By Dormi, 13 months ago, In English

Hello, Codeforces!

Welcome to the Codeforces LATOKEN Round 1 (Div. 1 + Div. 2) supported by LATOKEN, that will start on Jun/13/2021 18:35 (Moscow time). It will be a combined rated round for both divisions (and div 3). Please, note the unusual start time.

All problems were created by Aaeria, crackersamdjam, Dormi, Ninjaclasher, and Plasmatic.

We would like to thank:

You will be given 8 problems and 180 minutes to solve them. This contest features at least one interactive problem, so we recommend reading the guide of interactive problems before the contest.

We hope that statements are short and pretests are strong and that you find the problems interesting! Good luck, have fun, and we wish everyone high ratings!

The scoring distribution will be announced closer to the beginning of the round.

Thanks to LATOKEN, winners will get awesome swag:

  • 1st place = MacBook Air M1 + Job offering and Stock Options
  • 2nd place = Xiaomi Mi 11 + Job offering and Stock Options
  • 3rd place = Xiaomi Mi Notebook Pro 15" Enhanced Edition + Job offering and Stock Options
  • 4th-10th place = Job offering and Stock Options + Merchandise Pack
  • 11th-55th place = Merchandise Pack

Random 10 participants outside of top-55, who solved at least one problem and participated in rated Codeforces rounds before, will get a t-shirt!

A few words from the LATOKEN team:

We welcome you to the intellectual battle between the best coders from around the world! LATOKEN strives to make trading crypto as easy as having a social media account. We are always looking for the best of the best to join our team. As a result, LATOKEN is currently in Coingecko’s Top 15 with over 600 000 app installs within just one year from its launch and 1m+ users in total. The emerging blockchain based financial systems will open exciting opportunities, because crypto is fully global and mostly independent of conservative banks.

If you want to become a part of something bigger, work remotely with a team of international professionals and change the world, contact our recruiters for internships and employment opportunities in the LATOKEN team. Use our bot or fill out the form below and tell us a little about yourself.

Bot→
Contact Form →

UPD: The scoring distribution is 500 — 1000 — 1250 — 1500 — 2250 — (1750 — 1250) — 3250 — 3500.

UPD: Editorial

Congratulations to the winners:

  1. Benq
  2. tourist
  3. Radewoosh
  4. Endagorion
  5. ecnerwala
  6. heno239
  7. greenheadstrange
  8. gamegame
  9. Pyqe
  10. maroonrk

The prizes will be distributed in the next few days after we remove cheaters.

 
 
 
 
  • Vote: I like it
  • +638
  • Vote: I do not like it

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13 months ago, # |
Rev. 2   Vote: I like it +116 Vote: I do not like it

Do the contest or TLE real life

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

As a tester, the problems are great, as always. I hope you enjoy one of the first Canadian rounds!

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13 months ago, # |
  Vote: I like it +193 Vote: I do not like it

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

Mandatory "tourist, What will you do with your MacBook Air M1?" comment.

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13 months ago, # |
  Vote: I like it +180 Vote: I do not like it

I was just wondering why 1st-3rd placed participants are not being offered the job while 4th-10th are XD.

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    13 months ago, # ^ |
      Vote: I like it +124 Vote: I do not like it

    FAANG already hired them.

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

    They are the real thugs, they can live with winning cash prizes in big coding contests.

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

He is Valentin Preobrazhenskiy not Valentine.

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13 months ago, # |
Rev. 4   Vote: I like it +17 Vote: I do not like it
"Random 10 participants outside of top-55, who solved at least one problem and participated in rated Codeforces rounds before, will get a t-shirt!"

Am I going to be the one of those 10 ? rounding up number of people solving atleast 1 problem to 20000 as we have 3 hours given probaility is ( 1/(20000)), . It comes out to be (0.00005)

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    13 months ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Well almost, you need to multiply 0.00005 by 10, because "Random 10 participants outside of top-55...". So that's 0.00005*10 = 0.0005. But I don't think it changes anything)

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13 months ago, # |
  Vote: I like it -21 Vote: I do not like it

be annoy of the suck statement at last several round.

statements are short look forward to this round.

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

    I was wondering why your rating graph is like this!!!

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

      yep, I was graduated several year, now i want CF agin!!!

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13 months ago, # |
  Vote: I like it -24 Vote: I do not like it
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13 months ago, # |
  Vote: I like it -26 Vote: I do not like it

Unfortunately, the start time is 11.35 PM in my area :(

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    13 months ago, # ^ |
      Vote: I like it +47 Vote: I do not like it

    thank you for useful and interesting comment

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13 months ago, # |
  Vote: I like it +65 Vote: I do not like it

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13 months ago, # |
  Vote: I like it +34 Vote: I do not like it

As a tester I must say the problems are nice and interesting. May you all get a positive delta.

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

If you forgot to see the prizes section of the post.

Do visit again!

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13 months ago, # |
  Vote: I like it -21 Vote: I do not like it

very weird time :|

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

Not interactive again

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

Codeforces if True Love.

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13 months ago, # |
  Vote: I like it +94 Vote: I do not like it

Just switched from Instagram to Codeforces, a few days back, and trust me Codeforces is much better.

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13 months ago, # |
  Vote: I like it -33 Vote: I do not like it

note the unusual start time

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13 months ago, # |
Rev. 2   Vote: I like it +69 Vote: I do not like it

4th-10th place = *Job offering* and Stock Options + Merchandise Pack

I want this level of confidence in my life.

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13 months ago, # |
Rev. 2   Vote: I like it -24 Vote: I do not like it

This forces us to stay up late in night. :(

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13 months ago, # |
  Vote: I like it +21 Vote: I do not like it

I will sleep during the round.

Spoiler
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13 months ago, # |
  Vote: I like it -26 Vote: I do not like it

Another MacBook for tourist`

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13 months ago, # |
Rev. 3   Vote: I like it -23 Vote: I do not like it

Codeforces is the best ❤️

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13 months ago, # |
  Vote: I like it -74 Vote: I do not like it

Is it rated?

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

Sorry I'm getting back to CF after many years and now I see there is also a Div3. Will this round also be rated for Div3 folks?

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    13 months ago, # ^ |
    Rev. 2   Vote: I like it +20 Vote: I do not like it

    Yes, there is pretty much no div3 as such. Anyone who's a div3 is also a valid div2

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

    As now there's also a Div4 you're Div4. But all contests of higher divisions than your division are rated for you. Unless round has separate contests for separate divisions. Because then you can only participate in your division contest.

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

      his ranks say otherwise, he'll soon reach expert

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

        Thanks! I hope so too. I aim to reach Div1 sometime in the next few months. Let's see how it goes.

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13 months ago, # |
  Vote: I like it -47 Vote: I do not like it

Just switched from Instagram to Codeforces, a few days back, and trust me Codeforces is much better.

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

I guess top gear isn't popular here :/

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13 months ago, # |
  Vote: I like it -12 Vote: I do not like it

wannahavealife 1000-7 ? ? ? ?

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

it will be unrated for div 3??

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

A couple more years and I can finally win a prize. (Hopefully)

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13 months ago, # |
  Vote: I like it +156 Vote: I do not like it

Screenshot-20210611220724-857x462

Seems codeforces finally realized nobody reads this rules anyways, so they just removed it lol

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

This contest is rated for division 3 or not?

Any please replay

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    13 months ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    Yes

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

      Thanks for your fast reply

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

        So are you gonna participate or not? Just curious about the effect that being rated causes on participation.

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

          I will definitely participate in this contest .

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

    Every div3 member is also considered a part of div2 too. So hopefully it would be rated for everybody

    It will be a combined rated round for both divisions.

    @authors can you please help with this ambiguous statment maybe write it "rated for everybody" ?

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

interesting interactive problem. I hopeeeeeee

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

    Or another binary search problem for Problemset

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

RIP my hairs after 3 hours :(

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

    Wait.. You guys still have hair?

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

      wait, you guys still have +ve delta

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

        cuz we guys don't use alts.

        Lol.

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

          wait, there are alts??

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

            It's clearly not working for you.

            Lol sorry XD.

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

              Out of my place to conjecture this, but is moongazer your alter ego's account? Seems like instead of discussing ethical issues inside you discussed them here!

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

                Unfortunately, I am not that insane.

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

                  well, then tell me what's 1000-7

                  ohh... did you find it intimidating

                  but ya i am not your alter ego.

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

Eat->Code->Sleep->Repeat

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13 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Contest will clash with Roland Garros finals :(

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

    I mean, it's dealing with crypto, and crypto is highly unregulated...

    But anyways, free money for CF!

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13 months ago, # |
  Vote: I like it -57 Vote: I do not like it

congrats for new macbook tourist

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

Is this contest rated?

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

The scoring distribution is 500 — 1000 — 1250 — 1500 — 2250 — (1750 — 1250) — 3250 — 3500

(1750-1250) means?? can anyone explain please.

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    13 months ago, # ^ |
    Rev. 2   Vote: I like it +5 Vote: I do not like it

    There will be two parts to the problem. One easier version(F1-1750) and one hard(F2-1250).

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13 months ago, # |
  Vote: I like it -59 Vote: I do not like it

One More Macbook to tourist

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13 months ago, # |
Rev. 2   Vote: I like it -42 Vote: I do not like it

LOL!

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

so are the ratings of questions according to div 1 or div 2. Hope that this round will not be like the deltix round

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13 months ago, # |
Rev. 2   Vote: I like it +26 Vote: I do not like it

For some reason there is a scoring distribution update in the English version of this blog post, but there is no one in Russian version.

UPD: It's ok now.

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

Great

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

Don't forget to consider the UNUSUAL START TIME

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13 months ago, # |
  Vote: I like it +83 Vote: I do not like it

For real, what's the point of making combined round when there are enough problems to split round for both divisions? Nobody from div1 wants to solve first few problems as well as nobody from div2 even reads last few problems

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

Hope! this round will go better for me than the Deltix round

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13 months ago, # |
  Vote: I like it -21 Vote: I do not like it

Downvotes coming in from other Grandmasters on codeforces and radewoosh ! LOL ! Truth is bitter guys !

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13 months ago, # |
Rev. 2   Vote: I like it +27 Vote: I do not like it

The entire comment section in one not-somewhat-related meme:

picsart-06-13-05-14-58

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

Most People: This is definitely my chance to prove myself and get MacBook Air!! Tourist: This is definitely my chance to be first again!!

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13 months ago, # |
  Vote: I like it +17 Vote: I do not like it

I am betting that tourist will not win this round. It will be some other LGM.

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

it's div1 + div2 = div3 :)

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

I wonder why there was no official announcement in the blog of unusual timing and even more surprisingly no one in the comments sections pointed that out. I opened my laptop just now to notice this.

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

today junior tesla is expecting to solve 4 problems and upsolve extra 1 problem. Lets see.

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13 months ago, # |
  Vote: I like it +16 Vote: I do not like it

Thanks to Djokovic I can convince myself that my rating crashed because I was watching Roland Garros finale
while giving the contest.

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

Are B pretests weak, or did I solve by mistake? hopefully, it gets FST.:)

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13 months ago, # |
Rev. 5   Vote: I like it -7 Vote: I do not like it

Although the Problem C was very interesting here's a meme
IMG-20210613-220345

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13 months ago, # |
  Vote: I like it +34 Vote: I do not like it

Div 1+2 = rating loss

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

It takes eternity for queue to end

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

    Only to get "Idleness limit exceeded on pretest 1".

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

queueforces

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

There are no running submissions.

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

Is this still rated because the queue has been stuck for a long time now ?

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

I'm waiting for the results of my submissions from last 10 min.

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

queue too long.

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13 months ago, # |
  Vote: I like it +6 Vote: I do not like it

People who are in queue are gonna remain in the queue forever if the number of running submissions is zero.

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13 months ago, # |
  Vote: I like it -44 Vote: I do not like it

hope this contest will be unrated

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

In queue!! ;(

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13 months ago, # |
  Vote: I like it +382 Vote: I do not like it

I always wanted to do it, and today I finally did

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

    you have plenty of time to play with these queue times :D

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

If someone wrongly pastes commented solution of A in problem B's submission, and locks problem B alone, won't this be a problem as others in room who have not solved A can see solution of A by solving only problem B ? And what if someone has commented previous problems solutions for all submissions ? How does codeforces handle this ? MikeMirzayanov

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

    Nice doubt xD (No one can handle this I believe).

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

    Do you mean that somebody uploads their solution for A instead of a solution for B by mistake? But a solution for A is very unlikely to pass pretests for B. And it makes no sense to show the source code of failed submissions to hackers (I don't know if that's how it really works, because I have never tried hacking other solutions in non-educational rounds yet).

    Or do you mean that somebody submits a valid solution for B, together with a commented out solution for A in the same source file? Is anyone ever doing something like this? I do sometimes use my A solution's source code as an initial template for B. But by the time when I'm ready to submit it, no parts of A remain there.

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

      Is there any difference between submitting code with commented solution of other problem in it and pasting it in an open blog ( or social media, IDE, etc. ) ?

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

Anyone else having difficulty to understand the Problem C? Or am I the only one?

UPD: I got it. And I'm stupid. :(

UPD2: I think, I solved it.

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

    I perfectly understand it, the only problem is I am stupid and cannot figure out how to solve it ;)

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

      Same with me for Problem B. I guess it'll be a sad contest for many.

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

        Unfortunately it was that way for me with both B and C; I suppose I can kiss my pupil rank goodbye...

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

          Yes predictor is showing -90 for you

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

        I dunno why but when I first read problem B I was thinking of the 3 columns problem without any proof and then boom dunno how I got accepted L.O.L.

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

          Would you like to explain what is 3 columns problem?

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

            The subproblem I generate to consider problem B i guest?

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

            Consider the i, i — 1 and i + 1 position.

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

    I spent half an hour, read several times, and don't understand what it means by "only by swapping numbers in a column". and finally I ask a question to admin, and then I got it.

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

      Yes, I didn't notice the announcement first. I felt really dumb noticing it after few minutes.

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

interactiveforces...

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13 months ago, # |
  Vote: I like it +76 Vote: I do not like it

ME: Get's TLE on D :( .No worries, let's stress test by generating some random trees. Google please give me a random tree generator?
GOOGLE: Sure

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13 months ago, # |
Rev. 2   Vote: I like it +16 Vote: I do not like it

The person who explained B's sample test cases

Spoiler
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13 months ago, # |
Rev. 3   Vote: I like it +8 Vote: I do not like it

How to solve E? I think the soln is something like, do individual blocks on k, for last (n % k), if (n % k) is:

  • 1 — then no answer exists
  • any even number — 2 more queries
  • 3 — 3 more queries
  • any odd number more than 3, remove 3 elements in 3 queries and remaining even in 2.

Is this roughly along the correct lines?

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    13 months ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    You can think that you have n columns and in one operation you add +1 to exactly k of them. You need to obtain odd sum in each column.

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

    How can you go for 11 4? I couldn't get this.

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

      if n % k is odd and k is even, the answer wouldn't exist.

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

        Thanks, then why the hell my solution is failing. Btw this is the only case where answer ceases to exist, right?

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

          Not sure about when n % k is odd and k is also odd. How are you solving this?

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

            Just make a move on first $$$n - n \% k$$$ elements and last n % k elements and this is same as k -> odd and n % k -> even.

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

              you might save some moves by saving up n-(n%k) indices by repeating them in previous queries. like

              n=13, k=5

              q1 : 1 2 3 4 13

              q2 : 5 6 7 8 13

              q3 : 9 10 11 12 13

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

                Yup I know and implemented this too, the above move needs to be done when n/k == 1.

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

    Answer doesn't exist only when k has more powers of 2 than n I think.

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

      n = 6, k = 4:
      1 2 3 4
      1 2 3 5
      1 2 3 6

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

    That's part of it, but note that it's not always optimal to do $$$\lfloor n / k\rfloor$$$ blocks, sometimes you might want to do more or less. When I say more, I mean you can wrap around and cover an index twice. As an example, the optimal strategy for $$$n = 10, k = 7$$$ might look like this:

    ? 1 2 3 4 5 6 7      
    ? 8 9 10 1 2 3 4     
    ? 1 2 10 9 8 7 6     
    ? 3 4 10 9 8 7 6
    

    EDIT: After reading other comments it seems I did E in a very jank way, I did some strange bullshit with blocks of size $$$k$$$ that passed pretests so pray no FST.

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    13 months ago, # ^ |
    Rev. 3   Vote: I like it +32 Vote: I do not like it

    Make a graph with N + 1 nodes.

    Node i represents a state where i indices are used odd number of times.

    Here Node 0 is our starting point, and Node N is our desired end goal.

    Make an undirected edge from i to j, if you can perform an operation move from state i to state j. (can be bruted).

    Answer is shortest_path(0, N).

    Since we can move from one state to another state easily (again, brute), we get the answer after knowing the shortest path.

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

      Edit: Ignore

      Could you elaborate on how the number of edges can be bruted? Can't the degree of each node be upto O(k)?

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

        For a state i, brute the number of ones being converted to zeroes. (zeroes_to_ones + ones_to_zeroes = K).

        This way, you get all neighbours of i in O(k).

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

          Sorry I just realized my question was utter nonsense. I was asking how this didn't become $$$O(n ^ 2)$$$ even though degree could be upto $$$O(k)$$$ per node. I forgot $$$n \leq 500$$$ in the problem.

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

    If n%k equals 1 and k is odd, then answer exist. For example, for n = 6, k = 5:
    1 2 3 4 5
    1 2 3 4 6
    1 2 3 5 6
    1 2 4 5 6
    1 3 4 5 6
    2 3 4 5 6

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    13 months ago, # ^ |
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    Using bfs from the final state and keeping count of number of positions used odd number of times will suffice

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

Today I got one of the best feeling that is solution D got accepted in last minute.

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

    I got WA on pretest 3. Can you give some hints?

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      13 months ago, # ^ |
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      Were your queries more than N/2?

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13 months ago, # |
  Vote: I like it +17 Vote: I do not like it

what was pretest 3 of E?

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

Thanks for including some easy problems for amateurs like me. :)

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

https://codeforces.com/contest/1534/submission/119394653 — This same case passes in my machine in 2 queries, i tried multiple other cases too. But CF says here it took more than 2 queries. Did i miss something here?

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

D problem... Killed more dreams than Fear every will ... :(

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

Solved F1 two minutes before the contests ends, what an exciting contest for me.

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

Can someone tell me, Why I'm getting ILE in Div2D ?
SUBMISSION.
I mean, I literally tried every combination of printf and endl. But couldn't get it worked. I usually use endl as a flusher but here it's not working.

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    13 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    vi query(int r) {
        // cout << "! " << r << '\n';
        printf("! %d\n", r);
    

    The query format differs from what you print here.

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

+0 delta GG.

Should have not given the contest in the first place :(

Screenshot-290

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    13 months ago, # ^ |
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    Hey buddy , do you know how can ishow the delta points for some reason i can't see it ?

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13 months ago, # |
  Vote: I like it +6 Vote: I do not like it

I feel the problem statements could have been clearer/presented better, especially those with large statements.

At first glance, I saw "It can be proven that that if it is possible to recover the XOR sum under the given constraints, it can be done in at most 500 queries. That is, d <= 500", and went to write a code for atmost 500 queries..

Didn't help it was an interactive problem, where, if there's a WA I don't know if it's my IO format, or my solution is actually wrong.

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13 months ago, # |
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Was B anything related to finding median of array

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

    No, you just find histograms that are higher than both the neighbours and reduce them to until its height is equal to the highest of their neighbours

    Submission

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      13 months ago, # ^ |
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      spend about 1.5 hours thinking ,what might have gone wrong in my median's logic :'(

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      13 months ago, # ^ |
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      i literally did the same thing but got WA in pretest 2 :( . :(:(:((((((((((((((.

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

    I solved B by finding every peak in the array (places where the height is greater than before or after), then doing operations to get that peak to the max height of its neighbors.

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13 months ago, # |
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how to solve D?

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    13 months ago, # ^ |
    Rev. 3   Vote: I like it +21 Vote: I do not like it

    Every tree is bipartite graph. You can ask only about nodes in smaller part. To know if node is in smaller part u can use this fact: distance between nodes from different parts is odd. So u just ask about first node, counting (cnt) nodes that have odd distance, and if cnt < (n+1)/2, first node is in bigger part, otherwise it is in smaller part. And then just greedy alg.

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      13 months ago, # ^ |
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      can u explain these smaller/bigger parts in more detail. And how you are representing a tree as a bipartite graph. Would be appreciated

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        13 months ago, # ^ |
          Vote: I like it +4 Vote: I do not like it
        1. So, in bipartite graph every node belongs to one of two parts. Part that contains more nodes than other is bigger part, other part is smaller part.

        2. So, as I said if distance between two nodes is odd, nodes belongs to different parts. So, ask about first node, every node that has even distance to the first one, is in the same part with first node, otherwise, if distance is odd, it is in different part

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

    Here was my approach:

    Arbitrarily choose some node to be the root. Query that node. That gives you the rank of each node in the rooted tree. Our goal is to find the parent of each node.

    Iterate over the nodes in any order. Query each node whose parent is not yet known. We can use the result of the query along with the rank information to find that node's ancestors, siblings, and children.

    Notice that if we associate each query with the node and its parent, these pairs don't overlap. This proves that we won't exceed $$$\lceil\frac{n}{2}\rceil$$$ queries.

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

I think I missed D by printing a edge multiple times :(

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13 months ago, # |
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Did anybody find todays problems had more casework to do or is it just me.

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

The only time I am in top 500 is during system testing (T_T)

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13 months ago, # |
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How to solve B?I had some proof that decreasing a number to the maximum of two sides will work but got a wrong answer on pretest 2

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

    you only need to decrease a number if it is greater than both its adjacent element.

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

      I did so but still a wrong answer.

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

        Cut the blocks which are larger than both left and right ones.

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

        shouldn't you set mx = 0 at the very start? Cause it's particularly given that we can't go lower than 0 and also it can be proved that there's no benefit of doing that as well.

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

        I think this is becoz , u have not dealt with n=1 separately for input 1 1 2

        the answer should be 2

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

    Erase a block off of a histogram if it is strictly higher than the previous histogram and the one after it. If you do so, you will pay $1 for each erased block and decrease ugliness by 2.

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

    Make sure you only do this on bars that are strictly higher than both of its adjacent bars. Also be careful with the left and right end bars as they only have one adjacent bar.

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

      I take care of both this situations but it still doesn't work.Can you take a look at my submission if you have time?

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    13 months ago, # ^ |
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    Handle n=1 case separately. For n=1,ans will be arr[0]

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      13 months ago, # ^ |
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      Yep I see my codes doesn't work there . Thanks!

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

    Suppose we have our original array orig and and our new array arr. We want to minimize:

    $$$\sum |orig[i] - arr[i]| + \sum_{} |arr[i] - arr[i - 1]|$$$

    So.. if we update $$$orig[i]$$$ to some $$$arr[i]$$$, we need to make sure that it is going to decrease

    $$$|arr[i] - arr[i - 1]| + |arr[i + 1] - arr[i]| + |orig[i] - arr[i]|$$$

    as compared to

    $$$|orig[i] - orig[i - 1]| + |orig[i + 1] - orig[i]|.$$$

    And from here, you can formally prove that its optimal to update a value iff it forms a mountain.

    And this makes some intuitive sense too: obviously, updating a valley is a bad idea. And if we update something that is neither a valley nor a mountain, then its pretty pointless and suboptimal.

    Also case when $$$n = 1$$$ needs some special case (or at least, my implementation needed special care).

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

      Padding the array with one extra 0 element in the beginning and one extra 0 element in the end removes the need to handle special cases.

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13 months ago, # |
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how to solve problem C ?

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

    count the number of cycles formed from $$$a,b$$$ and then $$$answer = 2^{#cycles}$$$

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    13 months ago, # ^ |
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    2^(number of connected component)

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    13 months ago, # ^ |
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    Notice that when you swap a pair, you have to swap some more pairs to make the configuration solved. If you swap ai with bi, you have to see where bi is present in a[]. Keep on swapping such pairs until you get ai in b[]. All these pairs form a cycle together which says that if you swap any one of these, you have to swap all of these to keep the configuration solved. There are only 2 states possible for a cycle, straight or upside-down. Count the number of cycles and print 2^cycles.

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

    Try to inverse some column, u will notice that now there is same number in first row, then try to inverse it, and repeat it until it has solved configuration. Every column than u have visited belongs to one cycle. Every cycle is independent from other cycles. Every cycle have 2 states. So answer is just 2^(amount of cycles)

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

    In C, the columns build components. That is, if you swap one column, you also have to swap another one, then another one and so on, until a circle is complete.

    Find the number of such circles, and ans=2^number since foreach circle we can swap it, or not swap it.

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

    Construct a direct graph with edges from $$$a_i\ ->\ b_i$$$

    $$$answer\ =\ 2^{number\ of\ cycles\ in\ the\ graph}$$$

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

    u have to find the number of connected components. this is because u cannot flip all the columns independently, for instance if some columns are (2,3) (3,4) (4,2) , they all are dependent on each other therefore , if u flip (2,3) u must flip all the pairs of columns that have a common element (here (2,3) (3,4) (4,2) ) So , find total number of connected components (here one of them is 2-3-4) and return 2 raised to the power this number.(as all of them will have 2 ways independently)

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    13 months ago, # ^ |
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    C gave me deja vu for this problem: 1465C - Peaceful Rooks

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13 months ago, # |
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What was pretest 3 in problem D? I failed it 7 times!

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13 months ago, # |
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How to do D? I was trying to start with 1. then I divided all numbers according to their distance from 1. now i would go and choose from nodes at distance i,i+1 and which ever are lesser in number i would iterate over them and print them and find out their neighbours in upper/lower distance level and increment i by 2. this ensures that every time i choose half or less than half elements every time. Hence in total i never print more than n/2+1 elements

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13 months ago, # |
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Why is the code of other people not visible

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13 months ago, # |
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https://codeforces.com/blog/entry/91617?#comment-803516 YAY I was right. All Tourist fans owe me a buck xD. Congrats Benq

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13 months ago, # |
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119382434 Can someone please tell me why I'm getting runtime error on this?

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13 months ago, # |
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Anyone tried randomization in D and got it accepted?

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13 months ago, # |
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When Ninjaclasher forces you to do a problem with your own name in it

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13 months ago, # |
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How to solve B with ternary search?

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13 months ago, # |
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Thank you for the round! The problems were interesting.

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13 months ago, # |
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why G was so big constraint? My O(NlogN) solution get TLE (TT)

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13 months ago, # |
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Hey Guys Do you know how can i see the delta points columns in the contest ?

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

    install an extension called CF-predictor

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    13 months ago, # ^ |
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    use cf predictor chrome extension

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    13 months ago, # ^ |
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    You can use cf predictor extension. It gives the real time Delta changes during the contest. Just Google it

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

Does anyone know why I can't view other people's submissions? Apologies if the reason was mentioned somewhere already and I missed it.

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13 months ago, # |
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finally the rating loosing streak is back. So excited to have more negative deltas. Wish me luck.

Thanks for this round and making me feel how dumb I am.

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13 months ago, # |
Rev. 2   Vote: I like it -6 Vote: I do not like it

.

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13 months ago, # |
Rev. 3   Vote: I like it +21 Vote: I do not like it

Apparently many people solved $$$E$$$ with some casework. You can do also like: (which I think is intended solution).

Solution
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13 months ago, # |
  Vote: I like it +83 Vote: I do not like it

Big ups to the authors for the nice problems, and bigger ups for including the answers to the possible sample queries in the interactive problems where it's reasonable. That's such a nice thing to do <3

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13 months ago, # |
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i could't submit D even after solving it cuz even though i saw a sample problem of interactive problem i could not submit it without idleness, please hack my account and put me out of my missery , basically i am under water its too much raining

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13 months ago, # |
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will the number of cycles and number of connected components be equal in C?

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    13 months ago, # ^ |
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    Yes, basically all connected components will be pure cycles

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      13 months ago, # ^ |
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      2^(Cycles) is the answer

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      13 months ago, # ^ |
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      Please can you check my submission 119395523? I calculated 2^(connected comp) but it's giving WA on tc4.

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

        In your solve function, for clearing the g[i], i should be from 0 to n, i.e., for(int i=0;i<=n;i++) (you did i<n instead of i<=n)

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13 months ago, # |
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Is there a way to see how many people failed system test for a particular problem ?

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

    They will never show that as people will backfire for weak pretests. Although this contest had strong pretests. No one on my friend list got FST.

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

      I found a way, just go to the status bar and check all rejected submissions keeping the tests greater than the number of pretests.

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

What if i tell you, that in problem A you just need to check whether given matrix satisfies:

RWRWRW                   
WRWRWR        
RWRWRW 
WRWRWR

or

WRWRWR
RWRWRW
WRWRWR
RWRWRW
WRWRWR

Realized this in post-contest discussion with my friend.

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

Finally Tutorials are here. I was thinking of finally offering some goat to the devil. Thanks to problem setters for fast update.

PS: BEFORE YOU COMPLAIN NO TUTORIAL TO BE SEEN IN THE BLOG. TRY TO FIND IT FIRST

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

Fst on D coming :(

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13 months ago, # |
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Now it's time to wonder if I got a great place thanks to competing drunk or despite competing drunk.

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13 months ago, # |
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https://codeforces.com/contest/1534/submission/119394653

Can somebody try this out on their machine. I submitted this in contest and it gave WA on pretest-1 and i tried it on my machine multiple times and it worked well. I'm really frustrated as I don't understand how the same code in 2 machines can give different output (my strategy isn't random, the program will always query the same nodes for a particular tree everytime it runs).

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    13 months ago, # ^ |
    Rev. 2   Vote: I like it +5 Vote: I do not like it

    When your local submissions behave differently from codeforces submissions, it often indicates that your code may have undefined behavior. In your case, I think you are trying to access an element out of bounds of the done array: see 119406377 where the assertions I added fail. I think it is due to the line fire2.pb(mp(x,b+1));, where b can be up to n-1

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      13 months ago, # ^ |
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      Thankyou so much for taking out the time to look into this and getting back to regarding the issue. I definitely learnt something new today. I understand why this happened now. Have a nice day :)

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13 months ago, # |
  Vote: I like it +28 Vote: I do not like it

Thanks for the great problems and strong pretests!

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13 months ago, # |
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Looks like highest rating record on codeforces is going to be broken.

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13 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Two questions:

F: Does anybody remember a similar situation when solving a small version and then trying to upgrade it was a mistake? Normally it's rather a good way, but there the small version required only SCC while the big one doesn't (which may lead you to really long code).

H: Isn't an interaction with $$$10^5$$$ flushes rather slow/dangerous/generally not CF-friendly?

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

    I finally found a problem with a good upgrade in this F. Not like F1 has an interesting idea, just a straightforward "build graph, do SCC", where the main point of failure is building the graph in a wrong way, but there's an element of double checking — if F1 passes, you can expand on it to solve F2, while if your SCC is wrong and pretests for F1 don't catch it, then pretests for F2 might because it's way more than SCC!

    There's a good reason to go for SCC in F2 even if you don't have F1. As soon as you have it abstracted as a directed graph, compressing to a DAG gives you plenty of guarantees for the next step — actually figuring out the solution. It's not hard to write either, just reproduce those 20 lines of Tarjan from whichever online resource you want.

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      13 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      Well, I took my code to F2, erased the whole thing with SCC, did some obvious stuff and it worked, so I don’t find SCC useful here at all.

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        13 months ago, # ^ |
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        Well, I used SCC and didn't have to erase and rewrite, so I think it was a good path to take.

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          13 months ago, # ^ |
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          What’s your solution then?

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            13 months ago, # ^ |
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            Mark vertices that you need to cover, make scc, mark respective vertices in scc, unmark those that are reachable from another marked.

            Now you have set of marked vertices none of which is reachable from the other.

            I claim that you can sort them in some order that any source in SCC will cover segment of marked vertices, just sort them by column id of any marked representative in scc vertices.

            Now you have to figure out minimum and maximum reachable marked vertex from each source and cover array with minimum number of segments, which can be done greedily without actually finding segments: take first unmarked vertex, go through back edges and mark every node as visited, from each marked node go down and mark all nodes as visited as well. You will reach all sources that cover first marked vertex, which is precisely what greedy on array does. Then take next marked unvisited vertex and so on.

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

            First, rephrase the whole thing as: you've got a directed graph where some vertices are marked (the $$$a_i$$$-th sand in $$$i$$$-th column), find the smallest set of vertices such that each marked vertex is reachable from at least one chosen vertex.

            I compress to DAG, note that only topmost components should be chosen, then note that from the way the graph is constructed, no two topmost components can share a column, so they're easy to sort. Also, paths in the graph are actual paths in the grid, so the set of topmost components from which a given vertex/component can be reached is a contiguous interval.

            With a simple graph traversal, for each component, I find the leftmost and rightmost of the top components which are above it (minimum and maximum of their numbers in sorted order) and the problem reduces to picking the smallest number of points that cover all given intervals, which is well-known.

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    13 months ago, # ^ |
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    I thought there should be a natural extension in F. So I continued with the SCC formulation and reduced it to finding a minimum size set of vertices in a DAG such that every leaf is reachable from some vertex in the chosen set. Then I tried to solve it with some (wrong) flow models. Does anyone know if this problem admits a good solution in general?

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13 months ago, # |
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119382434 Can someone tell me why I'm getting runtime error?

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    13 months ago, # ^ |
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    In practice mode (i.e. after the contest), runtime errors on codeforces can be caught by the C++ diagnostics compiler automatically. I resubmitted and it shows that your char arrays a and c are too small: 119406134

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      13 months ago, # ^ |
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      Thanks for the information. Is there any way to overcome this error?

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

        Make the arrays bigger, e.g. char a[n+1][m+1], c[n+1][m+1]. Alternatively, use 0-based indexing in the rest of your code.

        An array int a[50] only has valid positions a[0], a[1], ..., a[49].

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13 months ago, # |
  Vote: I like it +36 Vote: I do not like it

Benq has raised the bar of highest cf rating to 3796 now. Beating Tourist . Congratulations Benq.

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

https://codeforces.com/contest/1534/submission/119344612 Why did this fst? Can't seem to figure out.

Edit -> Don't use global variables when you don't have to.

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13 months ago, # |
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Hi. Can someone explain me why I got WA on test 4 in B? The idea is correct but I can't find the mistake in my solution. Thanks

119386786

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    13 months ago, # ^ |
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    hist[i+1] causes an undefined behavior because the max n is 4e5 and the size of hist is 4e5, enlarge hist by 1 element and soln will pass

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      13 months ago, # ^ |
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      I gave hist a size of 4 * 100005 which is 4 * (1e5 + 5) so that shouldn't be the problem. Also, I just tried with hist[(int) 4e6] and again got WA on test 4.

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

        ah, I'm blind, sorry, you don't fill it with zeros for every testcase, that's the reason, it is global

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          13 months ago, # ^ |
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          Thank you, that was the problem, I can't believe that was the reason, well now I will never forget about that lol. Thanks for your help!

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

good to be back

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

can some one tell why I am getting WA on this submission my friends solved literally the same idea and they got AC I literally can't find any wrong I am getting wrong on pretest 2 my program is printing 01234501234512345623456734567845678956789100 how can the program prints smth like this plz help https://codeforces.com/contest/1534/submission/119401086

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

    I made a few little changes in the cases where n = 1 and n = 2, and it passed: 119404096

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13 months ago, # |
  Vote: I like it +19 Vote: I do not like it

Since the highest ever rating title was most likely taken (at least for now) from tourist by Benq, do you think he will take his revenge and first reach the 3800 mark?

Any bets who will be first over 4000?