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maomao90's blog

By maomao90, history, 9 months ago, In English

Hello Codeforces,

We are very glad to invite you to participate in Hello 2024, which will start on Jan/06/2024 17:35 (Moscow time). You will be given 8 problems and 2.5 hours to solve them. One of the problems will be divided into two subtasks. The round will be rated for everyone. There will be at most 2024 interactive problems, so please read the guide for interactive problems before the contest.

All the problems are written and prepared by me.

Spoiler

We would like to give our sincere thanks to:

The score distribution is $$$250 - 500 - 1000 - 1500 - 2250 - (1500 + 1500) - 4000 - 5000$$$.

Hope everyone will enjoy the round!

Congratulations to the winners!

  1. ecnerwala
  2. ksun48
  3. VivaciousAubergine
  4. gamegame
  5. cnnfls_csy
  6. maroonrk
  7. tourist
  8. Geothermal
  9. kmjp
  10. yosupo

Congratulations to the first solves as well!

UPD: Editorial

Announcement of Hello 2024
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9 months ago, # |
  Vote: I like it +157 Vote: I do not like it

As a tester I went from expert to specialist during the making of this round

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9 months ago, # |
  Vote: I like it +75 Vote: I do not like it

as a tester, I can happily tell you that this round is surely one of the rounds of all time.

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    9 months ago, # ^ |
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    Congratulations to everyone on the first competition of 2024! YAY!

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

    I'm 1330. Is this round too difficult? Don't wanan lose morale in the beginning of the year xD

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      9 months ago, # ^ |
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      Bruh, rating doesn't matter, I'm also 1330, And even if I lose 1100 rating I Would be happy bcuz of the experience I've gained, it's all about learning nothing more

      just enjoy the problems and chill, rating doesn't matter

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      9 months ago, # ^ |
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      why do you care about rating ? If you care about rating so much you can't improve in long run , see my graph I have lost expert but giving contests will only allow me to improve faster than people who are camping in certain rank

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    9 months ago, # ^ |
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    what does this mean, will it be harder than usual ?

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      9 months ago, # ^ |
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      nothing, it is simply one of the rounds of all time

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        9 months ago, # ^ |
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        Is the difficulty closer to Div.2 or Div.3?

        P.S. Anyway, good luck to everyone who participates!

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        9 months ago, # ^ |
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        Wow, just ended up on CodeForces, hope it'll be fun :)

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    9 months ago, # ^ |
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    I wouldn't normally expect that from a round but this is surprising truly amazing work guys.

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9 months ago, # |
  Vote: I like it -20 Vote: I do not like it

Please open our correspondence

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

Hope to have fun in $$$1^{st}$$$ contest of $$$2024$$$.

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9 months ago, # |
  Vote: I like it +365 Vote: I do not like it

as a tester

Screenshot-2024-01-02-191045

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

waiting for this contest...

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9 months ago, # |
  Vote: I like it -24 Vote: I do not like it

My last wish for Goodbye turned out true, so purely trying my luck, hope to become IM!

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9 months ago, # |
  Vote: I like it +198 Vote: I do not like it

Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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    9 months ago, # ^ |
      Vote: I like it +179 Vote: I do not like it

    Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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      9 months ago, # ^ |
        Vote: I like it +139 Vote: I do not like it

      Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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        9 months ago, # ^ |
          Vote: I like it +132 Vote: I do not like it

        Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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          9 months ago, # ^ |
            Vote: I like it +134 Vote: I do not like it

          Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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            9 months ago, # ^ |
              Vote: I like it +129 Vote: I do not like it

            Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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

              Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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

                Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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                  9 months ago, # ^ |
                    Vote: I like it +115 Vote: I do not like it

                  Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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                  9 months ago, # ^ |
                    Vote: I like it +103 Vote: I do not like it

                  Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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                  9 months ago, # ^ |
                    Vote: I like it +30 Vote: I do not like it

                  Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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                  9 months ago, # ^ |
                    Vote: I like it -160 Vote: I do not like it

                  Hi, Cars 1 is better than Cars 2 and 3.

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                  9 months ago, # ^ |
                    Vote: I like it -49 Vote: I do not like it

                  nobody cares

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                  9 months ago, # ^ |
                    Vote: I like it -53 Vote: I do not like it

                  I'm not sure about 2, but definetly better than 3 yes.

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                  9 months ago, # ^ |
                    Vote: I like it +37 Vote: I do not like it

                  Hormat 🫡 maomao90 🐱 for contributing to civil defence 👮 and protecting 🙏 us from people like iLoveIOI 🥶

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                  9 months ago, # ^ |
                    Vote: I like it -20 Vote: I do not like it

                  I agree with u, Cars2 was better

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

                  Hormat maomao90 for contributing to civil defense and protecting us from people like iLoveIOI

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                  4 months ago, # ^ |
                    Vote: I like it -15 Vote: I do not like it

                  Hormat maomao90 for contributing to civil defence and protecting us from people like iLoveIOI

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9 months ago, # |
  Vote: I like it -17 Vote: I do not like it

This is my first "Hello Year" contest.

I promise to solve at least 3 problems!

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    9 months ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

    There is nothing to do with your promise... Efforts are better than promises !

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9 months ago, # |
  Vote: I like it +35 Vote: I do not like it

As a tester, I can guarantee that this will be the best round of the first week of 2024

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    9 months ago, # ^ |
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    Lol, I can confirm that as it's the ONLY round in the first week of 2024.

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      9 months ago, # ^ |
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      I can guarantee that this will be the best "Hello" round in 2024 XD

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9 months ago, # |
  Vote: I like it +267 Vote: I do not like it

I was forced to test.

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9 months ago, # |
  Vote: I like it +172 Vote: I do not like it

As a tester, there is a non-negative number of problems in the problemset, and at least one person will win the contest.

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

As a iLoveIOI, peepeepoopoo

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

Hope this contest will be good, unlike last contest :)

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

Scoring distribution?

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

Deleted

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9 months ago, # |
Rev. 2   Vote: I like it +18 Vote: I do not like it

Hopefully there won't be any more googleable or oeisable problems in this round

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

    coordinator diff

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

    Can you tell me what problem(s) is(are) googleable or oeisable in Goodbye 2023(if there is any)? Are you referring to other contest instead?

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9 months ago, # |
  Vote: I like it +33 Vote: I do not like it

Please, don't be mathforces this time

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

    But math is fun...

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      9 months ago, # ^ |
        Vote: I like it -64 Vote: I do not like it

      Using algorithms is more interesting than doing math

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

        Sry to tell you but the fact is you can't be a good algorithmic programmer without being good at math

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          9 months ago, # ^ |
            Vote: I like it +78 Vote: I do not like it

          Best problems are when math, algorithms, data structure and implementation are in balance. When it's overly biased like OEIS lookup of single number input it's not fun.

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

          talkig of maths, I want to ask about the last contest"bye bye 2023", problem B.

          In the case whereb%a=0, why did we assume that the lowest divisor of b is equale to the lowest divisor of x ?

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            9 months ago, # ^ |
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            yep , i do also have the same doubt. someone please help bruh..

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

            When b % a = 0, you'd know that b = a p, where p is the smallest prime in x. Why? Let's assume that p was not the smallest prime in x, then a (b/p) would not be the second largest divisor as you'd be able to choose a smaller prime. Anyway, b is only missing this one prime, so x = b p. p = b / a, so x = b * b / a.

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            9 months ago, # ^ |
              Vote: I like it -6 Vote: I do not like it

            you will find the best answer here after min 13 https://www.youtube.com/watch?v=6vbL_jd5Ghw

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        9 months ago, # ^ |
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        Algorithms are math

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        9 months ago, # ^ |
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        But math is also a algorithm,right?

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

    Math is Life

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    9 months ago, # ^ |
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    Tfw yet another genfuc question shows up in Div1.

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

Interactive problems are awesome!!!

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

3amy Amrharb <3

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

Guys remember to not upvote the blog before the contest.

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

hope this will be better than "Goodbye 2023"

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    9 months ago, # ^ |
      Vote: I like it -76 Vote: I do not like it

    Nothing can be worse than "Good bye 2023" contest

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

      I don't want to say this but it's true

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

      Why are people downvoting you? Didn't you say the truth?

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9 months ago, # |
Rev. 4   Vote: I like it +26 Vote: I do not like it

Hope Hello 2024 != Good Bye 2023

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9 months ago, # |
  Vote: I like it +66 Vote: I do not like it

Hoping this contest brings the coordination back on TrAK

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9 months ago, # |
  Vote: I like it +42 Vote: I do not like it

I ran some code and managed to optimise the upper bound on the number of interactive problems to $$$7$$$ from $$$2024$$$.

I'll write a formal proof of my algorithm later and edit that into this post.

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

I hope this round is better than before.

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9 months ago, # |
  Vote: I like it +53 Vote: I do not like it

Fun fact: 2024 is divisible by 11 and 23

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

hope that we'll have $$$\mathtt{fun}$$$

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9 months ago, # |
  Vote: I like it +52 Vote: I do not like it
Spoiler

Sounds more promising than Goodbye 2023.

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

I hope to kick off 2024 by becoming Pupil after this round.

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

    Same

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      9 months ago, # ^ |
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      Same

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        9 months ago, # ^ |
          Vote: I like it -18 Vote: I do not like it

        You can submit 10 WAs on A and then resubmit it 100 times.

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

          Fortunately, WA penalty doesn't do anything unless you get ac, and you can't get less than 30% of the points for a problem no matter the penalty.

          Submitting 1234567891 wrong hacks should work though :))

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

18o3 sir as a tester orz...:)

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

Let's have fun in the first contest of 2024! Wishing everyone a positive delta!

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

    It's annoying how every blog announcement has like 5 of these "positive delta" wishes, even though it is impossible for everyone to get a positive delta

    Ahem, back to troll content: Good luck eveyrone! Hope you all get +200 delta in contest and reach new rank in contest!!1!!1 Hope i can reach my dream rating of 800

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

the last contest was "good bye rate" ... this contest going to be "hello rate" what do you think ?

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9 months ago, # |
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goodbye 2100, hello 2000

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9 months ago, # |
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We hope this will be better than the previous one

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9 months ago, # |
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18o3 orz

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

Hello 2024

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9 months ago, # |
  Vote: I like it -65 Vote: I do not like it

Clashing with LeetCode Biweekly. Skipping this one.

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    9 months ago, # ^ |
      Vote: I like it +150 Vote: I do not like it

    For the 69'th time, its not clashing with leetcode biweekly, leetcode biweekly is clashing with it

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      9 months ago, # ^ |
        Vote: I like it -6 Vote: I do not like it

      This is what happens when an unstoppable force meets an immovable object.

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      9 months ago, # ^ |
        Vote: I like it -91 Vote: I do not like it

      LC Biweekly always happens on the same day. CF contest happens randomly any day so you are wrong

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        9 months ago, # ^ |
          Vote: I like it +56 Vote: I do not like it

        Consistency of timings is not a measure of quality, if anything it is the reverse since the round nust happen even if the problems are not up to the mark

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

          "codechef starters = bad" ~ codechef admin

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

            I mean, nowadays cf rounds being scheduled before being ready xD, seemz like an universal problem

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          9 months ago, # ^ |
            Vote: I like it -64 Vote: I do not like it

          I feel like LC has better quality questions than CF. Most of CF rounds are Mathforces af like Good Bye 2023

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

            you are delusional, LC has the worst questions known to mankind, practically every single problem is stupid and standard and well known

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

      This has 69 upvotes. It's too perfect.

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    9 months ago, # ^ |
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    Just solve LeetCode Biweekly in 5 minutes

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      9 months ago, # ^ |
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      I aint that gud. Will Solve LC in an hour and then on to CF

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

18o3 orz tester

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

There will be at most 2024 interactive problems — what does it mean?

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    9 months ago, # ^ |
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    This means that there may be interaction problems in this round, and that you will need to learn how to deal with these kinds of problems ahead of time.

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      9 months ago, # ^ |
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      i got it, but i am confused about 2024 part

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

As-Salaam-Alaikum 2024

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

Hopefully, I was able to solve first 4 problems!

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9 months ago, # |
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as a noob ,I hope I can solve problem 1 within 1 minute and not get hacked

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9 months ago, # |
  Vote: I like it -15 Vote: I do not like it

As a tester problems are good...but I'm not a tester

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9 months ago, # |
  Vote: I like it -6 Vote: I do not like it

moo

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9 months ago, # |
Rev. 3   Vote: I like it -149 Vote: I do not like it

Do not be a second 74TrAkToR or marzipan again!

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

How humorous! I am looking forword to participating in this round!

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

Hyped up by the blog !

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

Please provide scoring distribution

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9 months ago, # |
  Vote: I like it +24 Vote: I do not like it

I wish 2.5 hours were 2 hours 50 minutes

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9 months ago, # |
  Vote: I like it -61 Vote: I do not like it

I feel the contest will not be very good

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

could a beginner participate this contest ?

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9 months ago, # |
  Vote: I like it -7 Vote: I do not like it

Can the Python be used while solving in here?

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    9 months ago, # ^ |
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    You can solve with any language that can be used to solve a problem in the problemset (Including python).

    But I wouldn't recommend using it as it's much slower than c++, you may need to further optimize your solutions in order to pass the tests.

    If you are planning to use Python submit using PyPy instead of Python which is usually much faster.

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

    For very early problems definitely. In harder problems, especially with more complex implementation, you can run into TLE, but at that point learning to code in a suitable language isn't the hardest part.

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

As a newbie i tried the sample interactive problem but solution was incorrect and i cannot see correct solutions of other people. Can someone please help me with this one

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

Hope it can make me excited instead of the frustrating "Good Bye 2023".

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9 months ago, # |
  Vote: I like it -28 Vote: I do not like it

74TrAkToR won't be the coordinator of this contest. Yay!

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9 months ago, # |
  Vote: I like it -11 Vote: I do not like it

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

Guys, I am a complete beginner to programming. I have started learning basics of C++ from sololearn.com. If there is anyone who is hearing me, who is candidate master or master. Please can you help me? I want guidance for CP.

I want to dedicate 1 year for doing CP full time. I want to utilize this time to get maximum output.

Thank you.

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

Auto comment: topic has been updated by maomao90 (previous revision, new revision, compare).

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

i can not wait to start it!!!

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

I hope this contest has the opposite number of votes as Goodbye 2023.

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

I hope I can solve the first 3 problems

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

will we see "happy new year" instead of "accepted" in this round?
MikeMirzayanov

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9 months ago, # |
  Vote: I like it -6 Vote: I do not like it

omg mao zedong round orz orz orz

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

I don't usually love contests with 500 point for problem B

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9 months ago, # |
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can someone help me with this question

A brave Knight "A" has an array of monsters to face, and will use a combination of might and magic to defeat as many as possible. In this challenge we'd like to know if the knight is successful at defeating them all, and if not, how many monsters are defeated. A can see the monsters and their order ahead of time. Despite being evil monsters they will politely queue and challenge A in the current order. Knowing this, A can plan what to do so that it is optimal.

The first monster will always be defeated by A's squires while A prepares for battle For each other monster there are two possibilities :

1.If the current monster is weaker than the previous one (i.e. monsters[current] < monsters[current-1]), The enemy surrenders — what goblin would face someone who has just defeated a dragon?

2.If the current monster is stronger than the previous one (i.e. monsters[current] > monsters[current-1]), then A has two options :

2.1) Might! A fights the monster taking damage (reducing hit points by the difference between the current and the previous monster). 2.2) Magic! A can drink an invulnerability potion, defeating the monster without taking damage.

Write a function that takes as initial parameters A list of monsters in order of how A will face them, with their strength as values; A’s initial hit points; A’s amount of invulnerability potions. And returns The 0-based index of the last monster A defeated.

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9 months ago, # |
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Hope this time I can finish at least 5 problems.

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9 months ago, # |
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Hope to become CM this round!!!

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9 months ago, # |
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I hope for a positive delta

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9 months ago, # |
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9 months ago, # |
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I wish it's somehow better than Gb2023 lol

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    9 months ago, # ^ |
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    That bar is so low you could use it to play limbo.

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    9 months ago, # ^ |
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    IMO first ever CF round would have been better that Gb2023. At least people might have learnt about maybe Dijktra or knapsack rather than just coding math operations without understanding significance.

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9 months ago, # |
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Wish a good perf and an enjoyable round.

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9 months ago, # |
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As a not tester, i can tell u this round it's much better than Goodbye2023

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9 months ago, # |
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Happy new year frands .

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9 months ago, # |
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how does score of questions related to our rating

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9 months ago, # |
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Geothermal will win codeforces round Hello 2024

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9 months ago, # |
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Will OEIS will be helpful in this round also? :P

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9 months ago, # |
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Hope, 2024 will better perform than 2023.

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9 months ago, # |
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I have deregistered even , I had registered for the contest

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9 months ago, # |
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ImbalanceForces

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9 months ago, # |
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is time limit for C too tight ?

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    9 months ago, # ^ |
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    if not, please share the approach after the completion of the round.

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      9 months ago, # ^ |
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      Greedily considering this problem. We set the last number in sets $$$a$$$ and $$$b$$$ to $$$x$$$, $$$y$$$ (where $$$x$$$ and $$$y$$$ are the maximum values initially).

      Assuming we add the number $$$z$$$ to the set:

      1. $$$x>z, y>z$$$: Add $$$z$$$ to the set represented by the smaller number in $$$x$$$ and $$$y$$$.

      2. $$$x>z, y<z$$$: Add $$$z$$$ to the set represented by numbers greater than $$$z$$$ in $$$x$$$ and $$$y$$$.

      3. $$$x<z, y<z$$$: Add $$$z$$$ to the set represented by the smaller number in $$$x$$$ and $$$y$$$.

      Then we solved the problem within the complexity of $$$O(n)$$$.

      submission link

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9 months ago, # |
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kringe round it was too bad!

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9 months ago, # |
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unable to solve C,i better not be a retard

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    9 months ago, # ^ |
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    Every contest should be unrated when "YOU" can't solve problem C.

    Nice JOKE.

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    9 months ago, # ^ |
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    I think that its about calculating the longest ( non increasing subsequence) but I couldn't figure out an approach except for the n^2 one

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      9 months ago, # ^ |
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      There's an nlogn way to find LIS, but it's not required for the problem. I solved this by greedy

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      9 months ago, # ^ |
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      C is not about LIS. But LIS is solvable in o(NlogN). I tried really hard though, to prove LIS way of solving C, but i can't. This sort of problem is really pain in the ass i gotta say.

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        9 months ago, # ^ |
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        I made a mistake I meant to say (Longest Non-Increasing Subsequence) not LIS

        did you try to calculate that and it gives you WA too ?

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      9 months ago, # ^ |
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      I don't think calculating the longest non increasing subsequence is the right approach as there are multiple possible such sequences and it is not necessary that all of them will give the same penalty

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    9 months ago, # ^ |
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    apologies.

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9 months ago, # |
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Prove_with_ACforce

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9 months ago, # |
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I've participated in codeforces contests for 7 years, and I still can't solve Div2C. I don't know how much I've progressed in past 3-4 years. Maybe its time for me to quit this game now.

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9 months ago, # |
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C was nice. I love it when I prove a solution during the contest

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    9 months ago, # ^ |
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    it was about finding the ( Longest Non-Increasing Subsequence) right ?

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

      No, just try to start with the end. Then, you can compare each new item with the last added item in each of the two subsequences

      The rest is casework

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

        ok thanks, but can it be solved if we found the longest non-increasing subsequence?

        the (Longest Non-Increasing Subsequence) penalty will be 0 and then we calculate the penalty of the remaining numbers in the set. do you think that this is a valid Solution?

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          9 months ago, # ^ |
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          You mean the longest non-increasing subsequese which will give us penality 0, right?

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            9 months ago, # ^ |
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            yes , I'm sorry

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              9 months ago, # ^ |
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              No worries!

              The longest non-increasing subsequence will not work

              Consider this test case:

              10

              7 4 1 6 2 3 5 8 1 9

              If we take the first subsequence as the longest non-increasing subsequence it can be

              7 6 5 1

              The other will be

              4 1 2 3 8 9

              Which has penality of 4

              But consider this solution wich have penality of 3 only

              1 6 3 9 8

              7 4 2 5 1

              The first has penality of 2 and the second has penality of 3

              which is less than the (longest non-increasing) solution

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                9 months ago, # ^ |
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                thanks, some comments on the editorial section are talking about the correctness of this approach you can place this counter example there too

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                9 months ago, # ^ |
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                No it will be 7 6 5 1

                4 1 2 3 9 8 This will have penatly 3.

                u interchanged them.

                I know LDS wont work because you can take this sequence :

                27 28 29 100 99 98 97 96 20 19 18 30 27

                When by LDS

                100 99 98 97 20 19 18

                27 28 29 30 27

                The penalty :3

                The better would be 100 99 98 97 96 30 27

                27 28 29 20 19 18

                Penalty :2

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                  9 months ago, # ^ |
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                  Yes, you are correct I changed the two numbers while testing

                  Thanks for clarifying

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9 months ago, # |
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Goodbye, 2024.

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9 months ago, # |
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I tried so hard to solve C. I tried varies of approaches to deal with it, but still failed. But I didn't give up. I tried to drew a lot of examples, tried to use dp, binary search, ternary search, graph, extended euclidean, ford fulkerson algorithm, .... And finally, I realised that I still unable to solve C.

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    9 months ago, # ^ |
    Rev. 2   Vote: I like it +4 Vote: I do not like it

    Kind of the same... Sent 2.5 different solutions and tried maybe 5 approaches and all WA 2

    UPD: actually I had correct idea but just initialised arrays wrong... Now I am specialist for the first time since 2014...

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    9 months ago, # ^ |
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    problem C reminded me of my skill issue

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    9 months ago, # ^ |
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    Same goes with me ,First I tried with LDS(using binary search) then soon realised the question might not be that much complex.

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    9 months ago, # ^ |
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    I think the solution will be greedy, the basic argument is every element will either go array a or b

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9 months ago, # |
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wow what IS d? new year, new pain

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    9 months ago, # ^ |
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    I found out that if x is the biggest element must have a brother that's equal to x-1 (both leaves). From that i think you can delete those two elements from the array and substitute them with their father (that has value of x-1) and solve again. I tried this with some data structures (double linked list and priority queues, very ugly) but got WA on pretest 2. I spent like 1.5 h on this :(

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      9 months ago, # ^ |
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      "I found out that if x is the biggest element must have a brother that's equal to x-1 (both leaves)."

      I guess, not "x is the biggest", but x is the deepest.

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        9 months ago, # ^ |
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        if x is biggest, then it won't have children. Though this doesn't mean it will have a leaf brother, at least one maximum should have a brother that's a leaf. Maybe it becomes a problem if there's two possible brothers, i'm not sure if both choices lead to a tree or not

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      9 months ago, # ^ |
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      I saw that too, but what if X has both neighbors equal to X-1? Which one do you merge it with? What if it has one or both neighbors equal to X? I didn't really see any breakthrough in this line of thought

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        9 months ago, # ^ |
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        WOW I fucking misread the statement. I thought we could freely color the edges 0 or 1, turns out one edge HAS to be 0, and one edge HAS to be 1 for each non-leaf. I'm going to kms

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        9 months ago, # ^ |
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        If you do this until you can't:

        • Choose an inner node
        • merge it with its child with 0 edge

        The result will be a tree, not necessarily binary. The Dfs over leaves is now equivalent to dfs in tree, but with one tweak. We need to decide after which children (or it's possible at the beginning) we insert the inner node into dfs_order. Everything is possible, which means if (e.g.) choose the root, then there will some subtrees dfs_order concatenated on the left, and some subtrees dfs_order concatenated on the right. This can be checked with RMQ and recursion, but instead of deciding where to break the concatenated blocks in the root, we will decide them in the children. So at the end we need to check if the root is unique.

        240558002

        (oh i see you misread but maybe this will be helpful to someone else)

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        9 months ago, # ^ |
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        In that case if you have [... x-1, x, x-1, ...], then you merge any neighbor to x and always get [... x-1, x-1, ...]. Anyways this is the editorial sol

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      9 months ago, # ^ |
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      i did exact same thing and 5 minutes after the contest i realised, that there must stay only 1 element and it must be equal to 0, otherwise it's a "NO".

      240601323

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9 months ago, # |
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CCCCCCCCCCCCCCC!help!

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9 months ago, # |
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is problem c dp?

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9 months ago, # |
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is D DSU?

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    9 months ago, # ^ |
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    Yea, I did a DSU based solution; however, something like linked lists might be easier to implement

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9 months ago, # |
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    9 months ago, # ^ |
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    Wow, I came up with the same solution without realizing that the complexity becomes linear if done recursively.

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9 months ago, # |
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Really nice problem D! A bit hard for its position though?

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What's the idea for D?

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    9 months ago, # ^ |
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    Firstly, there must be only one $$$0$$$ in a valid sequence. Next, give some examples on the draft paper. You will find that if $$$x (x>0)$$$ appears in the sequence, then $$$x-1$$$ must have appeared.

    Let $$$x$$$ be at position $$$t$$$. Combined with the drawing, it can be found that $$$t$$$ must be within a interval $$$(l, r)$$$, satisfying the conditions of $$$\forall i \in (l, t) \cup (t, r), a_ i>=x$$$, and $$$a_l=x-1 \vee a_r=x-1$$$.

    I'm not very good at expressing its proof in words, sorry!

    Finally, we use dfs and binary search to solve this problem. Pass three parameters $$$l, r, x$$$ into the dfs function, representing the current interval $$$[l, r]$$$ and the value $$$x$$$. Record the position of each value in the array $$$t$$$, find the value $$$x-1$$$ in the interval $$$[l, r]$$$, split the entire interval into several small intervals, and recursively solve the problem.

    Happy New Year!

    Here are some examples as a reference:

    5
    6
    1 0 3 2 3 1
    6
    1 0 3 3 3 1
    5
    1 0 3 2 1
    5
    0 1 0 1 1
    7
    0 1 2 3 4 5 2
    
    Yes
    No
    Yes
    No
    Yes
    
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      9 months ago, # ^ |
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      Additionally, if you use bfs instead of dfs, some optimizations can achieve $$$O(n)$$$ complexity!

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9 months ago, # |
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The webpage lag 15 minutes after the start of the competition caused me a lot of trouble.

Anyway, the problems all look interesting, thanks!

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9 months ago, # |
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Today contest: Problems A & B is which ratings predict plz..? problem C question is kinda hard for me, I mean understanding the concept! what should I do to resolve this? also C: predict ratings? how much...

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9 months ago, # |
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problem d is interesting but so hard, didn't solve :(

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9 months ago, # |
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$$$\frac{Div. 3 \ + \ Div. 1}{2} \neq Div. 2$$$

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9 months ago, # |
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great contest but did not manage to perform as expected !!

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9 months ago, # |
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How to practice for problems like C (guessable but not trivial greedy problems)?

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    9 months ago, # ^ |
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    Practicing Greedy problems might help.
    Practice reasoning based on "Proof by contradiction"

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9 months ago, # |
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Accidentally submitted F2 to F1 & got -350 score...

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9 months ago, # |
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Great contest! Thanks.

Hello 2024 != Good bye 2023

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9 months ago, # |
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Wow, it was such quickforces. My last accepted submission was on 0:11 .

My ideas to D:

  1. DFS. Assume we are on vertice. We remember what is above in stack. Either sum on path is $$$a_i$$$, so we calculated it, leave the leaf; or it is inner vertice, do exactly two dfs-s. If we are out, and array $$$a$$$ is no used fully, append one vertice above and start from it with only one dfs call. Incorrect.
  2. Greedy. Go from left to right. When we are on some position, we can do $$$-1$$$ on segment with this position as left border, and all different right positions. Use maximum right border every time. Incorrect model.
  3. Stress. Try to find pattern. Could not find.
  4. Split array on $$$0$$$-s to components. Size of each component is at least 2^max on compoment. What is next?
  5. Go from small values to big values. For all segment of values at least x its length has to be at least 2^length. Incorrect.
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9 months ago, # |
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Im failure after taking this contest :(

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9 months ago, # |
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Even though I couldn't solve the problems, I liked the problems as they are short and nice.

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9 months ago, # |
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Interesting contest, third problem was as easy, as it was hard.

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9 months ago, # |
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Accepted/Tried

How brutal the C test is. (The pretest btw)

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9 months ago, # |
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cloudflare is SHIT

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9 months ago, # |
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Now I know how difficult to welcome the new year is.

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9 months ago, # |
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I wonder how many people proved solution of C.

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    9 months ago, # ^ |
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    welcome to codeforces

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    9 months ago, # ^ |
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    In my opinion, high-level coders will prove this because they can demonstrate it almost instantly.

    The proof of this problem is a typical one. Consider iterating from 1 to n. Suppose the current two subsequences end with a and b, assuming a <= b. Suppose the current number is x. Consider two cases:

    1. You put x after a number greater than or equal to it. In this case, if both a and b are greater than x, choose a. Otherwise, choose b.

    2. You put x after a smaller number. In this case, it will choose to put x after a.

    We consider that if we can make two choices in the current step, it must hold a < x <= b. If we choose 2, it becomes x, b, and penalty++. If we choose 1, it becomes x, a. We can imagine that the penalty is like a free ticket, which can change any number into INF at any time (including changing a into INF instantly), so 'a' with one less penalty is strictly better than 'b' with one more penalty.

    I don't know how others did this problem, but I only realized the answer to this problem after the proof.

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    9 months ago, # ^ |
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    Proof by AC is the most powerful method

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    9 months ago, # ^ |
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    Didn't really prove it, but my reasoning went kind of like this. There's no real reason to take a penalty if not strictly necessary. If i raise one of the two sequences when not required I might also have done this later for the same cost. Then I just thought about how to keep a good state and figured out the best greedy moves after some tries. Then i proved by AC.

    Proof is a big word, but having an idea what you're doing is good enough usually

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9 months ago, # |
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More and more vegetables,what should I do?

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9 months ago, # |
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First two problems were satisfying. Solutions are short and pretty

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9 months ago, # |
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Is F1 difference + segment tree?

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E seems classic, but I can't solve it. How to solve E? QAQ ~

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240509224 240515600

Identified and rectified a discrepancy in Problem A submissions; both versions passed the pretests and shared the same logic. However, there was a point reduction of 50 points. :"(

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9 months ago, # |
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Someone Kindly share the solution of A using recursion. Thanks.

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    9 months ago, # ^ |
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    u can just solve it through if a + b is an odd or not

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      9 months ago, # ^ |
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      Yeah. Missed that simple observation :(.I am trying to understand how recursion works. Confused how to implement this as there can be 6 cases I think?

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    9 months ago, # ^ |
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    you don't need recursion observe that in every turn total coins will decreased by 1. when will it become 0 ?

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Deserves the first contest of 2024. I really enjoyed it. :) Plus, thanks for the flash-fast editorials.

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9 months ago, # |
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C was the hardest problem of all time

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Why do I perform well in shxt rounds and brick the good rounds, weird

spent 1h writing F for nothing

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9 months ago, # |
  Vote: I like it -11