vovuh's blog

Codeforces Round #634 (Div. 3)
By vovuh, history, 2 years ago, In English

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Hello! Codeforces Round #634 (Div. 3) will start at Apr/13/2020 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have 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. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours 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 a 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 MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

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UPD: Also thanks to ma_da_fa_ka and infinitepro for testing the round!

UPD2: Editorial is published!

 
 
 
 
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2 years ago, # |
  Vote: I like it +45 Vote: I do not like it

vovuh logic

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

vovuh hello , I guess my rating will down to less than 1600,but I've registered before,will I be rated in this case?

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

Wow wow, another Div.3 but this time I will not get rate :D

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2 years ago, # |
  Vote: I like it +7 Vote: I do not like it

I good opportunity to get my rating back from today's Div2 lol

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

What does "open hacking" mean, sorry I am new? How can you hack a solution?

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    2 years ago, # ^ |
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    After each contest, there is a time of 12 hrs where if we find someone's solution incorrect, then we can give input the case at which the submission fails, if hacked you are rewarded with some points sometimes.

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      2 years ago, # ^ |
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      Not after each contest LOL

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      2 years ago, # ^ |
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      is there a way to sort solutions by programming languages? Now, I have to go through all the submissions to find Java submissions.

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

        On the right hand side of status page,you can find language option under status filter and select language Java.

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        2 years ago, # ^ |
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        status > status filter > languages

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    2 years ago, # ^ |
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    1)After this contest (the same for following div.3 rounds and educational rounds), there's a period of 12-hour hacking phase.

    That means, you can hack anyone you want(while in normal rounds you can only hack your "roommate" during the contest).

    2)If you find something wrong in other's solutions(but passed the pretests), you can give an input which will make the solution fail (wa,tl,...)

    In div.2 you get 100 points from each successful hack and lost 50 points from each unsuccessful one

    In div.3 there's no prize or penalty for hacks.

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      2 years ago, # ^ |
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      Simply That Problem Will be Marked as Unsolved

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      2 years ago, # ^ |
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      What's "roommate"? Sorry, newbie.

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        2 years ago, # ^ |
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        In a normal round, you are assigned to a specific room, with ~50 people if I'm not mistaken. During the round, you can attempt to hack their solutions, but only if you lock your own solution first (this is because you will be seeing others' codes, it wouldn't be fair if you could change your own code).

        A successful hacking attempt will give you a 100pt boost, while an unsuccessful hacking attempt will take away 50pts from you.

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          2 years ago, # ^ |
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          and what does locking the solution mean? If hacking starts after the round then how can one change his code?

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2 years ago, # |
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Hope it won't go too mathematical ;/ questions about string is more interesting;)

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2 years ago, # |
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**BACK TO BACK CONTEST **

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2 years ago, # |
  Vote: I like it +77 Vote: I do not like it

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    2 years ago, # ^ |
      Vote: I like it +79 Vote: I do not like it

    But comparison of other websites code forces server is good

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      2 years ago, # ^ |
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      It's not just good it is pretty awesome when compared to others. Some of them had to cancel their contest just because the servers weren't able to handle the load and the submission queue was getting overloaded.

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    2 years ago, # ^ |
      Vote: I like it +63 Vote: I do not like it

    Did you ever participate in codechef cook off or luch time...they can't handle even 5000 participants properly...whereas codeforces handles nearly 20K participants...comparitively codeforces servers >>>>> any other cp server

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      2 years ago, # ^ |
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      Can't you guys take it as a meme...

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        2 years ago, # ^ |
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        Exactly!!!

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          2 years ago, # ^ |
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          These guys become really too serious.

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            2 years ago, # ^ |
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            Their serious and sensible reply let me know that CF's servers are not bad, very good instead. Next time I'll be more patient.

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    2 years ago, # ^ |
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    The girl is the middle would rather be saying that "Give the lengthy problem A and engage participants."

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    2 years ago, # ^ |
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    Did you ever participate in atcoder contests It servers cannot hold even 5000 participants. Codeforces servers when it is +5000 submissions at time it will get some time but when it is less it works good!

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      2 years ago, # ^ |
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      it was just a meme bro... I know CF is really good.. I apologize for posting this...fine?

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After every contest i felt disappointed.But believes that practice makes a man perfect. Best wishes for all. <3

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

Hoping to see "Yet Another Problems".

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

finally a div 3 contest :(

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2 years ago, # |
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Can "Hacking A Soluton" also be included in Div 3 ?? Just an idea.

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    2 years ago, # ^ |
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    When div3 was first proposed then it was said that a beginner should waste time in hacking instead of solving problem. So contest time hacking is not available in div3 rounds. And there is 12 hours hacking phase for improving hacking skill. But i think 12 hour is too long. It can be reduced.

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      2 years ago, # ^ |
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      Is Hacking Possible in Virtual Contest??

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        2 years ago, # ^ |
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        yes if contest is still in progress hacking

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      2 years ago, # ^ |
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      I think 12 hour just for global, maybe someone finish the problems then have to go to bed.When they weak up, they can hack.

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        2 years ago, # ^ |
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        Actually after 6 to 7 hours nothing for hacking. Almost all the hackable solution are already hacked in this time.

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2 years ago, # |
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Thank you, as always I am excited for another div.3 :)))

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2 years ago, # |
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I should try to increase my rating in this contest....

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2 years ago, # |
  Vote: I like it -16 Vote: I do not like it

Hello, is this rated?

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

I'll try to skew my propagating wave at least this time. Pray for strong pre-tests

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2 years ago, # |
← Rev. 2   Vote: I like it +143 Vote: I do not like it

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Problem set in recent codeforces contests seems tough for me. Looking to get some easier problem this time.

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Div 3 means higher participation. I hope the testing is fast!

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

hopefully some short problem sentence

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Looking forward for the contest .✌️

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2 years ago, # |
  Vote: I like it +24 Vote: I do not like it

Why setter is yelling at tester?

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    2 years ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    Because of my invaluable suggestions :)

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      2 years ago, # ^ |
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      reveal unto us, are the problems sexy?

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      2 years ago, # ^ |
      ← Rev. 2   Vote: I like it +16 Vote: I do not like it

      I think you should explain about your suspicious behaviour in the past contest. Because it looks like you've cheated before.

      your suspicious behaviour

      Apologize in advance if I mistakenly understand it.

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        2 years ago, # ^ |
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        I am not being rude, but I don't think so I need to explain you anything :)

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          2 years ago, # ^ |
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          The thing is, your code is totally same with another code in the past contest.

          I don't want to be an annoying guy who repeat one thing everywhere. I just want to know what happen. If it isn't a cheat, I will sincerely apologize. If you've already be punished for that, I won't mentioned it anymore.

          I will keep annoyying you if and only if you really cheated and haven't been punished so far.

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Hoping for stronger pretests in A. Got hacked (unfortunately) in the educational rounds and it had been a while since I had solved 3 T_T and I missed it thanks to the A.

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    2 years ago, # ^ |
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    You should be more careful with some corner cases

    That's more important and valuable than strong pretests.

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  • I hope it will be like this for second time!!
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I hope to see Math problem~~

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you make me a green i dare you

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good luck

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2 years ago, # |
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Hope the system test won’t be too long.

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2 years ago, # |
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It's good to see vovuh back but I expect that codeforces will do something in order to handle such large number of submissions during the contests and I guess this time it won't be queueforces :)

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I hope that the gap between the div.3 rounds will be decreased , its the best round

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

Hope my rating will go up.

I'm so vegetable......

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2 years ago, # |
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Hope the statements will be shorter -_-

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

Thanks vovuh again for bringing such contests in this pandemic situation we highly thanked to codeforces and its community for bringing such contest in this period to make us somehow busy and involved.Thank you codeforces community !

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

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    2 years ago, # ^ |
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    The color spectra doesn't quite match Codeforces band XD

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

I am becoming Expert Blue after this round.

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2 years ago, # |
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Is it rated?

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2 years ago, # |
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Isn't it more logical to not allow experts or above from submitting solution for first 20 minutes rather than making problem A hard/confusing for div3 rounds atleast. Many participants don't even submit solution if they find problem A confusing or tough for it's bracket.

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My rating is just 1600, so helpless

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    2 years ago, # ^ |
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    lol similar is case for guys having rating just 1900

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Good Luck Have Fun!

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what happens when i hack my own solution ?

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    2 years ago, # ^ |
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    pts — your points

    So, if you hack task A, then:

    new_pts = pts — points_from_task(A) + hack_points

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    2 years ago, # ^ |
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    If successful:

    You will lose a problem solved

    If unsuccessful:

    Nothing changes.

    So it's obviously meaningless to hack yourself unless you've found some mistakes in your code after the contest.

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2 years ago, # |
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Good luck to you all

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2 years ago, # |
  Vote: I like it -23 Vote: I do not like it

What you see: Codeforces Round #XYZ (Div. 3)

What I see: Another unrated contest :'(

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    2 years ago, # ^ |
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    architb_12 why the hell are u giving this contest ,if u have such a mentality,and posting this nonsense doesn't show that u are cool

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      2 years ago, # ^ |
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      pizza_hut what mentality? I am just saying that I am sad that the contest is unrated. It is much more fun when it is rated. What is your problem with that?

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        2 years ago, # ^ |
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        architb_12 whats the problem when contest is rated for 70% of users on codeforces

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          2 years ago, # ^ |
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          I think you have completely misinterpreted me. I meant that it is more fun to give a contest when it is rated for you.

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          2 years ago, # ^ |
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          hey, first of all, learn to give respect to others (doesn't matter he is highly rated or low) and don't talk rudely. He was just saying that this contest is unrated for him and of course this was unrated for him because he has done efforts to be at that position where Div3 even Div2 is unrated for him. So, please don't post such non-senses. Even today tourist gave the Div3, can you stop him...No nah?

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2 years ago, # |
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Can the frequency of Div.3 contests increase ?

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let's hope that i get one question correct on this site tutututu

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2 years ago, # |
  Vote: I like it +61 Vote: I do not like it

This is what happens when Tourist attends Div 3.

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    2 years ago, # ^ |
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    How can one read the problem, figure out the solution and write down the code in 1 minute.....

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    2 years ago, # ^ |
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    he submits all the answers after that he reads the questions

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What's the point of having 50000 testcases, if giving verdict still takes you > 10 minutes?

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    2 years ago, # ^ |
      Vote: I like it +72 Vote: I do not like it

    I'm sorry, what about you? I don't see any of your submissions were judged for more than 1.5 minutes.

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      2 years ago, # ^ |
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      F took > 10 minutes. It really didn't affect me. I understand its because these rounds don't have pretests, I guess.

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  Vote: I like it +38 Vote: I do not like it

Wow the servers have REALLY IMPROVED. MikeMirzayanov orz

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  Vote: I like it +75 Vote: I do not like it

tourist solved 7 problems in 22 minutes and still atheists exist. orz

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

Tests today was rlly fast, thx)

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2 years ago, # |
  Vote: I like it +16 Vote: I do not like it

A nice and balanced round !!

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2 years ago, # |
  Vote: I like it +7 Vote: I do not like it

How to solve E2 and F?

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

Easy +rating, nice contest)

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

I was logged out suddenly while submitting please solve this problem it occurs most often during contests with me!

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2 years ago, # |
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How do you do E2? I spent like 45 minutes binary searching for the second endpoint(and I think it was probably the right way since you have blocks of xyx or just one large block of length x), so you can use binary search to find the optimal value for the second endpoint, but it wouldn't work.

EDIT: Just realized that I thought a_i<=26 for E2 (as it was in E_1) so that's why I was getting wrong answer

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    2 years ago, # ^ |
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    Observation:

    The contribution of a number is limited by number of occurrences of prefix and number of occurrences of suffix, In other words, if you have a prefix of 7 occurrences and suffix of 2 occurrences, Then you're only contributing to the final answer with just 4(2 from the left side which has 7 occurrences, and 2 from the suffix), So contribution = min(prefix,suffix) * 2.

    Knowing the above, We can just try for each number x, A possible prefix that has 1 occurrence.. 2 occurrences .. 3 and so on, you can fill the middle with an element that occurs the most, you have only 200 distinct element, just go over them all.

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      2 years ago, # ^ |
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      Oh my this is neat. I mean the idea about checking only prefix and suffix. Thanks!!

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2 years ago, # |
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Video Editorials + Facecam

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How to solve E2. Any idea??

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    2 years ago, # ^ |
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    store position of every element in a vector v[203].for each number between 1 to 200 run two pointer from start and end of its position vector and between them find maximum time occuring integer frequency using prefix sum 2-d array.

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      2 years ago, # ^ |
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      can you elaborate on the prefix sum 2-d array? how are you storing in them?

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my approach for E1 was to check for the next max frequency of a number between the extreme positions of all the numbers having max frequency ...what's wrong in this??

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more than 60k accepted submission and aprox 26k participants.wow amazing.

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2 years ago, # |
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Why i can't hack someone? It's write me "Illegal contest ID" or "Неверный идентификатор соревнования"

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Fast servers but speedforces :(

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How did you all solve E2? I used Mo's and I wonder if that was overkill.

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    2 years ago, # ^ |
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    Oh man the alphabet is only size 200, Mo's was definitely overkill.

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    2 years ago, # ^ |
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    Yeah, I used Mo's too, don't know how others got that right

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    2 years ago, # ^ |
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    complexity- (2*10^5)*200 I will tell very briefly... for each index i let x=arr[i] and till now(including i) let it(x) has appeared fth time so if (Frequency of x in whole array)/2 >= f then i found index j of x from the end of array such that in segment arr[j...n] occurrence of x is also f and hence I iterate through all the number 1<=n<=200 an found the no of occurrence O= of such n between (i,j) . so length of three blocks palindrome will be f+f+O.

    Do this and update the result if you got a larger length.

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    2 years ago, # ^ |
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    Here's my approach:

    1. Loop through all $$$k$$$ possible "symbols". This will be the symbol forming the prefix and suffix (aaa...aaa)
    2. Put two pointers on the first ($$$i$$$) and last ($$$j$$$) occurrences of the symbol.
    3. Find what's the best score for the middle part (...bb...) between $$$i+1$$$ and $$$j-1$$$ using a precomputed table of counts for each prefix and each symbol.
    4. Move pointers towards each other to the next two occurrences of the symbol from #2. At every time, you know exactly how many of these symbols there are to the left (= to the right) of the middle part, so the score for this partition is middle part score + 2 * symbol count.
    5. Repeat until they meet, keeping track of the maximum score.

    $$$O(n\cdot k^2)$$$ in total, but the operation #3 is rare in practice, so it passes the pretests ¯\_(ツ)_/¯. Got MLE on the first attempt though, lol. Don't define int long long where it's not necessary.

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      2 years ago, # ^ |
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      Excellent solution. I thought O(N * K^2) would fail, but yours is super simple, so I guess it has a very LOW constant factor.

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

      That is $$$O(n \cdot k)$$$ in total if you have position pre-calculation, not $$$O(n \cdot k^2)$$$. That is because $$$\left(\sum \text{occurrences of symbols}\right) = n, \text{not } n \cdot k$$$.

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

        but for each prefix and suffix of given size and letter, he is going over all 200 letters to find biggest centre block. So I think its really $$$nk^2$$$ with small constants.

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

          I think prefix table can be calculated for vector<int>(200), then the max_element(for "b") can be found in O(k) time. Since we traverse over O(n) two pointer pairs (i,j), this gives total time O(nk).

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

      FYI this approach is almost optimal; but at step 2 you should initialize $$$i$$$ and $$$j$$$ to the two middle occurrences of the symbol (if the amount of symbols is odd, just ignore the middle one), then move them away from each other, towards the end.

      The benefit of this is that you can start with an empty set of frequences $$$f[200]$$$, then initialize that to the counts of each symbol between $$$i$$$ and $$$j$$$. Keep track of the most frequent symbol whenever you update ++$$$f[\cdot]$$$. And then do --$$$i$$$, ++$$$j$$$, and update $$$f[]$$$ as you go. You can always keep track of the most frequent letter in the middle part this way, and complexity is $$$O(n)$$$ because you do ++$$$f[\cdot]$$$ at most $$$n$$$ times.

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

      For E2, I coded a O(nklog(n))time algorithm using the same idea as yours except using binary search instead of prefix sums. I realised that the complexity is too large. However, for testcase 9, which has 134 repeated 2e5 times, my solution should simplify down to O(nlogn). However, when I ran it on my system, it took more than 30 seconds and gave me TLE on codeforces. Can you please help? Here is my submission https://codeforces.com/contest/1335/submission/76624993

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

        I also tried the same thing but got tle on test 15 .76625017

        Later i think to remove binary search and precompute freq table but memory limit exceed on test 9 .76625886

        can anyone help ?

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

          Replace ll from int in your program

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            2 years ago, # ^ |
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            wtf , It worked .

            how they can set such a memory limit.

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      2 years ago, # ^ |
      ← Rev. 4   Vote: I like it +1 Vote: I do not like it

      E2 complexity analysis:
      $$$\displaystyle \sum_{c=1}^k freq_c \cdot k = \displaystyle k \cdot \sum_{c=1}^k freq_c = O(k \cdot n)$$$

      Sum over $$$c$$$ for choosing the number in $$$1^{st}$$$ block. $$$freq_c$$$ for iterating over all possible lengths of $$$1^{st}$$$(and $$$3^{rd}$$$) block. Multiplied by $$$k$$$ for choosing the number in $$$2^{nd}$$$ block.

      $$$\displaystyle \sum_{c=1}^k freq_c \neq k \cdot (freq_c)_{max} \neq k \cdot n $$$

      $$$\displaystyle \sum_{c=1}^k freq_c = n $$$

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    2 years ago, # ^ |
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    What is Mo. I thought of doing binary search but failed to do so.

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    2 years ago, # ^ |
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    Could you please explain how you applied Mo's?

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

      The problem can be reduced to answering a bunch of queries of the form: What is the most frequently any element appears in the range [L,R]? If the alphabet is large, you could attack this with Mo's Algorithm.

      Unfortunately, it slipped my mind that since the alphabet is so small (only up to $$$200$$$ symbols), you could more simply create 200 prefix sum tables instead.

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        2 years ago, # ^ |
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        I didn't quite understand from your code how you're generating these queries. Could you please elaborate how you're generating these ranges?

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        2 years ago, # ^ |
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        I did, but one must be careful to not use long long, since it used to much memory. We have to use a $$$32$$$ bit type.

        76620592

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          2 years ago, # ^ |
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          My solution is similar to yours but it is giving TLE on test 2. I am not able to figure out the reason.

          My Submission

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        2 years ago, # ^ |
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        Is Mo's algorithm the only way to do it?

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2 years ago, # |
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Really good round! I loved problems D and F. Had an idea for F but gave up after 20-25 minutes of implementation after I realised I didn't account for something mentioned in the question. Had about 20 minutes for E1 and E2 (I regret not having attempted E1 and E2 before F). I believe I'd have been able to solve E if I made a wiser decision earlier but shit happens, lol. Also, ideas for E, anyone?

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

    You can enumerate a and the number of a, and then enumerate what's b, it seems 200*200*n, but you can find the really time complexity is 200*n

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      2 years ago, # ^ |
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      Keven Can you explain your solution? Is it similar to this: https://codeforces.com/blog/entry/75908?#comment-602884

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        2 years ago, # ^ |
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        200*200*n, first 200 enumerate a, n is the number of a, so first 200 times n is n, time complexity is 200*n

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      2 years ago, # ^ |
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      can you explain me why it is not 200*200*n?

      I got AC on E1? but i assume that complexity was 200*200*n, now i sent same code in E2 and i got AC

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

        In your solution you have y = v[a].size() - 1; which is actually not always n. In fact it would make the make the factor of 200*n to just n.

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

can anyone point out which test case is giving wrong answer in this code for question B of this contest. https://codeforces.com/contest/1335/submission/76554481

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    2 years ago, # ^ |
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    i missed corner case when we have to take 26 unique characters.my bad

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

In problem F, if N and M's range can be confirm instead of N*M <= 1e6 will be better

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

How do you guys solve D?

I finally figure out that we must change 9 postions which are respectively distributed in the nine squres.

So I enumerate the column and the row (maybe like N-queen Problem) and get the answer....

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    2 years ago, # ^ |
      Vote: I like it +58 Vote: I do not like it

    Author's solution is to replace all 2 with 1.

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      2 years ago, # ^ |
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      Wow, this solution is very cool!

      I think replacing 3 with 1 / 2 / ... is also ok.

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      2 years ago, # ^ |
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      Why 2?

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      2 years ago, # ^ |
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      wow that's innovative, instead of figuring out which cell to edit.

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      2 years ago, # ^ |
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      Yeah, I also changed every 9 to 8. It took just less than 4 minutes for me to solve this. Fastest D problem ever for me.

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

    I place a number 9 - currentNumber if its non zero else 1 in every cell, such that whose row,col and block was not visited before through any cell. To check that I use simple hashing.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    Idea
    Code
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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I did the same xD

              mat[0][0]=max((mat[0][0]+1)%9,1);
        mat[1][3]=max((mat[1][3]+1)%9,1);
        mat[2][6]=max((mat[2][6]+1)%9,1);
        mat[3][1]=max((mat[3][1]+1)%9,1);
        mat[4][4]=max((mat[4][4]+1)%9,1);
        mat[5][7]=max((mat[5][7]+1)%9,1);
        mat[6][2]=max((mat[6][2]+1)%9,1);
        mat[7][5]=max((mat[7][5]+1)%9,1);
        mat[8][8]=max((mat[8][8]+1)%9,1);
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        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Shhh, don't post anything remotely related to the word "same" :p They'll think we cheated

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          2 years ago, # ^ |
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          lol should I delete it?

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            2 years ago, # ^ |
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            Nah, hehe. It's cool. Now, it'll look like: Yeah, they cheated, but if they're talking about it openly, they might not have....

            Jokes apart, I'm just having fun. GL for future rounds

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

    a[1][1] = a[2][1], a[2][4] = a[1][4], a[3][7] = a[2][7], a[4][2] = a[3][2], a[5][5] = a[4][5], a[6][8] = a[5][8], a[7][3] = a[8][3], a[8][6] = a[7][6], a[9][9] = a[8][9] This was my idea. Later, got that replacing all 2's with 1's was good enough.

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

Funny way to solve F without detecting cycles: notice that if two robots get to the same cell, then they will always move to the same cell from then on. This means that we can find for each cell where the robot starting in that cell will be after some $$$2^{22}$$$ moves using binary lifting.

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

    This is author's solution :)

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

      Hi, this is my first contest. Can you please share the link to the author's solutions? I am not able to locate it. Thank you!.

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

        I'll post them with the editorial. Please, be patient.

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          2 years ago, # ^ |
            Vote: I like it -42 Vote: I do not like it

          The contest was not at all balanced, I must say. Even you can see the rating of people and compare the no. of questions they solved. Starting 4 questions were not of much use as almost half of the contestant solved the first 4. Also, question D looked like it was meant for April Fool's 202 but was posted by mistake?

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            2 years ago, # ^ |
              Vote: I like it +34 Vote: I do not like it

            Why do you posting this from fake account? :)

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              2 years ago, # ^ |
              ← Rev. 2   Vote: I like it -21 Vote: I do not like it

              You ignored my comment... What am I saying wrong according to you? You should accept the mistakes. :)

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

                Which mistakes? Prediction mistakes? Maybe, I need to clean my magic ball to see the future better.

                The only (and very doubtful) "mistake" here is the "gap" in $$$5400$$$ accepted solutions among official participants (which means that E1 was solved by 1/4 of people who solved D). This is not a big gap.

                I could show you examples of really big gaps but I don't think I need to prove you something anymore. If you don't realize that this prediction with difficulties is good enough, I have nothing to say to you.

                And, if you post your opinion from fake account, it seems like you don't want to anybody know who are you actually :)

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

                  The last point is quite obvious.. :) I just wanted to say that these weekly contests are supposed to test our algorithmic (and not puzzles) skills. Thank you for replying..

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

                  Yeah, you are right. As far as I remember all the Div.3s I have given were perfectly balanced and excellent problemsets and all of them were made by you:)

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

                  Well, I disagree with you :) Most times there are some mistakes that you couldn't notice, but this time the contest went fine. There also were some minor mistakes but not as fatal as usual. This is the rare case when almost everything is fine and I'm glad to see that.

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

                  Thank you for the round! No queue, no weak pretests.I found problem E2 very interesting and problem D a bit hard for me but interesting too.Please make more div 3 contests, we really appreciate you :)

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

      That's actually a really cool approach. Have you also tested a cycle-based solution? I am trying to implement a fairly straightforward search and it keeps running out of memory.

      Looking at 32 pages of MLE results in Status, I think the memory limits for E and F could have been a little bit higher.

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

        The approach with extracting cycles and doing some dp is much harder to implement, so I didn't even tried to do that. I could but didn't do this.

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

          I ended up flipping the dimensions if N > M, and surprisingly this helped (76628096). According to the test results, it uses up 254.5 out of 256 MB, lol.

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            2 years ago, # ^ |
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            Well, I'm sorry to hear that. I really thought to increase memory limit, but my solution uses ~200MB (the two-dimensional vector of size $$$nm \log nm$$$) so I thought that the 50MB will be enough (because this is the only array which should have such size).

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

              Hello ; can you please explain me why does N*M*M solutions pass for problem E2 ?

              tourist has written a code with that time complexity ; here is his submission — https://codeforces.com/contest/1335/submission/76520445

              And I am not able to understand how does that happen ; curiously ; I even did try to check the test cases ; and I see test case 9 which should break such solutions.

              So clearly ; I am not understanding something ; can you please help me with this.

              Thanks !

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

                The solution of tourist has the complexity $$$O(nk)$$$ where $$$k$$$ is the size of the alphabet. If you take a look at the two outer cycles, you can notice that they just iterate over all characters of the string and this sum is obviously $$$n$$$.

                Maybe, when I post the editorial, it will be more obvious to understand why this is $$$O(nk)$$$. Just be patient.

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                  2 years ago, # ^ |
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                  Oh ; I see ; I apologize for being a bit impatient ; and I thank u for the reply.

                  I understand the complexity analysis of that solution now ; Thanks !

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

              vovuh please look at my submission for E2 this it got MLE 252700 KB but now I am submitting it got accepted this with 252800 KB (more space than previos)... :'( and because of that I was not able to submit E1 also during contest...

              very sad!!! :'(

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

    Thanks for teaching me something. I learned binary lifting for RMQ/LCA but didn't consider reusing the table for k-th ancestor queries.

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

I got E 5 min late -_-

Nice round!

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2 years ago, # |
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help me please which problem i have with this code on problem B? https://codeforces.com/contest/1335/submission/76592383

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

    your code is giving wrong output in this test case :- 1 10 10 10

    your output contain only 9 distinct character but reuqired distinct characters is 10

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    2 years ago, # ^ |
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    Idea is wrong.....

    My idea
    Code
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    2 years ago, # ^ |
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    You have taken 'g' twice in your array

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2 years ago, # |
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How to solve E?

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

How to think fast like tourist... :(

When he was solving each question in about 1 minute, I was drawing on mspaint.exe and looking for the solution :)

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

    practise,practise and practise .smartwork +hardwork

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

      In normal practice I could solve A and B with 1000 difficulty in less than 2 hours. :/

      Probably my brain crashed because of taking too much practices on last night lol...

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        2 years ago, # ^ |
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        they are doing coding for many years now , if you continue doing this you will be master in upcoming years

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        2 years ago, # ^ |
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        Do you know this website? https://vjudge.net/ I think is a good sit for practice. There are all kinds of projects to practice on.

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

    Well pen-paper seems more reasonable for me

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      2 years ago, # ^ |
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      I have my own drawing pad & stylus pen so I don't wanna waste them too haha.

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

the best contest ever for me

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

Problem D

Can somebody please help me in finding out the problem of this submission. Apparently, the inputs are not being properly taken, instead some absurd garbage values are being shown. I was stuck with this for the last 40 minutes of the contest and still could not figure it out.

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

    You declared the array as integer, hence it will read whole row at once (not character by character as it is supposed to be).

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    2 years ago, # ^ |
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    Input is not n*n integers, it is n strings in n lines.

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

    You can't take the input as int sudoku[10][10]. Make it char sudoku[10][10] and change your code accordingly.

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    2 years ago, # ^ |
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    That memset tho...

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    2 years ago, # ^ |
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    cin >> sudoku[i][j]; this takes takes whole row as input. Not a single integer

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

https://codeforces.com/contest/1335/submission/76590246

https://codeforces.com/contest/1335/submission/76509810

both solutions are identical and the person who submitted it from different accounts, intentionally did so to hack (and get points if hacking gets you extra points, not sure tho) !

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    2 years ago, # ^ |
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    Maybe that's the reason why he's still gray.

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

    Successful Hacks don't change anything except the victim will lose a problem solved

    Hacks with this method are......meaningless, honestly

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2 years ago, # |
  Vote: I like it -48 Vote: I do not like it

people making such D should be banned from making future contests

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

    Can you ban MikeMirzayanov?

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

    People making such comments should be banned from making future comments

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    2 years ago, # ^ |
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    I did read the "you can change any..." as "you can swap any numbers in the grid". Still have no solution for that problem.

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    2 years ago, # ^ |
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    Can you please make some better to and provide them to problem setters so that the people who works so hard for making this problem can take some rest.

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      2 years ago, # ^ |
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      Haha , why should i do that. I only want to solve problems not make them. But if some problem is bad and i say that it's bad dosen't mean i need to make problems.

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        2 years ago, # ^ |
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        Here is a Chinese proverb for you：“你行你上啊”. no can no BB!

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    2 years ago, # ^ |
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    It's your problem...

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

~~~~~
How is the answer for 
6
1 5 1 1 1 1
5?
Test case 2 E

2! According to me it should be 6!``

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    2 years ago, # ^ |
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    1st and 3rd part must have the same length x

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    2 years ago, # ^ |
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    1 1 1 1 1 is the longest palindrome. What is your 6 length palindrome?

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    2 years ago, # ^ |
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    Number of a's on both sides of b's has to be equal as per the problem.

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    2 years ago, # ^ |
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    It should be 5. The longest 3 block palindrome here is 11111

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    2 years ago, # ^ |
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    1 1 1 1 1 is the answer and here both x=0,y=5,x=0

    basically every answer should be a palindrome with max 2 types of elements, and the 2nd type of element should be sandwiched b/w 1st type of elements , hope it gives you a little idea ! PS: thanks ashkANOn for pointing out the mistake.

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    2 years ago, # ^ |
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    remember that the 3 palindrome should be of form x y x. the length of first x and second x is the same

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2 years ago, # |
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It's really bad for me. 1335B - Construct the String said that " to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. " e.g. For test case: n, a, b are 22 17 12 my code output: abcdefghijgkkkkkkabcde

is it wrong?

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    2 years ago, # ^ |
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    My construction is:

    abcdefghijkllllllabcde

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      2 years ago, # ^ |
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      Oh, it's a really silly mistake. Thanks for helping

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    2 years ago, # ^ |
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    The first substring "abcdefghijgkkkkka" has only 11 distinct characters.

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    2 years ago, # ^ |
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    Yes it is wrong for the prefix of size 17. I think you are trying to print this "abcdefghijkllllllabcde".

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2 years ago, # |
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Can someone show me your solution to C?

My solotion is to binary-search the answer and check whether it is legal.

But my solution runs nearly 900ms.

Is there some easier and faster solutions?

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    2 years ago, # ^ |
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    Here. Mine runs in 61ms. Idea could be figured from the code but hit me up in PM or here if you need help understanding something.

    Code
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      2 years ago, # ^ |
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      Can you please briefly explain your solution?

      I have a little difficulty in understanding it....

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        2 years ago, # ^ |
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        Sure.

        I map every skill to the number of times it appears in the input. I do this because I'd like to find the skill value which is present in max number of people. This is what the variable Same holds. The size of the map will give the total number of different skill levels. However, one of the skill levels must be used to compose the team with same skill levels. So, the number of distinct skills are Map.size() — 1. And then it's a matter of evaluating which team size is maximal.

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          2 years ago, # ^ |
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          Btw, the only check you need to do is whether the biggest count equals the size of the map or if it is greater. If they’re equal, return min(biggest count, map size — 1). If not, return the min(biggest count, map size)

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          2 years ago, # ^ |
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          Thanks bro!

          I have understood your solution and finished the problem in 62 ms.

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        2 years ago, # ^ |
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        I can explain mine if you want to :)

        Solution

        Code: 76607507

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          2 years ago, # ^ |
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          Thank you, but I have finished the problem in 62 ms(^v^)

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            2 years ago, # ^ |
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            76583554 Can you please tell me why my answer is wrong in the second testcase. I have used the same logic as you. Please help.

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          2 years ago, # ^ |
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          Can you please tell me why my answer is wrong in the second testcase. I have used the same logic as you. Please help.76583554

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

Hi, can I be unrated for Codeforces Round #634 (Div. 3). Reason is with Problem D with Java. I submitted the code https://codeforces.com/contest/1335/submission/76566765 during the contest without PrintWriter but TLE'd on test case 2, although my solution is O(81*10^4) which should be under the time limit. After the contest I submitted the exact same code with PrintWriter https://codeforces.com/contest/1335/submission/76614841 and got AC. @vovuh vovuh

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2 years ago, # |
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Best contest ever for div 3. Thank you god vovuh very much and hope everyone can become expert !

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2 years ago, # |
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I didn't read D correctly and thought you needed to keep the "sudoku structure" ( just swap numbers around, no replacing ), so I came up with a backtracking solution and I just kept wondering how was it a div3D and how did so many people manage to solve it lol.

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2 years ago, # |
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Strong Pretests and Intersting Problems, Fast Test and Good Network.

Thanks to vovuh & MikeMirzayanov

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2 years ago, # |
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Plz change your time of contests in india.

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2 years ago, # |
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What is the intended complexity for E2?

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    2 years ago, # ^ |
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    O($$$N$$$ * $$$200$$$)

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    2 years ago, # ^ |
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    O(200*200*nlogn)

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      2 years ago, # ^ |
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      Not for E2

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      2 years ago, # ^ |
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      I can be done in O(N*200) analyzing from the "center" for each distinct value. Centers for value V means two distinct pointers (a, b) where the amount of V before a is the same as the amount if V after b. Then keep jumping a backward and b forward, both at the same time, but the jump is actually to the previous/next V value. While you do this, count the frequency of values in the middle of a and b, and keep the one with maximum occurrences, and add that to (amount of V before a)*2 keeping maximum for each time you move the pointers. This is linear for each value V

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        2 years ago, # ^ |
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        Nice solution!, initially I had thought of this method, but I tried to move the pointers closer instead of farther and it led to some complications in maintaining the maximum

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        2 years ago, # ^ |
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        Ah! I realise now that what I did was itself O(200*N). I did the complexity analysis wrong. Thanks!

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2 years ago, # |
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Where can I report this self-hack? https://codeforces.com/contest/1335/submission/76616926

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

In F,Why did you keep 1 as white and 0 as black ?

If your intention was to delay submission time by 2 minutes, congrats :D

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2 years ago, # |
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Could you please help me understand why my code for Problem C doesn't work? It seems pretty straightforward and I have seen other solve it my way, however my code fails on test 4. I have no idea why. Here is my submission.

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    2 years ago, # ^ |
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    I don't understand your code, but it seems to fail on the case "1 1 1 1 1 1 1...". The answer should be taking the whole array (n)

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      2 years ago, # ^ |
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      My code fails on case 4, do you mean when all the numbers are the same? In this case the answer is 1. How my code works: 1. Counts all the different numbers 2. Finds the biggest amount of repeating numbers. 3. Takes the smallest of the two (+- 1) 'difs' — different numbers 'maxN' — biggest amount of repeating numbers 'curN' — current amount of repeating numbers

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    2 years ago, # ^ |
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    Don't compare two Integer(object type) using ==. It compares reference not value.

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2 years ago, # |
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I am getting automatically logged out when i press submit button for the first time in a contest.This is happening regularly with me from past 5 to 6 contests. I am still a specialist and those extra 15-20 seconds don't matter to me right now, but this might be a bug in the server that may need a correction. Please have a look MikeMirzayanov. Thanks!

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    2 years ago, # ^ |
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    Maybe your cookies are not working.

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      2 years ago, # ^ |
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      but,this is not happening with me on other coding platforms! If it were cookies problem, then it would have occured every time i hit submit, or at other platforms too! It does not happen on the light codeforces website even!

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    2 years ago, # ^ |
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    I have faced the same issue when clicking from the problem itself instead of going to the submit code tab

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    2 years ago, # ^ |
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    I'm also facing some problem while submitting my first solution of previous 2-3 contests. I get some html 403 forbidden error with some token value!

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2 years ago, # |
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In E1 test 2 set 107 is "4 1 4 1" and the expected answer is 4?

Am I missing something?

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2 years ago, # |
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Why are some of the submissions containing this part?

if (t == 'something') cout << "OK" << endl;

Does it give any benefit?

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    2 years ago, # ^ |
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    They do that so they can get easy hacks from another account

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