Markadiusz's blog

By Markadiusz, 3 years ago, In English

Hello Codeforces!

Asymmetry and I are glad to invite you to Codeforces Round 743 (Div. 1) and Codeforces Round 743 (Div. 2), which will be held on Sep/18/2021 17:35 (Moscow time).

Each division will have 6 problems and 2 hours to solve them. All problems were written and prepared by Asymmetry and me. The round will be rated for both divisions.

We would like to thank:

We've put great efforts into preparing this round and we hope that you will enjoy it.

Good luck!

UPD: We would also like to thank Ari for testing the round and KAN for translating the statements into Russian.

UPD2: Here are the scoring distributions:

Div. 1: $$$500$$$ — $$$1250$$$ — $$$1750$$$ — $$$2500$$$ — $$$2500$$$ — $$$3250$$$

Div. 2: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$2250$$$ — $$$2750$$$ — $$$3500$$$

UPD3: We are sorry that the round became unrated due to the long queue. We hope that you enjoyed the problems anyway.

Winners

Congratulations to the winners!

Div1.

  1. tourist

  2. Benq

  3. VivaciousAubergine

  4. ecnerwala

  5. Molewus

Div2.

  1. ChenRui

  2. rainboy

  3. OguriCap

  4. _October

  5. Dallby

UPD4: Editorial is out!

  • Vote: I like it
  • +878
  • Vote: I do not like it

| Write comment?
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3 years ago, # |
  Vote: I like it +270 Vote: I do not like it

Since Monogon had ascended into the coordinator realm, I became the new VIP tester.

btw, as a VIP tester, I tested

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3 years ago, # |
  Vote: I like it +179 Vote: I do not like it

As a tester, I wish you sunny weather ^^

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3 years ago, # |
  Vote: I like it +253 Vote: I do not like it

As coordinator I want to thank the VIP authors Asymmetry and Markadiusz

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

Looking forward to participate

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3 years ago, # |
Rev. 3   Vote: I like it -122 Vote: I do not like it

Challenge:hit 200 downvotes under this comment (bye bye contribution)

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3 years ago, # |
  Vote: I like it +145 Vote: I do not like it

As a coauthor, I wish everyone best of luck and a positive skill delta after the contest.

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3 years ago, # |
  Vote: I like it +46 Vote: I do not like it

As a tester, I wish everyone happy rating :)

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

Don't wanna miss the contest but suffering from anxiety and depression for the past few days.Hopefully I will overcome anxiety and depression soon and start playing in contests

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    3 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    Mental health is much more important than competitive programming, so take it easy, hoping for some well wishes from a fellow stranger.

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

      Yeah,I am going to meet my psychologist today for that reason.Being mentally strong is important in cp too.Because you can have all the skills in the world but if your mind is unstable during 2 hour contest I Don't think you can deliver your best.Anyway,I am going to consult my psychologist soon and come back strongly.And thanks for your kind words.Cf has really been like a family to me

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

        I hope you become well in the coming days and I wish the best of luck for you in the contest

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

          Thanks for your support.Unfortunately I am not stable enough to give today's contest but hopefully I will be giving every contest in cf from 23rd September.I wish you all the best in today's contest

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

            Thanks turzo_sawroop , I'm still working on a strategy to reach a better level but I think it will take some effort and time to reach a higher rank

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              3 years ago, # ^ |
              Rev. 4   Vote: I like it 0 Vote: I do not like it
              • solve 1/2 1200 rated problems

              • solve 1/2 1300 rated problems

              • solve 1/2 1400 rated problems

              Do these three things everyday

              • solve 2 1500 rated problems

              Do this in every week

              • Don't miss any contest

              Do all these in the next 2 months and be confident and always believe you can be as good as anyone.

              Hopefully your rating will be between 1200-1300 after that

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

                From my experience, I think you just need learn the basic technique to get rating >1600. Do too much easy problem doesn't make your skill better.

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

kickstart round is also scheduled from 17:00 UTC on the same date, can u guyz plz prepone the round?

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3 years ago, # |
  Vote: I like it -89 Vote: I do not like it

As someone who wants to hit -69 contribution mark, I request y'all to downwote for high ratings. thank you!

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3 years ago, # |
  Vote: I like it +122 Vote: I do not like it

As an old, almost retired contestant who grew up in the same school as the authors, I'm thrilled to see what my younger comrades prepared :)

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

Interesting division 2 scoring distribution. I expect (and hope for) a sizeable jump between C and D.

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

Early scoring distributions! :D

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

I am a newbie and started giving contests on this platform from this month itself. Excited for this another upcoming contest!! All the best everyone!

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

    I am a specialist and started giving contests on this platform from last year itself. Excited for this another upcoming contest!! All the best everyone!

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

      I am a pupil and started giving contests on this platform from this year itself. Excited for this another upcoming contest!! All the best everyone!

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

        I am a future expert and started giving contests on this platform 20 yrs ago. I was excited for this then upcoming contest!! All the best everyone!

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

Target to solve A, B & C asap to get a good rank, D seems to be much harder based on rating distribution.

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

Noice looking contest good luck :> HEHEHE HA

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

I hope a good result for everyone !

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

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

Very good stuff worked to prepare this contest I guess, thanks

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

As your father,I want some contribution

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

I just want to say that 743 is a prime number.

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

Wish Me Cyan !!!

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

"An Unexpected error" is showing up whenever i am submitting my solution!!! Anyone else facing the same issue????

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

Queue :(

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

too long queue :(

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

A contest after ages and what we see is a Queue

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

unrated please

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

QueueForces T_T.

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

such a long queue , it has to be unrated .

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

15 minutes since A is in queue.....

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

too long queue:(

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

while true: dequeue()

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

please extend the contest by 15 mins at least due to the long queue.

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

    don't make it unrated because the problems are really good.

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

      just cause you solved , this is unfare to wait for 20 min to get wa verdict

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

        i never said it is fair to not make it unrated but an alternative is to extend it since a contest is much awaited by everyone.

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

          imagine waiting for B and then getting wa after 30 min for some silly stuff , meanwhile some guy solved after 30 min and submitted in first go , I dont think extending contest will justify now

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

This should definitely be unrated!

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3 years ago, # |
  Vote: I like it +40 Vote: I do not like it

The contest must be unrated . Such a long queue

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3 years ago, # |
  Vote: I like it +29 Vote: I do not like it

I think there should be a clear guideline in Codeforces when the round should be unrated. Because queue in first few minutes can severely affect the rating in case there is a difficulty gap in the problems.

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

Why so long queue??

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

Too long queue, please, make this round unrated

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

The queue is too long today? i submitted a solution 5 mins back and still it's not judged yet.

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3 years ago, # |
  Vote: I like it +31 Vote: I do not like it
Queueforces ;-;
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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

MikeMirzayanov what type of queue is used for submissions, my submission at 7 min is not yet judged but some user's 12 min sol are judged. Please help!!!!

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3 years ago, # |
Rev. 2   Vote: I like it -7 Vote: I do not like it

If unrated, my possibility of becoming CM will be ruined for this time.

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

Unrated please!!

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

Why queue is so long?

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

Is it unrated?

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

20 mins just to get wrong answer so sad

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

The contest should be unrated cause of long queue!

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

I am waiting for more than 30 minutes, still my solution is not judged .

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

Disgusting queues

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

Such a bad queue :(

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3 years ago, # |
Rev. 2   Vote: I like it +19 Vote: I do not like it

Me: submits

Codeforces: Click this

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

Imagine submitting before 30 min and getting WA. :))

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

It'll be unrated, there's no point participating. There's no better way to compensate for never getting results on pretests.

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

queueForces

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

it's unrated no doubt just have fun solving problems

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

Nice round. It's a pity that this round is unrated(

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

Great questions, sad the round will be unrated.

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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

Second person to solve C only for round to get unrated xD

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

I thought the problems were really nice, so I'm disappointed that the round went unrated.

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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

will it be unrated for div 1 also?

still 0 announcement

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

When I solved A and B under 15 minutes, the round gets unrated. Of course, why not :(

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


Thousands of submissions are in queue and only one of them is running. why?

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

B and C were so beautiful, very sad it's unrated.

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

this is not expected from such big organization please fix this for once I did so good in the contest (a and b) this fast (i'm newbie) :(((

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3 years ago, # |
  Vote: I like it +131 Vote: I do not like it

Results of other OJ:

Result of Codeforces:

lol

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

Feels sad man , when first contest you give goes unrated :(

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

nowdays Long queue is common problem in cf. please do something MikeMirzayanov :(

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

Imagine waiting for a Codeforces round for 1 week... and this.

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3 years ago, # |
  Vote: I like it +38 Vote: I do not like it

Unlucky comeback after ~2 years of inactive. F.

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

unraited...

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3 years ago, # |
  Vote: I like it +90 Vote: I do not like it

How to get rid of queues on CF?

The answer is...
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3 years ago, # |
  Vote: I like it +24 Vote: I do not like it

It was the first contest I liked in probably the last 12 months and seeing it getting unrated this way, I feel sad for the authors very much. Markadiusz and Asymmetry I liked the problems a lot and hope to see you guys with another contest soon!

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3 years ago, # |
  Vote: I like it +52 Vote: I do not like it

Problem 1C seems to be the same with Problem C in 2020-2021 ACM-ICPC, Asia Kunming Regional Contest.

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

cucked my comeback

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

The problems are nice! Because it is unrated now, I can sleep early today LOL :)

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

This is still a good problem collection and worth trying out. I will still try to do my best even if it's unrated.

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3 years ago, # |
Rev. 2   Vote: I like it +52 Vote: I do not like it

1-gon's first round got unrated due to the long queue. Now Asymmetry and Markadiusz set their first round coordinated by 1-gon, and the round got unrated due to the long queue. RIP

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

logic behind b?

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

    Since a[0]!=b[0] we need to make those two elements so that a[0]<b[0]

    That is, we iterate all a[i], and foreach one we find the smallest possible j so that a[i]<b[j] That pair of i,j need i+j operations, we search for the min such pair.

    We can find the smallest j for a given i by binary search on the prefix arrays of a[] and b[]. make preA[i] the smallest value in a[0..i], and preB[j] the biggest value in b[0..j]. Binary search the smallest j where preA[i]<preB[j].

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

      We can avoid binary search since we know the numbers in a[] are 1, 3, ..., 2n-1. Create an array c[] such that c[k] gives the index j for number k i.e. if a[i]=k, we get j by c[k]. To fill c[], we can iterate over array b and store index j for all numbers smaller than the current no in b.

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

        Doesn't this require sorting, which yields linearithmic time complexity either way?

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

          No, the numbers are relatively small compared to the length of the input (<=n), so a simple loop, described above, is enough.

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

            Oh I completely misunderstood that comment. I got it now, thanks!

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3 years ago, # |
  Vote: I like it -49 Vote: I do not like it

give editorial ASAP.
I have to do a lot of homework.

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

    well I think: 2020 -> 0022 -> 0020 -> 0002 -> 0000

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

    2020 -> 2002 (swap 3rd and 4th element) -> 2001 -> 2000 -> 0002 (swap 1st and 4th element) -> 0001 -> 0000

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

The problems were good. (especially Problem B...able to do in O(N)) PS: All the best for Kickstart buddies. :)

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

Any thoughts of problem E? It seems to be a dp problem, but I didn't figure out the recursion formula.

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

For problem C, we first figure out if it is possible to finish the whole book by SCC (strongly connected component). If size of some (possibly 1) SCC are greater than 1, then it is impossible to complete the book. After that, we use 2 min heap to store pages with indegree 0. First heap stores current traverse. Second heap stores next traverse.

Link for my submission

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

    Using SCC is a bit overkill and definitely a waste of time for this problem.

    The goal (when checking if understanding the book is possible) is to check if there is a cycle. We can do that with a DFS (or multiple, in case of a DAG, which we are dealing with here) which, instead of marking a vertex visited, can set two different kinds of flags.

    The DFS works as follows:

    1. Set flag "in progress" for current vertex
    2. Consider all neighbours, for each 2.1. If its flag is "in progress", then a cycle is detected 2.2. If its flag is "unvisited", then run the DFS on the neighbour.
    3. Set flag "visited" for current vertex

    We can count the result with a DFS too, using dynamic programming on a DAG. Every vertex with no outgoing edges receives value 1, because that chapter will be understood during the first reading of the book. Every other chapter will be understood after all its prerequisite chapters are understood, so the result for the chapter is the maximum of the results for the prerequisite chapters... but slightly modified to account for the order of the chapters in the book.

    I'm not the best at explaining so maybe my code will help.

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

      topological sort :|

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      3 years ago, # ^ |
      Rev. 5   Vote: I like it 0 Vote: I do not like it

      I actually used a different approach. I made two heaps : the first one is the "main" heap, the second one is the "pending" heap

      The "main" heap's purpose is to store the progress chapters you can learn right now given that you're already learn all required chapters. Meanwhile, the "pending" heap's purpose is to store the chapters which you have learned all the required chapters but the chapter number is less than the chapter we currently on

      So let's process the chapters one by one starting by those which have no requirements. (E.g. the ones which have 0 in-degree)

      Each time we're processing of the current chapter (let's say we name it U) we do this :

      • check all other chapters V which has U as requirements

      • subtract V's degree by 1

      • if V's degree become 0 then : A. If V > U, then this chapter comes after U, then put it on the "main" heap OR B. If V < U, then this chapter lies before U, since we can't read it backwards, put V to "pending" heap

      on the end of each processing, if our "main" heap becomes empty, put all chapters in "pending" heap to "main" heap and increase the number of read by 1 (because this means we will start re-reading the book)

      To check if it's impossible, just check if after we're done with all the processing, there exists at least a chapter we haven't processed before

      My implementation

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

        This work is O(nlog(n)) though due to the heap! But it's still a fine approach.

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

How to solve div2 D?

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    3 years ago, # ^ |
    Rev. 3   Vote: I like it +36 Vote: I do not like it

    First of all, xor of all the elements of the array should be $$$0$$$ (xor of all the elements of the array isn't changed after any operation).

    From here, I am assuming that xor of all the elements of the array is $$$0$$$.

    If n is odd, it's always possible to change all the elements to $$$0$$$.

    Note that if $$$a_{i}=0$$$ and $$$a_{i+1}=a_{i+2}$$$, $$$a_{i}\oplus a_{i+1}\oplus a_{i+2}=0$$$. This is the main idea of my approach.

    Now select indices $$$n-2,n-4,...,1$$$. It's easy to observe that now the first element of the array will be $$$0$$$ and $$$a_{i}=a_{i+1}$$$ for all even $$$i$$$ less than $$$n$$$. First element will always be $$$0$$$ because it is the xor of all the elements of the array Now select $$$1,3,...n-2$$$. After these operations, all the elements of the array will be $$$0$$$ and total number of operations performed is $$$n-1$$$.

    If $$$n$$$ is even, you can split it into two subarrays of odd length. So if you can find an $$$i$$$ such that $$$i$$$ is odd and xor of first $$$i$$$ elements is $$$0$$$, you can change all the elements to $$$0$$$, otherwise you cannot.

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

      Oh holy crap this is brilliant. And here I got stuck with some greedy ifological strategy which... I'm reasonably confident would work but I didn't manage to implement it.

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

        Yeah, such approach can work too but I had to consider many different cases.

        Basically: we iterate from left to right, either A keeping all elements behind 0 or B leaving all elements behind 1 (for simplicity ensuring that the current element is 1 too after previous operations).

        A: if applying the operation to the current $$$i$$$ will result in zero xor, we do it. Else if the $$$i$$$th element is 1, we ensure that $$$i+1$$$ is too; we know that $$$i-1$$$, if it exists, is zero, and we can zero $$$i-1..i+1$$$, and if it doesn't, we switch to B.

        B: if applying the operation to the current $$$i$$$ will result in zero xor AND there's an even number of 1s behind us, we zero $$$i..i+2$$$ and 1s behind us two at a time. Otherwise we ensure that the next element is 1 (working through a few cases it can be shown that if it's 0, the current xor must be 1).

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

    First you must have an even number of 1s in the array, otherwise it's impossible to change all elements to 0s.

    Then group all the 1s in pairs: (1st, 2nd), (3rd, 4th), ... we will solve the problem by changing each pair of 1s to 0s repeatedly:

    1. if there are an even number of 0s between two 1s, we firstly change those 0s to 1s by performing operations on i such that $$$a_i=1, a_{i+1}=0, a_{i+2}=0$$$ repeatedly. Then perform operations on i such that $$$a_i=0, a_{i+1}=1, a_{i+2}=1$$$. Note that assume the indices of the two 1s are $$$l$$$ and $$$r$$$, we must have either $$$a_{l-1}=0$$$ or $$$a_{r+1}=0$$$ in order to solve this case.
    2. if there are an odd number of 0s between two 1s, similar to case 1, we first change all zeros to ones except for the rightmost zero. Now we have $$$1\dots 101$$$, then we perform an operation on 101, then do the similar thing on case 1 but from right to left. Note that there are no extra requirement to this case.

    So the strategy is firstly find one pair that can be solved then solve the remaining pairs from that pair in both direction. If none of the pairs can be solved, it's impossible to change all 1s to 0s. Submission: 129221560

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3 years ago, # |
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How to solve div2 B ?

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    3 years ago, # ^ |
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    since array a is odd numbers and array b is even numbers.
    so lexicographically smaller/greater is defined at index 0.
    so only comparing values at index 0 is enough
    
    
    let cost[x] = # swaps to move a value v < x in array 'a' to index 0.
    
    ans = 2n
    and iterate over i = 0 to n - 1:
      ans = min(ans, i + cost[b[i]]); // i <- #swaps to move i_th number in array b to index 0
    
    if iterating is over.
    the ans will be the answer
    
    ----------------
    how to create cost array?
    intialize cost array with max value = n
    for (int i = 0; i < n; ++i) {
        cost[a[i]] = i;
    }
    for (int i = 1; i < 2 * n; ++i) {
       cost[i] = min(cost[i], cost[i - 1]);
    }
    
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3 years ago, # |
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My segment tree solution for Div1 A

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    3 years ago, # ^ |
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    can you just briefly explain what you are storing/updating in the nodes?

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3 years ago, # |
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Can anyone help me to find a counter case? 129205285

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    3 years ago, # ^ |
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    1
    4
    3 2 3 4
    0
    2 2 4
    0
    

    Correct answer is 3 yours gives 2.

    I have used a similar approach with the topological sort, except instead of storing the chapters to be read in the next iteration in vec, i just added n to the chapter and pushed it into pq. 129229753

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      3 years ago, # ^ |
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      Thanks a lot, but i was so dumb that i forget priority queue stores greater elemnet first. I thought it keeps smaller element first. You just saved a lot of time mine.

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3 years ago, # |
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I have tried a video editorial for problem C :book https://youtu.be/KWkkZEGffZw . Have a look at it..and if have any doubts let me know in comments

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3 years ago, # |
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Can I change the statement that a[i] is (0/1) to [0,100000] on Problem 'Xor of 3'.Is there a solution?

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    3 years ago, # ^ |
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    Maybe you can solve the task in each bit, but the operation times may exceed $$$n$$$.

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      3 years ago, # ^ |
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      I think We can do that as well in n-1 operations.

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    3 years ago, # ^ |
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    I think my solution doesn’t use the 0/1 condition so probably yes.

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3 years ago, # |
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editorial is also still ``in queue'' LOL

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3 years ago, # |
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For Question B, Can Someone please tell me why my approach is wrong? https://codeforces.me/contest/1573/submission/129231370

My approach is that our task is to make a[0] < b[0], this can be done by two ways, either bring the closest number which is greater than a[0] in 'b' to the 0th position, let's say it is at j-index, then the answer will be 'j', or bring the closest element from 0 in a which is less than b[0], and the final answer is the minimum of both!

Any counter test-case which fails this approach would be highly appreciated!

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    3 years ago, # ^ |
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    You are only considering making the swaps on either a or b. In some cases it is beneficial to do the swaps on both arrays.

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3 years ago, # |
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Is div1 B supposed to make us feel pain?

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3 years ago, # |
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waiting for editorial ...

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3 years ago, # |
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orz Molewus for the round being unrated.