pwned's blog

Codeforces Round 987 (Div. 2)
By pwned, 6 weeks ago, In English

Hello Codeforces!

We (pwned, ACGN, HappyPacMan, maomao90, AverageDiv1Enjoyer, and trunkty) are thrilled to invite you to Codeforces Round 987 (Div. 2) on Nov/15/2024 15:35 (Moscow time) (note the unusual time)! After more than two years of preparation and nearly thirty problems on our shortlist, our contest is finally here for your enjoyment!

In this round, you will help Penchick as he embarks on a world trip through the Philippines, Indonesia, and even through the Australian beaches and attempt six problems in two hours!

The score distribution will be as follows: $$$500-1000-1500-2000-2500-3000$$$.

This round would not be possible without the support of everyone behind this round:

Lastly, we thank MikeMirzayanov for creating the incredible Codeforces platform, where many of us have engaged in friendly competition, honed our problem-solving skills, and forged lasting friendships throughout the years!

From our keyboard to yours, we (and Penchick) wish you good luck, positive delta, and an exciting competition experience!

Note: There is at least one interactive problem, so please read the guide for interactive problems if you are unfamiliar with them.

UPD: Editorial has been posted, go check it out!

UPD: Congratulations to the winners!

Div. 1 + Div. 2:

  1. tourist, as expected, fully solving all problems in 53 minutes
  2. antontrygubO_o, who did so just 2 minutes later
  3. noimi
  4. A_G
  5. Pyqe
  6. StarSilk
  7. dyppp
  8. kotatsugame
  9. busamate
  10. tarjen

They, along with 8 other contestants, are the only AKers (full-solvers) in the whole contest!

Div. 2 only:

  1. LGlcx, who full-solved in 100 minutes!
  2. Sky_Maths
  3. G2Esports
  4. boboquack
  5. ac_de_taffy
  6. Jack.YT
  7. Alex2184
  8. MrPizza
  9. priyanshu.p
  10. CC_cccc

We hope you all enjoyed the round!

Want more of Penchick?
  • Vote: I like it
  • +537
  • Vote: I do not like it

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6 weeks ago, # |
← Rev. 5   Vote: I like it +22 Vote: I do not like it

pwned orz!

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6 weeks ago, # |
  Vote: I like it +42 Vote: I do not like it

Omg “highschool CPers” CF round!!! Expecting an interesting problemset!!

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6 weeks ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it

:penchickpride: pwned orz!!

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6 weeks ago, # |
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:pinoypride: penchick pwned orz!!

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6 weeks ago, # |
← Rev. 3   Vote: I like it +13 Vote: I do not like it

I waddle for the fish!

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

pwned orz!!

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

As a taster, I found the problems very delicious

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    6 weeks ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    So true, the flavortext is very delicious! Especially the diverse cuisines and dishes featured in the round!

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      5 weeks ago, # ^ |
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      I admire you, testers. It looks like this round will be cool. I'll give it my all this round.

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6 weeks ago, # |
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hi orz ppl @pwned @culver0412

i will definitely join! :penchickcheer:

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6 weeks ago, # |
  Vote: I like it +64 Vote: I do not like it

As a tester, maomao90 is gay

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6 weeks ago, # |
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As a setter, I'm surprised Penchick hasn't had jet lag yet.

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6 weeks ago, # |
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pwned orz

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6 weeks ago, # |
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excited for penchick-pilled round !!!!! (pwned orz)

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6 weeks ago, # |
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As a tester, I'll let you in on some facts:

Facts

GL&HR!

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6 weeks ago, # |
← Rev. 3   Vote: I like it +24 Vote: I do not like it

as a participant, here is an obligatory penchick image that nobody has posted yet

penchick-squak

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

After more than two years of preparation and nearly thirty problems on our shortlist

Couldn’t be more excited to get into this

images

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

Excited to participate, pwned orz!

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

As a tester W round

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

    As a unique problem idea-er, W round.

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6 weeks ago, # |
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as someone giving their advice in creating interesting problems, please upvote

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

As a participant, I love Penchick!

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    5 weeks ago, # ^ |
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    As a participant, i love high rating

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5 weeks ago, # |
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As a participant, I read all the comments and wondered why only one peng' has orange beak and why are the 2 rubik's cubes having 2 different "**shades of green**" ?!

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5 weeks ago, # |
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duckforces

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5 weeks ago, # |
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Hoping for the best Div.2 Round! Btw, the penguins are cute <3

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5 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

penguinforces

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5 weeks ago, # |
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So excited,hope that the problems are as cute as the announcement and the penchicks.

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5 weeks ago, # |
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As a tester, good luck and

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5 weeks ago, # |
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Penguiiiiiiiiiins

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5 weeks ago, # |
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"Why is there no Div 4? The last one was 1.5 months ago..."

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5 weeks ago, # |
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You for being an awesome member of the CF community! The "You" of this sentence is surprising!!!!!!!!

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

As a tester, after creating six problems for us, Penchick is now resting peacefully.

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

i will get another positive contribution with this comment

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

As a participant, I am not a tester.

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

Hope to comeback CM after this round.

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5 weeks ago, # |
  Vote: I like it +22 Vote: I do not like it

omg NeroZein mentioned

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

Masters Masters Everywhere.

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5 weeks ago, # |
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I think "note the unusual time!" should be bold and large font like

note the unusual time!
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5 weeks ago, # |
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pwned orz!

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5 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

Oh,this Penchicks are so cute!

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

As a tester, did I come late? I kept asking Penchick for the delicious feast and almost forgot the time for this round.

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5 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

Finally, a plush toys photo session in round announcement <3 Thanks, guys!

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5 weeks ago, # |
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Penchick

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5 weeks ago, # |
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I'm back :)

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5 weeks ago, # |
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The way right penchick is saying Good Luck

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5 weeks ago, # |
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Very cute!

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5 weeks ago, # |
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what are penchicks?

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So many people registering for this contest! Hope everyone +ve rating!

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5 weeks ago, # |
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very good unusual time, very scary interactive problem, like from porcelain

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

NO way C is a real problem

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5 weeks ago, # |
  Vote: I like it +19 Vote: I do not like it

This is true SpeedForces.

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

i choked really hard today. C is the worst C i ever met in a div 2 too

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

    Anybody forgot $$$3^2+4^2=5^2$$$ in contest? Note: I spent an hour remembering that lol

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5 weeks ago, # |
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i just spent 30 minutes searching for a mistake just to discover that i wrote cerr instead of cout

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5 weeks ago, # |
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is D a graph problem?

edit : can you give some hints

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    5 weeks ago, # ^ |
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    Yes.

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      5 weeks ago, # ^ |
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      How did you solve it as a graph problem? I used

      spoiler
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        5 weeks ago, # ^ |
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        I used a segment tree can you share your logic

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          5 weeks ago, # ^ |
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          Can you explain your Solution

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            5 weeks ago, # ^ |
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            just as said in the problem you can travel to larger elements in the left

            so first i calculated the prefix maximum for the array this will tell you what is the largest element that you can travel if only first operation is allowed now i created a segment tree that can give maximum in any range of array , then traversed the array from right to left

            if i am at an index i what is the value that i can reach anything smaller in the left from 1, prefixmax[i]-1 (as going to the prefix maximum will increase the scope of traversal in left ans the result will be even larger) so just calculate the maximum for this modify answer[i] with max(query(1,prefix[i]-1),prefix[i])

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          5 weeks ago, # ^ |
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          Spoiler
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      5 weeks ago, # ^ |
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      I used dsu and set. Iterate from left to right, and for current val, all the left side larger than the val should be connected with current val (and we can do this recursively with dsu.find).

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    5 weeks ago, # ^ |
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    At least my solution has nothing to do with graphs.

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5 weeks ago, # |
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plz provide enough test cases

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    5 weeks ago, # ^ |
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    Sometimes there is a deliberate shortage of provided test cases since otherwise the pattern to the problem would be obvious. Few test cases often indicates you just need to find a pattern yourself by working through some examples.

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      5 weeks ago, # ^ |
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      ok understood thanks for your explanation . I try to find it by practicing more problems.

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5 weeks ago, # |
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What is the solution for odd in C ?

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    5 weeks ago, # ^ |
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    Spoiler
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      5 weeks ago, # ^ |
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      n>=27*

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      5 weeks ago, # ^ |
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      I realised this that 3,4,5 is the smallest pythagorean triplet but could not construct it

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        5 weeks ago, # ^ |
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        1 10 26 can have same fillings as they have diff of square of 3,4 and 5. Also, we can set 27 and 23 with same fillings. Now we are only left with consecutive even length numbers {2,9}, {11,22}, {24,25} and {28, inf}. Set every 2 consecutive numbers with any same fillings [2,3] [4,5] [6,7].....

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      5 weeks ago, # ^ |
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      how for n==25

      my code get AC for n>=27

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        5 weeks ago, # ^ |
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        Yeah, it was a mishap in my mind. Edited to 27.

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    5 weeks ago, # ^ |
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    there is a fixed pattern of length 27 291641087 in odd cases and then the remaining part becomes even.

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    5 weeks ago, # ^ |
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    Asking for the same :(

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    5 weeks ago, # ^ |
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    "1 3 3 4 4 5 5 6 6 1 2 7 7 8 8 9 9 10 10 11 11 12 12 13 13 1 2"

    works for 27 and beyond

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      5 weeks ago, # ^ |
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      In your solution, 3 3 are adjacent while there distance is 1, How so?

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        5 weeks ago, # ^ |
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        1 is a perfect square

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          5 weeks ago, # ^ |
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          OK, I should visit KOTA(suicide Center) now :(

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            5 weeks ago, # ^ |
            ← Rev. 3   Vote: I like it +4 Vote: I do not like it

            Just hit the gym, watch some anime and you will be fine.

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              5 weeks ago, # ^ |
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              Real hit the gym!!

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                5 weeks ago, # ^ |
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                Your physique tells everything about you. Respect ++ Although I have seen one your solution in Edu section .. maybe in Binary Search course if I am not wrong.

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        5 weeks ago, # ^ |
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        I literally implemented with the similar doubt. Actually the problem becomes quite complicated if the solution excludes similar adjacent elements.

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          5 weeks ago, # ^ |
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          Exactly :(

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            5 weeks ago, # ^ |
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            I challenge you to solve the problem when adjacent elements aren't allowed (i.e. you cannot use $$$1$$$ as a square)!

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              5 weeks ago, # ^ |
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              Yes, I have solved this exact problem in the contest. Here are some observations I've made; please correct me if I'm wrong:

              1. if n is ODD then answer -1
              2. else we can solve for all mod of 8
              • for n%8==0 we can use 4 distance
              • for n%8==2 we can use one 9 at 0th then use the old 4 distance
              • for n%8==4 we can use two 9 at 0th and 2nd index and then old 4 distance
              • for n%8==4 we can use three 9 at 0th, 2nd and 4th index and then old 4 distance
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    5 weeks ago, # ^ |
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    1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 10 11 11 12 13 13 1 12

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    5 weeks ago, # ^ |
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    Observe if we can have a Pythagorean triplet, where x^2 + y^2 = z^2 then the odd case can be done easily, and the smallest Pythagorean triplet = (3,4,5) so the gap should be minimum 25, so if start with 1, then last value should be >=26, and the smallest odd number for which it's satisfied is 27. So for all odd values >= 27 we can have a valid possible ordering.

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    5 weeks ago, # ^ |
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    if(n>=27):

    1 on indices 0, 9, 25

    2 on indices 22, 26

    other as with even n: cur = 3 if i and i+1 is still empty then: a[i] = cur; a[i+1] = cur; cur++

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5 weeks ago, # |
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Thank you very much for the contest.

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5 weeks ago, # |
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what was the distribution technique for question 3rd... I thought of pairing in 2 because their difference will be 1 which is a perfect square but for odd N, i paired index 1 10 and 26 as their difference is also perfect sqaure.

Anyone ? whose solution got AC ? Thanks !!

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    5 weeks ago, # ^ |
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    with this approach you get a dist of 2 between idx 25 and 27 since there is gonna be an odd number of elem between 10 26.

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    5 weeks ago, # ^ |
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    you have to make sure [1, 10, 26] = 1 lets say, then [23, 27] = 2, so if n>=27 and odd this will work..

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5 weeks ago, # |
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Is this correct for F? I was not able to submit it due to the fucking internet

My solution
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5 weeks ago, # |
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Did thought about pythagorean triplet, and still choke at problem C.

Double the pain again.

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    5 weeks ago, # ^ |
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    So can we find out a valid filling when n<101 and n is odd?

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    5 weeks ago, # ^ |
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    Same brother, I didn't took the n=27 test case, so got wrong

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5 weeks ago, # |
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the B problem has an issue with python solution did you ever test it with python? i always get TLE

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    5 weeks ago, # ^ |
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    We tested it and it works on our side. Not sure what went wrong with your solution :(

    I submitted your solution in python 3 and it works. 291715567

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5 weeks ago, # |
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LETSGOOOO! I solved 4!

Nice contest! C was a cool problem. Thanks for the round pwned ACGN HappyPacMan maomao90 AverageDiv1Enjoyer and all testers!

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5 weeks ago, # |
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I submitted E in the last 10s of the contest but unfortunately got WA.

Anybody got an idea why this got WA on pretest 23?

Code
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    5 weeks ago, # ^ |
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    floating point inaccuracies can accumulate; floats are not accurate enough for this problem, you need an exact way to solve it.

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      5 weeks ago, # ^ |
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      Man that's such a pity! I could've solve it >_<

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5 weeks ago, # |
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Why does cpp23, cpp20 and cpp17 give different results for these solutions??

291657006 291656787 291655971

Code

nvm, overkilled it, realising it now that it could've been solved in a much more easier manner T_T

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    5 weeks ago, # ^ |
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    Haven't looked into the code yet, but if the same code gives different results in different compilers, there's a good chance that it's undefined behavior and some hidden errors.

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      5 weeks ago, # ^ |
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      yeah, tried looking for that, couldn't find so pasted the code here

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5 weeks ago, # |
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I really thought I would end up screwing up this round because I spent way too long on B, but I pulled off C and D quickly enough :D

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5 weeks ago, # |
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Did anyone else get Runtime error pretest 61 on E? Sys tests aren't happening and the suspense is killing me T-T

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5 weeks ago, # |
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My solution:

[D]

Observation1: note the maximum value as mx, and the number to the right of mx will definitely reach mx.

Observation2: if a[i]>min(mxpos,n), all a[i,...n] can reach mx.

So we can maintain a variable pos and continuously execute the following process:

pos=posmx initially.

  • find min i (1<=i<pos) such that a[i]>min(a[pos,...,n])

  • update pos:=i, or break if no such i

So we found a suffix that can reach mx. Then recursively solve the sub-problem.

[E]

Consider reverse operation: add nodes from the given tree to make it a perfect binary tree.

DP. Note dp[x] as the minimum depth at which subtree x becomes a perfect binary tree.

Observing the process of adding nodes (from leaves to roots) is actually a classic Huffman tree.

So we can use greed+priority queue to process dp [son[x][i]] to obtain dp[x].

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5 weeks ago, # |
← Rev. 3   Vote: I like it -8 Vote: I do not like it

Great problem C, really like it! Thanks for the round :)

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    5 weeks ago, # ^ |
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    Yeah, good one. Myself completely missed the 9,16,25 case

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      5 weeks ago, # ^ |
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      You are still missing.. Thats actually [1, 10, 26], [23, 27]

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5 weeks ago, # |
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Why is there such an important change in E so much later into the contest??? I was already working on the solution according to the problem statement by then, trying to write this complicated re-rooting dp code, and after submitting towards the end of the contest notice this change. How is this fair?? Please make this contest unrated.

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    5 weeks ago, # ^ |
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    The roots of the trees have always been well-defined in the statement. The binary tree is rooted by definition.

    The clarification was made to point this out, and didn't change the statement.

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      5 weeks ago, # ^ |
        Vote: I like it -19 Vote: I do not like it

      What do you understand from this statement — "Since Penchick and Chloe are good friends, Chloe wants her tree to be isomorphic to Penchick's tree"? I don't know why whould you change a well known definition for your convenience. Please make clear problem statements.

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        5 weeks ago, # ^ |
          Vote: I like it +23 Vote: I do not like it

        We did not change a well known definition for our convenience. The two trees are rooted, and rooted isomorphism is defined exactly as we did in the footnote.

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

    Footnote 2: A full binary tree is rooted tree, in which each node has 0 or 2 children.

    Combined with

    Footnote 1: A rooted tree is a tree where one vertex is special and called the root. The parent of vertex v is the first vertex on the simple path from v to the root. The root has no parent. A child of vertex v is any vertex u for which v is the parent.

    Is sufficient to define that the root of the binary tree is the top-most vertex.

    The root of Penchick's tree is defined as 1 in the statement "... with vertex 1 as the root".

    In footnote 3, isomorphism of rooted tree is clearly specified "Two rooted trees, rooted at $$$r_1$$$ and $$$r_2$$$ respectively, are considered isomorphic if there exists a permutation $$$p$$$ of the vertices such that an edge $$$(u, v)$$$ exists in the first tree if and only if the edge $$$(p_u, p_v)$$$ exists in the second tree, and $$$r_1 = p_{r_2}$$$."

    Hence, there are no issues with the original statement.

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

D can be solved with dsu. For each point, merge it with the point to the left of this point and with the highest height, and also merge it with the point to the right that is farthest away from it and with a height smaller than it. The answer corresponding to the set is maintained while merging. It can be shown that this merging is optimal.

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

    Can you please explain how did you develop this intuition? I thought about DSU but was not able to construct connected components in less than O(n²)

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5 weeks ago, # |
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Is there any reason for the memory limit to be smaller than the size of the stack for max N in python? I had a very inelegant solution to D that failed due to stack not fitting the memory.

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    5 weeks ago, # ^ |
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    Actually, there's no easy way to expand the stack size of Python on Windows, as the Python executable is already compiled with the default stack size (1MB, IIRC).

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      5 weeks ago, # ^ |
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      eee you can, sys.setrecursionlimit(10**5) increases memory usage so I suppose also the stack limit, but with 256MB you can fit just about 10**5 and N is 5 times more in D.

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        5 weeks ago, # ^ |
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        That function increases the internal recursion limit, but it does not expand the stack size of Python runtime.

        For example, running the following code with custom invocation will give you a stack overflow error (both in PyPy and Python). This means Python can't handle recursion calls nested just 10000 times, which indirectly shows Python runtime's stack size remains small even if you call sys.setrecursionlimit.

        import sys
N = 10_000
sys.setrecursionlimit(N + 100)

def f(n):
    if n == 0:
        return 0
    return f(n - 1) + n

print(f(N))
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5 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

C and D were interesting

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5 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

All problems here are bangers!! E is my favorite, but F is cool as well

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

comeback CM now (probably)

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5 weeks ago, # |
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Good Problems..

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5 weeks ago, # |
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How is it that suddenly 2000+ people solved C problem in the last 20-30 mins? Hope the cheaters are found and removed!!

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5 weeks ago, # |
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This was a really fun contest, with lots of interesting problems!

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5 weeks ago, # |
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Will I get back to 1900 after rollback?

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5 weeks ago, # |
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https://codeforces.net/contest/2031/submission/291676661

Can anyone let me know why my solution for B is wrong? I cannot find any edge case for this.

Please let me know why my solution is wrong, thank you very much

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

    When $$$n=1$$$, you didn't read the input array, which causes the testcases afterwards to be shifted.

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      5 weeks ago, # ^ |
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      I spent so much time trying to figure out what is wrong with the algorithm LOL; never forget to read everything from cin before outputting the answer

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      5 weeks ago, # ^ |
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      Thank you very much! I'd never make the same mistake!!

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5 weeks ago, # |
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So many new accounts on top positions :(. One has to improve if they want to simply keep their rating (otherwise they will be in an equilibrium rating lower than their “true” rating. Maybe this is not a problem if the amount of Smurfs is constant between contests, but I feel like this one has quite a lot)

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5 weeks ago, # |
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Making $$$2 \le k \le n$$$ i.e., asking subsequence of length atleast $$$2$$$ will make the problem $$$F$$$ as troll D2C ig ;)

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

nice round

only flaw is the statement of E is a bit misleading, I assumed the usual definition of isomorphic and got stuck (it seems that in such meaning it can be solved with tree decomposition with matrices, but I'm unable to make it clear)

my rating still did increase tho

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5 weeks ago, # |
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sto sto orz orz

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5 weeks ago, # |
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Good contest, thanks.

Also the problem F is amazing.

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5 weeks ago, # |
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Congrats to programpiggy for finally reaching M!

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5 weeks ago, # |
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Thanks for the contest! Participated virtually. Love problem C!

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5 weeks ago, # |
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Loved solving the 3rd problem. Interesting problem.

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5 weeks ago, # |
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How the heck did I get 9.8k official rank when there are only 8.3k official participants?? And I saw a guy in the standings list who is rated like 1k something, has the same rank as me, still got a +6 rating change meanwhile I got -46. Whats up with this

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4 weeks ago, # |
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There is a cheat alert sent to me says that my code is similar to other codes (about 12 or 13 person), I didn't do that but of course my code will be similar it's an idea and it's my implementation for it of course another person made the same. So please let me understand what's your point of view. finally sorry for inconvenience.

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