aitch, simonlindholm, grphil, V--gLaSsH0ldEr593--V, GreenGrape, Nebuchadnezzar, _kun_ and me. Thanks arsijo and _kun_ for their help in the round's coordination, and also majk, pashka, Jeel_Vaishnav, Ashishgup and UnstoppableSolveMachine for testing the round!

Good luck!

Congratulations to the winners!

  1. tourist
  2. Um_nik
  3. TLE
  4. mnbvmar
  5. sunset
  6. wxhtxdy
  7. ksun48
  8. cki86201
  9. snuke
  10. fateice

Editorial.

Announcement of Codeforces Global Round 1
 
 
 
 
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2 weeks ago, # |
  Vote: I like it +132 Vote: I do not like it

Among reds and yellows, there is GreenGrape too. :D

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2 weeks ago, # |
  Vote: I like it +18 Vote: I do not like it

I hope the difficulty gap between problems will be proper.

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2 weeks ago, # |
Rev. 2   Vote: I like it -71 Vote: I do not like it

Just love the unstable contests. GreenGrape is here so we can hope for hell of an unstable contest!I mean contest with weak pretests O_o

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

    maybe the purpose is to encourage us to hack others...

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

      I don't understand why people give down vote in almost everything. Everyone can tell their opinion. Now a days, if someone is asking for help they also get down voted.

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

        There is no disadvantages(at least I think) for downvoted comments unless you are posting something spam, illegal, NSFW, or whatever. So don't keep those downvotes in your mind :)

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2 weeks ago, # |
  Vote: I like it +56 Vote: I do not like it

how many problems will be there?

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

What is the number of problems? And round duration???

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2 weeks ago, # |
Rev. 2   Vote: I like it +222 Vote: I do not like it

Me in this round.

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2 weeks ago, # |
  Vote: I like it -22 Vote: I do not like it

It's not like I am going to get any GP score, but it seems a bit unreasonable to rank 4 out of 6 contests, especially with the fact that rounds are not announced weeks ahead of the schedule for participants to allocate time slot for the contest.

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

    Ranking 4 out of 6 is a good idea because of the reason you wrote.

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

      They clearly meant that 4 out of 6 is still unreasonable.

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      2 weeks ago, # ^ |
      Rev. 3   Vote: I like it -9 Vote: I do not like it

      Not if they are announced only half a week beforehand IMO, at this point I can't even tell if the time slot is fixed. I'd imagine I can't participate 1 or 2 of them later into the year since they are midnight rounds for me, and another 1 or 2 of them because it clashes with other activities that I could have possibly adjusted my schedule for it had it announced earlier.

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        2 weeks ago, # ^ |
          Vote: I like it +18 Vote: I do not like it

        But "4 out of 6" only helps you, compared to "6 out of 6".

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

          I don't mean it's dragging me back, but it's just not good enough to be viable for everyone.

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

            What I agree with is that they should announce rounds earlier if we must compete for some total score.

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

Nice contest after chinese new year :D Good luck all !

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2 weeks ago, # |
Rev. 2   Vote: I like it -16 Vote: I do not like it

wishing ** Good luck** to everyone who is going to participate. - !NOOB :|

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

    Thanks, everyone who did devote me :)

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2 weeks ago, # |
Rev. 5   Vote: I like it -9 Vote: I do not like it

Why am i getting negative contribution without doing anything.

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

I'm not so sure if this is just me or the case for others too, but I feel like weekends would be more convenient for such rounds, people might be more busy during weekdays. Just an opinion I might be wrong :) Best of luck to those participating in this one!

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

    On saturday there will be atcoder.

    On sunday will be opencup.

    So, it is quite reasonable why it is not on weekends.

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

      Oh true! Seems quite logical now. It's impossible to please everyone anyways so that's alright. Thanks!

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      2 weeks ago, # ^ |
        Vote: I like it +54 Vote: I do not like it

      Counterpoint: The Atcoder round isn't for reds and Opencup isn't public either.

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2 weeks ago, # |
Rev. 3   Vote: I like it -21 Vote: I do not like it

There used to be a time when questions at the level of C were thought-provoking. Now the contests are like type first 3 fast and sometimes the transition in difficulty is so drastic that as you move from 1 level to another, the difficulty increases manifold. I believe A should be solved by 80%, B should be solved by 60%, C by 20% and D by like 7-8%. And C level questions need to level up.

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

    I thought today's C was quite challenging and B was not so straightforward to implement.

    If you are able to solve C's easier than before, it's because you are getting better.

    None of the first three problems were typing questions, because they all required some thought. (Easy or difficult, depending on your level).

    A typing question is like this is the scenario — implement it.

    All the first 3 questions required some deductions.

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

      This comment was made before the contest.

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      2 weeks ago, # ^ |
      Rev. 3   Vote: I like it -12 Vote: I do not like it

      None of the first three problems were typing questions

      Lol. C was a typing challenge and you wrote so much code. No wonder you took so long to solve it.

      Is that what you call "competitive" programming? LMAO.

      See my submission here

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        13 days ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        Your solution is essentially the same.

        I prefer writing code which is clean and makes the logic clear.

        Get a life, man :)

        Surely stalking what time I make code submissions and tracking down every single comment I make on here isn't the highlight of your life.

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          13 days ago, # ^ |
          Rev. 2   Vote: I like it -12 Vote: I do not like it

          I prefer writing code which is clean and makes the logic clear.

          Rewriting functionality that the STL already provides for you doesn't make your code cleaner. It makes it worse.

          LMAO. Shame on you to call yourself a programmer.

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2 weeks ago, # |
  Vote: I like it -22 Vote: I do not like it

What is the reason for randomly giving T-shirts to 20 people, instead of top 50? Just wondering...

And what is the random generator?

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

Well shit. I can't participate again because the time is too early.

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

How many problems and duration ?

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    2 weeks ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    7-9 problems for 2-3 hours

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

Getting One T-shirt can make your motivation level rise to the moons! Best of luck everyone.

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

Will it be in ACM-rules? Or in CF rules?

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

There was(thanks for the correction!) a 'и' in the English post. Now that I learned Russian a little, I do know it means "and" in Russian, but for those who don't know Russian, it's a bit confusing (especially when it's in the problem statement -- which I've encountered long ago, though it was clear from the context that it means "and" ;)

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    2 weeks ago, # ^ |
      Vote: I like it +92 Vote: I do not like it

    Codeforces has its own variant of the language. Its people can gradually learn extra words like «и» и «Разбор».

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      2 weeks ago, # ^ |
        Vote: I like it +113 Vote: I do not like it

      Great, I've finally learned something other than сука блять.

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

Should we expect an interactive problem?

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

seem difficult for practicer like me to compete t-shirt...

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

Feb 7, 2019: Day of emancipation div2 and div3....

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2 weeks ago, # |
Rev. 2   Vote: I like it -122 Vote: I do not like it

Hi everyone!

I got my old PC fixed just in time to do the contest. Excited to do this contest and this is such a nostalgic feeling. Let us enjoy it together, everyone.

Also, 1 like for getting my main PC fixed. or my PC dies. Your upvotes are really needed. Who wants a poor PC that went through such a harsh life of having 50 chrome tabs open every moment to die? Your likes are needed people.

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    2 weeks ago, # ^ |
      Vote: I like it -36 Vote: I do not like it

    Oh, at this rate you guys are gonna even kill my old computer and I won't be able to do the contest. Keep it up everyone, 100 downvotes and my battery will blow up.

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

      Too late guys, I already finished the contest. There's no point in blowing up my battery anymore.

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

    Protip: you can put a whole persistent OS on a USB drive, and boot from it anywhere (except where BIOS is locked). You don't need your own PC at all.

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      2 weeks ago, # ^ |
        Vote: I like it -13 Vote: I do not like it

      Oh that's actually really cool. If I had something like that, I could be using an updated Chrome with an updated IOS which I could really use for a VPN. I could also use my main vimrc. But I still needed a repaired PC to use at my home. If I didn't have one, I would either give up the contest, or try to use a smartphone which is kinda ridiculous (has anybody done that?), or do the contest at a place with a PC which would work actually, but I don't know of a place like that off the top of my mind.
      tl dr; thank you for the protip.

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

It really is Chinese-friendly. I hope that I'll be not so green after this round.

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

Good luck everyone :D

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

Happy Rose day to all...

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

Will the round be sorted by penalties or will points be awarded?

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2 weeks ago, # |
  Vote: I like it +35 Vote: I do not like it

10,000+ registration is now become common for codeforces ):

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

Are the problems sorted by ther difficulty ?

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

Wish you good pretests and high rating!

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2 weeks ago, # |
  Vote: I like it +46 Vote: I do not like it

Traditional randomization scripts!

The second script is used to feed random places into the first script. Seed will be the score of the top1 contestant of today's contest, length is the number of people with ranks 31..500 (likely 470) and nwinners is 20.

get_ranklist.py
randgen.cpp
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2 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

I think wery0 liked this round wery much. Thanks, problemsetters (especially Mr GreenGrape) for good tasks and fast editorial. Also, I want to say thanks to MikeMirzayanov for great(grape) platforms: Codeforces and Polygon!!!!

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

How to solve E

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

    Look at what happens to aj - aj - 1 (i.e., the difference array) after an operation.

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      2 weeks ago, # ^ |
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      Can you elaborate more, I did similar approach but obviously wrong since I got WA :D

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

        I think sum of squares of difference array remains constant

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          2 weeks ago, # ^ |
            Vote: I like it +21 Vote: I do not like it

          when you make an operation, you actually make a swap in the difference array so you just have to check if the two arrays are the same

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

            Oh yeah, so that would be wrong. The correct approach would be to sort the difference arrays and check if both are same.

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

              by "the two arrays are the same" i meant the two difference arrays. You obviously have to sort them because if not any good test would have two identical vectors ;)

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

                Sorry, I was not saying that your approach is wrong. I was pointing out the mistake in my approach where I said that the sum of squares of difference array remains constant

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

        Seems to me that you can shift the difference array arbitrarily around, since you can swap any two positions. Also, the first and last position of c and t must be the same anyhow. Hence, you just need to check if t can be made.

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

      Did we just have to check if there exists cj - cj - 1 = ti - ci - 1 such that j ≥ i?

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

        Actually we just need to sort the difference arrays and check if they are same.

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

      Wow! That's a great observation.

      How did you notice that? :)

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2 weeks ago, # |
Rev. 2   Vote: I like it +8 Vote: I do not like it

Many people used pow function in A which will overflow but those ran fine. Can someone provide a test case on which these overflow solutions will be wrong if any?

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    2 weeks ago, # ^ |
    Rev. 2   Vote: I like it -9 Vote: I do not like it

    [deleted]

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    2 weeks ago, # ^ |
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    91 11 1 1 1 1 1 1 1 1 1 1 1

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

    I hacked two solutions with pow using 99 99 1 0 0 0 ... 0.

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      2 weeks ago, # ^ |
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      Is overflow treated as error? Values mod2 will be fine even after overflow, right?

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

        pow doesn't operate on integer. You will get a floating point number, that is very unprecise and converting it to an integer will return something bad.

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

          Feelsbad then, bcs i saw 3 people using this :(

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

      Is this for just the precision problems with the function pow?

      If yes, so what about the result being not fitting in long long? how to hack it, please?

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        2 weeks ago, # ^ |
        Rev. 4   Vote: I like it 0 Vote: I do not like it

        When pow(x,y) is very large then it will become inf which will make result inf.In C++ inf%2 is 0 which means they have defined it as even.So using any big number for which answer is odd will break the normal solution.

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    2 weeks ago, # ^ |
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    Luckily overflow results in values modulo 2^31 or 2^63 so it doesn't change the parity.

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    13 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Overflow solution will not be wrong because overflow will preserve parity.

    (Because it happens in cycles of 2^{32} or 2^{64})

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2 weeks ago, # |
  Vote: I like it +20 Vote: I do not like it

A was more interesting than usual. :)

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How to solve D?

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

    Did not solve it, but if im not mistaken you can decrease each value to ={1,2,3}. And then apply DP.

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      2 weeks ago, # ^ |
        Vote: I like it +57 Vote: I do not like it

      Nope. You should set that value to {1,2,3,4}. Consider the following testcase: 12 4 1 2 3 1 2 3 2 3 4 2 3 4.

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

    Just note that you don't need to remove the same consecutive triplet more than 2 times, because otherwise you could have just gotten the same result by removing each number separately.

    This in turn allows for a DP solution. Just go through the numbers in sorted order and keep track of how many consecutive triplets the previous number and the previous previous number have been a part of.

    EDIT: Some clarifications. The dp state is how many you have left of the previous number and how many you have left of the previous previous number. And because you only want to remove the same consecutive triplet at most twice you can assume that you have at most 2 of the previous previous number.

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

Very nice problem C!

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

    how to solve it?

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      2 weeks ago, # ^ |
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      Hint: Brute force f(a) for small values of a and look for a pattern.

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        2 weeks ago, # ^ |
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        i don't know how to deal 2^even-1

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          2 weeks ago, # ^ |
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          just brute force it for all the 2^even-1

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

          I guess you can prove that the largest divisor is always the answer

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          2 weeks ago, # ^ |
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          After I brute forced them, I hard coded them into my solution.

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        2 weeks ago, # ^ |
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        where i can see the all solutions

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          2 weeks ago, # ^ |
            Vote: I like it +54 Vote: I do not like it

          For a = 2k - 1,the binary form is 11111....Consider a 1 ≤ b < a,we will find that ,so the gcd is .Its greatest value is a's largest divisor(not including a itself).Iterating divisor can be done in .

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

      I am not submitted yet but what i get is basically what XOR is — odd number of ones gives you set bit.example 0001 1110 gives you 1111 so take the first case -all the bits is not set

      so first you need to check whether the given number having all the bits is set or not if they are not then just add those bits which are not set. example — a given number is a= 5 — its binary representation is (101).its given that you need to take a number which is less than current number(a) so you just need to take b as sum of those whose bit is not set in current number (a) a=5 binary = 101, b=2 its binary=10 when you xor it you will get all bits set means when you xor (a^b) (101 ^10)=you will get 111 and the and of those ( and the and of (a&b) is zero so when you compute gcd of(aXORb ,a and b) you will get a^b itself as (a and b) is zero so here is the first case

      2 nd case -

      suppose all the bits is set example 15 its bits representation is 1111 so you just need to check what highest number is divisible by a example

      just run a loop max int out=1; for(int i=2;i*i<=in;i++){ if(in%i==0){ if(out<in/i)out=in/i; if(out<i)out=i; }

      } cout<<out<<endl; and thats all .You will get your maximum gcd ..

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

how to solve B?

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

    Note that if we are allowed n pieces of tape (when k = n), we can end up using a minimum total length of n cm. (We would use 1 cm for each broken segment).

    Consider the case where k = n-1. Then we need one piece of tape to cover (at least) 2 broken segments. We want to minimize the total length, so we want this piece of tape to cover the two adjacent segments that are closest.

    In general, if we have k pieces of tape, then we will have to merge (n-k) adjacent segments.

    Let's define the adjacent difference of segments i and (i+1) as (b[i+1] — b[i] — 1).

    So for b[i] = 3 and b[i+1] = 4, the adjacent difference would be 0 and we can tape over both with 1 piece of tape of length 2.

    We can generalize this by sorting all adjacent differences then taking the sum of the (n-k) smallest ones and adding n to that sum to get our final answer.

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

      I'm very bad at ad-hoc problems but this particular one seems to offer two general observations that are worth for me to learn: (1) applying the pigeonhole principle to constrain the solution: "if you can use only n-1 tapes than at least two of the broken segments must be joined together by one tape"and more importantly (2) adding up the results in a special order: in this case, adding the in-between lengths separately from the n target segments. I was of course trying to add up all contiguous things together.

      Great problem. If anyone can think of ad-hoc problems with a similar flavor, perhaps where you have to apply idea #2, I would love to know.

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2 weeks ago, # |
  Vote: I like it +87 Vote: I do not like it

Best problem C I've ever seen~

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

Can someone explain why binary searching for the answer (minimum length) with a greedy function wrong in case of B problem? I kept on getting WA on pretest 7. A hint would suffice.

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

    What is your greedy function?

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

      For a fixed length, i did a sliding window, starting from a broken point and covering as many points as I can until the total used length is <= maximum length in current call to the greedy function. Then I would start from the next index and do the same until either all points are covered or can't be covered.

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

        How do you decide whether to start a new piece of tape at the next broken point vs extending the current piece of tape to cover the next broken segment?

        I added the description of my solution to B here.

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          2 weeks ago, # ^ |
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          Understood the mistake. Thanks.

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            2 weeks ago, # ^ |
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            Can you please give some test case where the binary search approach will fail... I also wrote solution using Binary Search but WA on pretest 7

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              2 weeks ago, # ^ |
              Rev. 2   Vote: I like it +6 Vote: I do not like it

              10 500000 4

              1 30003 61255 101250 141246 171244 202492 242483 282475 305289

              Correct Ans: 185309 with these indices forming groups [1,2,3][4][5,6,7,8][9,10]

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2 weeks ago, # |
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B was harder than C for me. LOL

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    2 weeks ago, # ^ |
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    i found c to be easier than a and b !

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2 weeks ago, # |
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how to solve B?no idea :(

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    2 weeks ago, # ^ |
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    you need to find k minimum difference pairs from neighbourhoods

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    2 weeks ago, # ^ |
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    the problem is almost similar to atcoder beginner contest 117 problem C which i solved just today .

    find the maxx sum = arr[last] — arr[first] you need to find the gap between each array elemets , sort them , and then start sweeping them k times . subtract difference[i] each time from maxx sum .

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2 weeks ago, # |
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Well...I think the pretest of problem A is too weak because I passed the pretest while writing x&1 wrong as x^1...

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2 weeks ago, # |
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E is one of the best problem E ever on Codeforces, until I realized that one can fail system test because they forget to check if c[1] ≠ t[1] or c[n] ≠ t[n].

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2 weeks ago, # |
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Does this count as cheating? I saw a submission for C where the user created a map of all the possible input a's to the answer. Probably these answers were brute forced earlier and logged?

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2 weeks ago, # |
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What is test case 35 in D? Many people including me are failing on that

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2 weeks ago, # |
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Problem E was on AtCoder: link.

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    2 weeks ago, # ^ |
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    good that's why i like mokoto , his mind is awesome , problem b was also similar to abc 117 c

    thanks to mokoto i solved it !

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    2 weeks ago, # ^ |
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    That's how I solve pE within 5 minutes lol

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2 weeks ago, # |
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Any idea how to solve D .

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2 weeks ago, # |
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even the submission that i didnt have time to submit is WA .. now I can rest in peace

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

I love this contest, where the difficulty is in ideas not just complex code.

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2 weeks ago, # |
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When can we konw the final result ?

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2 weeks ago, # |
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Just for curiosity, why are there no hacking session for this contes_- _t? The contests I participated since were rated for Div2 so... just curious.

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    2 weeks ago, # ^ |
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    Hacking sessions are only available for Educational Rounds and Div.3 contests. ;)

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    2 weeks ago, # ^ |
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    thanks :) glad to check ACs right after!

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2 weeks ago, # |
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So I was using ideone for this contest since I'm not home and I didn't want to install codeblocks on this pc and I got a message that bruce_knight copied my code for B. The message said I could be getting banned for this ? What are my options ?

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2 weeks ago, # |
Rev. 2   Vote: I like it -29 Vote: I do not like it

Even tourist couldn't solve H ?
There must be something wrong with tourist or that question...

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2 weeks ago, # |
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I think there are some problems that have appeared in another place. For example: E: BZOJ5071  I'm sorry that the language for this problem is Chinese, but this problem is similar to Problem E (My English is not good)

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    2 weeks ago, # ^ |
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    that is the strategy of tourist

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2 weeks ago, # |
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Can someone explain the logic behind E?

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    2 weeks ago, # ^ |
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    Consider the difference array d, i.e, di = ai + 1 - ai, then any operation is swapping two elements in the difference array.

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2 weeks ago, # |
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In api expected rating change shows +34 but in rating changes it shows -4

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2 weeks ago, # |
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I had almost solved problem G but there was not enough time.... It's an interesting problem to me though.

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2 weeks ago, # |
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This was my 2nd contest solved A,B. Initially afraid of submitting the solution but finally submitted it and got accepted. Aiming to solve more in coming contests. Thanks to the organizing team.

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2 weeks ago, # |
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Can F be done with centroid decomposition ?

For each leaf ,go up to root via centroids storing in each centroid ,distance of leaf from it and its index. For each centroid we keep this list sorted on the basis of indices. While answering a query, we go up to root from vertex v via centroids, at each centroid, we need to query min dist of leaf from this centroid whose index lies in L to R(simple RMQ).Take this minimum across all centroids in my path to root .

Is this approach correct ? will it fit the TL ?

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2 weeks ago, # |
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I think the rating changes for today's contest were not correct. As the "CF Predictor" extension showed something else and the rating changes were different.

Kindly look into the matter.

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2 weeks ago, # |
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Problem C was a great question. The only case you had to take care of was when a was of the form 2^x-1. The trick involved was gcd(a&b, a^b) = gcd(b, a-b) = gcd(a, b). So we need to find the greatest proper divisor of a. Kudos to the writers!

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2 weeks ago, # |
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Can D be solved using a recursive approach without running into "Memory limit exceeded"?

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2 weeks ago, # |
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Hey,guys.I have a problem on 1110D — Jongmah. I passed the test 1 locally on my machine,however get WA on the net. Here is running ID #49593747 Help me please!!!QAQ

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2 weeks ago, # |
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When will the editorial be published?

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    2 weeks ago, # ^ |
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    Well,I have already seen it in the Recent actions

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2 weeks ago, # |
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Thanks to this contest, I have become a candidate master. I'm so happy.

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11 days ago, # |
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KAN, Why 2 links to the same editorial are added to contest page?