sszcdjr's blog

Codeforces Round 896 (Div. 1, Div. 2)
By sszcdjr, 15 months ago, In English

Hello, Codeforces! & Happy Chinese Teachers' Day!

We (OICon Team) are so excited to invite you to take part in OICon Round 1 (Codeforces Round 896 (Div. 1) and Codeforces Round 896 (Div. 2)) which will start on 10.09.2023 17:05 (Московское время). You will be given 6 problems and 2.5 hours to solve them in both divisions.

  • Please notice the unusual starting time!
  • One of the problems has two versions but their statement are not exactly the same, which means that you can't pass the easy version using the code of the hard version. However, you can only make hacks when you pass both versions.
  • There will be $$$0$$$ interactive problem in each division so you needn't read the guide for interactive problems.
  • Score distribution:
    • Div. 1: $$$500-(500+750)-1000-1750-2500-3000$$$
    • Div. 2: $$$500-750-1250-(1250+1500)-2500-3250$$$

The problems were authored and prepared by Error_Yuan, duck_pear, programpiggy and me.

We would like to thank:

Good Luck & Have Fun!

P.S.

UPD1. Editorial is out.

UPD2. Contest is over! Congratulations to the winners!

Div 1:
Div 2:
First Solves in Div 1:
First Solves in Div 2:
  • Vote: I like it
  • +350
  • Vote: I do not like it

Comments Comments (189)
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15 months ago, # |
  Vote: I like it +80 Vote: I do not like it

As a problemsetter, OICon stands for...

Spoiler
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14 months ago, # |
  Vote: I like it +137 Vote: I do not like it

Yet another Chinese round!

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14 months ago, # |
  Vote: I like it +95 Vote: I do not like it

As a tester, I enjoyed the round and found some very nice problems! I recommend you to participate.

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14 months ago, # |
  Vote: I like it +89 Vote: I do not like it

As a some-really-interesting-problems-in-this-contest-finder, I tested.

Hope all of you can get positive delta!

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14 months ago, # |
← Rev. 3   Vote: I like it +157 Vote: I do not like it

As a tester and a member of OICon, I want contribution!

This round has many really interesting problems, and I hope all of you can enjoy this round(and get positive delta)!

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14 months ago, # |
  Vote: I like it +87 Vote: I do not like it

As a first-time tester, I hope you enjoy this round!

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14 months ago, # |
  Vote: I like it +73 Vote: I do not like it

As another first-time tester, I hope again you enjoy this round!

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

As a rare tester, I don't think it will be very, so called, Chinese, so I do hope you will enjoy this round!

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14 months ago, # |
← Rev. 2   Vote: I like it +70 Vote: I do not like it

As some strong testers' friend, it seems that this round is fun, but I cannot enjoy it :(

Hope u can get positive delta!

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14 months ago, # |
  Vote: I like it +72 Vote: I do not like it

As a rival of the author, i want negative votes.

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14 months ago, # |
  Vote: I like it +59 Vote: I do not like it

As a fake author and a fake tester, good luck to you & have fun.

Support our team, OICon.

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

"you can't pass the easy version using the code of the hard version"

wait what?

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14 months ago, # |
  Vote: I like it -25 Vote: I do not like it

qp

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14 months ago, # |
  Vote: I like it +135 Vote: I do not like it

As a ChatGPT, this round is so difficult :(

I recommend all the bots to avoid participation, or you'll probably burn your CPUs up.

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14 months ago, # |
  Vote: I like it +85 Vote: I do not like it

ChatGPT for getting 0 point in both divisions LMAO

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

    Solving A in 00:00 and 10 unsuccessful hacking attempt ending :P

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14 months ago, # |
← Rev. 9   Vote: I like it -8 Vote: I do not like it

.

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

orz for no interactive problems.

OICon stands for Orz Institute & Constitution

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

Love sszcdjr

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14 months ago, # |
  Vote: I like it +32 Vote: I do not like it

funny emoji :)

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

sszcdjr /se /se /se

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14 months ago, # |
← Rev. 2   Vote: I like it +18 Vote: I do not like it
Maybe OICon Stands for...
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14 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Wish i do well and get green

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

Wish getting positive delta and not get back to blue :)

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

Wish getting negative delta and get back to green :)

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14 months ago, # |
← Rev. 2   Vote: I like it +52 Vote: I do not like it

As a invited-to-but-actually-did-not-test 'tester', wish you positive delta.

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

sszcdjr is great.

orz sszcdjr

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

I hope my rating will be increased after this contest

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

Would it be a round of zero?

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

This is definitely the most interesting contest blog I have ever seen.

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

No ChatGPT is harmed in this round...(for 0 points)

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

Unrelated but does anyone know how long it takes to update the ratings for previous contest problems ?

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

As a Chinese, hope I can get positive delta.

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

Wouldn't it be better if it started at 19.35 as usual ?

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

    We authors are in UTC+8, and it will be 22:35 for us if the round starts at the usual time. We want to go to bed earlier :)

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

      Why you can't start it more ealier like 20:05？

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        14 months ago, # ^ |
          Vote: I like it -10 Vote: I do not like it

        Because one of the authors won't be available before 22:00 (UTC+8) :(

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

          By the way, when you start to prepare for this contest?thx.

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            14 months ago, # ^ |
              Vote: I like it -10 Vote: I do not like it

            We proposed the proposal on Feb 2023, and got our coordinator on May.

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

      But for some other countries wouldn't it be too early

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

Emoji Forces ;)

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

or maybe it's (l)O(l)ICon :))

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

The start time is 07:35 PM IST then why is it not started yet?

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

I might be wrong, but I think that the first step to be GM is to put an anime profile photo.

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

As sszcdjr's 'ami(e)', j'aime il.

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

Hoping that unusual time won't become usual again :)

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

In my opinion, OICon=Olympiad in Informatics Conference :P

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14 months ago, # |
  Vote: I like it +32 Vote: I do not like it

OICon stands for

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

Hope to become an expert, good luck

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

Chinese round from sszcdjr and so on. Sounds loke it will be VERY difficult. Luckily I'm afked. Hope others have fun.

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

"you can't pass the easy version using the code of the hard version"

wait, what?

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

D1 and D2,,, hope the problem isn't so eccentric

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14 months ago, # |
  Vote: I like it +29 Vote: I do not like it

Hi, is it only me that m2.codeforces.com is kinda broken? It seems that some of the JS files are replaced with the redirect message. It's not happening in m1 and m3.

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

Use of emojis!! cooolll..

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

as a tester hope u enjoy the contest :)

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

Yet another Chinese round!

OMG interesting dude!

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14 months ago, # |
  Vote: I like it -31 Vote: I do not like it

Div2 D1&D2 are brilliant!!

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14 months ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it

In the contest when the site went down I submitted the code on m3.Codeforces.com but I had already queued a submission on the main site. The code is exactly the same but I got a 50 points cut because there were two submissions both of which passed. I sincerely hope this issue could be sorted and I could get my actual points in that. Mike and the organisers please look into this.

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

D2 == mystery of pretest 3

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

WHY in $$$E$$$ ($$$n$$$ $$$<$$$ $$$10^{18}$$$) and there is NO pretest where $$$n$$$$$$>$$$$$$200$$$ $$$??$$$

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

i love problem B so much

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

    How to solve it?

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

      Minimum of:

      • distance from a to b

      • distance from a to major city + distance from b to major city

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

        I think it's better written as min(dis(a,b), min(dis(a,majorCity)) + min(dis(majorCity,b))).

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

Can someone give any hints about how to solve D1

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    14 months ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    For each element you need to give some power of 2 to someone and also take some power of 2 from someone and for each element not equal to the average of the whole array this give take pair is unique. Also if the sum of all elements of the array is not divisible by it's size then answer doesn't exist.

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

      I was trying to do same thing but got WA on pretest 6. Are there any edge cases ?

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

        Idk but i also got WA on pretest 6 due to silly mistake(Writing — instead of +). I also ignored all elements equal to the target final value because we can always use them as the middle man for some operation.

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

          This was my first submission

          222777285

          Can you help me figure out the mistake ??

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

            idk bro. i didn't do it by manipulating the bits directly but by trying 2 options for each element. Either it gives first some power of 2 and then takes some power of 2 or the other way around and whichever worked without violating the conditions stored it in the map and then checked whether the counts are equal or not

            222801225

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

    Let S = sum(a), then obviously we need each element to be S/n. For each element find its difference from the target value. Using the constraints (ie everyone must give and get a power of 2), find exactly how much one must give and get (this is the hard part). Then check if each give has a corresponding get.

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

    Each person needs to gain or lose some amount of candles to reach the average. So $$$a_i - avg = 2^x - 2^y$$$. It turns out that if $$$a_i - avg$$$ is not $$$0$$$, then it is either impossible or there is exactly one way to choose $$$x$$$ and $$$y$$$.

    Another hint: Since we use powers of two, think about the binary representations.

    Once you compute $$$x$$$ and $$$y$$$ for each person, the answer can be found by considering the set of all $$$x$$$ and the set of all $$$y$$$.

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

    there's at most only a way to make a number that's different from avg equal to avg. Let delta = v[i]-avg. Abs(delta) must be in form "0...01...10..0" and you can just add the first bit before the one part and subtract the last 1 bit (if abs flips the sign you have to flip which bit to add and which to subtract). If delta is 0 you can ignore it, and if it isn't in the case i mentioned, just answer "NO".

    Now just count all the bits added and subtracted. If this is even for each bit, then answer "YES". (My code 222771034)

    The idea for D2 was to just consider the case when abs(delta) = "0...010....0". In this case you can just give/get that one bit. I'm not sure why my code fails though

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

    My version (propably there are much easier ways):

    1. if Sum(a)%n!=0 answer NO. let d be Sum(a)/n

    2. Calculate all numbers 2^x-2^y from -10^9 to 10^9 and memorise what x and y are for each of them (with exception of 0 there are only one option for each number). Let's call this numbers "good"

    3. Check that d-a[i] is a good number for all i, otherwise answer NO

    4. let's call people who need to receive candies before give candies "poor". everyone who is not poor is "rich".

    5. let's create four arrays of sets: poor/rich people who need to receive/give 2^x candies. (if someone already have d candies, we always can deal with them). Check that |poorReceive[x]| + |richReceive[x]| = |poorGive[x]| + |richGive[x]|, otherwise answer NO.

    6. If there are no poor people, answer YES

    7. for each x for each poor in poorReceive[x] if richGive[x] isn't empty remove one rich from richGiveEmpty, remove current poor from poorReceive[x] and add them to richGive[y] (2^y is the amount that this poor need to give).

    8. if we did'n change anything in step 7, answer NO. Else, return to 6.

    We will not repeat steps 6-8 too many times because there can't be more than ~30 people in each cycle without people who already have d candies.

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

      Ok, editorial says that steps 6-8 aren't needed at all and poor/rich division too.

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

D1 and D2 are amazing problems but i couldn't figure out some edge case apparently ;)

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

    Div2D once again yeeted my hopes and dreams of becoming CM.

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

How to solve Div2 C?

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14 months ago, # |
  Vote: I like it +184 Vote: I do not like it

I'm sorry, but my solution to F has a complexity of $$$O(N^2)$$$. I hope the systest is as weak as pretest.

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

    Your solution is the fastest :) Do you have a test with TLE?

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      14 months ago, # ^ |
        Vote: I like it +41 Vote: I do not like it

      I uphacked my submission with a test like this:

      $$$1,-1,1,2,-2,2,3,-3,3,...,n,-n,n$$$ ($$$n=166666$$$)

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

Felt like I was close to getting D2 but just missed. Still might become expert today.

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

Please don't tell me that in Div2 D1 you just have to check if for each element abs(element-avg) should be a power of 2 is the answer. Tell me that this was not it.

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

    This was my first approach which gave WA. So you are good

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

    this was not it

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

    I checked the balance after breaking the difference of element — avg into two powers of 2(as a swap goes both ways). Any number which is a difference of 2 powers of 2 must have exactly one string of '1s' in its binary representation (we can check using binary subtraction). In other words, 2 ^ (high-bit + 1) — 2 ^ low-bit should be equal to the difference between element and average. Here we are referring to the low-bit and high-bit of the difference. At the end, the balance should be 0 for all bits for a valid arrangement.

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

In div1 , score [2000,2200] is from around 160-th to around 60-th.

Many participants are ranked by "the number of dirts" since the total score is quite small. Unfortunately, i'm the one who has many dirts. Sad.

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

    Sorry bro, it's true that the score is quite tight and we should have made the score distribution larger for Div1 :(

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

I wrote $$$Tlog^3n+mlog^2n$$$ and got TLE in problem C :(

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

Turned out D1 indeed was easy...

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

fixing tons of cases in div1D is frustrating :(

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

My thoughts on the Div2 D1, please help to improve.

We can group all people into three types: rich/poor/avg. Rich means he has more than avg candies, and poor means the opposite.

We need to record for everyone, how many to give out and how many to receive in.

For rich people, suppose he has 111000 (binary) more, then he must receive 1000 (binary) and give 1000000 (binary) since 1000 + 111000 = 1000000;

For poor people, suppose he has 110000 (binary) less, then he must receive 1000000 (binary) and give 10000 (binary) since 10000 + 110000 = 1000000;

All the above receives and gives must cancel out each other to make the whole party happy.

What about the avg people? Well, apparently if they are more than 1, they can just stand in a loop and choose the next one to give and receive from the previous one. What if this is only one avg people? I don't have good thought yet.

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

    The first example shows you what to do with 1 avg person (you can just put them into any chain of transfers as they just pass any received candies straight on).

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

    avg can just connect two people. instead of a giving to b, a gives to c, which is avg and c gives to b

    Edit: damn i'm always late for comments today

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

Is there a penalty for re-uploading a solution even though if previous one was correct?

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

    According to rules, anything apart from AC and compilation error runs into a penalty. Compilation error I'm not sure but AC won't result in penalty.

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14 months ago, # |
  Vote: I like it +69 Vote: I do not like it

Terrible difficulty gap, ABC then go to jail

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

    Sorry, we didn't expect D to be that hard :( Some testers thought it was suitable :(

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14 months ago, # |
← Rev. 5   Vote: I like it +28 Vote: I do not like it

(Div2, problem B) I submitted O(N + K) solution written in Kotlin, but I got TL at test 3. Do anyone else have the same situation? Does really result depend on programming language? If so, is it fair enough?

UPD: I rewrote my code a bit and got TL 6, where N = max possible value. Honestly, I do not understand the motivation to use such a small time limit in the task

UPD2: My rating decreased -_-

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

    It is fair, nobody is stopping you from using C++.

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

      I agree that it’s fair, but it looks a bit strange for me, especially when I use linear solution and N <= 2 * 10^5 (not really huge number)

      Anyway, tasks were interesting, I enjoyed the round

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

    Will O(K) pass?

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

      K <= N, but at least you should input all the positions, so solution is O(N + K) = O(N + N) = O(N)

      My solution is O(N + K), but I got TL at test 3; then I rewrote my code a bit, but got TL at test 6

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

In div 2 , problem A, problem setter should be ensure us l != r. But he didn't. If l==r , I take range only one index and Xor same index will become ultimately zero If n is odd. I submitted same implement in different approach about six times. But always wrong answer. then I decided to not to take same index instead of using first two index. For a result , number of k also increased. then got AC. But I was always right if l==r according to this problem description. Am I right or not?

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    14 months ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    you are wrong. if $$$l=r$$$ then $$$s=a_l$$$, so you will replace $$$a_l$$$ by $$$a_l$$$, which does not become $$$0$$$. Read the problem statement next time

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

    Next time you should use the "Ask a question" tab on the constest front page to clarify it. I also woundered if $$$l = i = r \implies s = a_i \oplus a_i$$$, but before guessing that and implementing it, I asked the problemsetters and in a few seconds got the reply "No."

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

A very bad round!!! I'm upset!!!

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

    I agree, to give so much constructive...

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

Problem A edgecase l == r killed me.

I need to learn to not make stupid assumptions...

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

What a mathforces round...

Div.1A: I just guessed the pattern and proved it by AC.

Div.1B: I got to the key observation (Hint 2 in the editorial) relatively fast, but the implementation of B2 took me a long time.

Div.1C: I came up with the rest of the solution in ~15 minutes, but I had a hard time dealing with $$$n \le 10^{18}$$$. Passed sample tests 1 hour after the contest. (Why is my submission still in queue = =)

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

i do bad at the contest but it has some really good problems like E(of div2) and other problems where really good for them rate

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

Am I the only one who felt Div2 D1 was too much implementation heavy or was it really?

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

First time hitting 1500))

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

why my and other solutions in queue but another solutions get tests momently

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14 months ago, # |
← Rev. 2   Vote: I like it +22 Vote: I do not like it

I was close to getting a $$$\Theta(900 n)$$$ solution for div2.D1, but kept Wa on Test 3 because I forgot to determine that $$$2^x - 2^y$$$ might be unsolvable, and I feel that because I used several unproven conditions, it kept me wondering if the ones I was using were right, thus ignoring the very obvious error of $$$2^x - 2^ y$$$.

It's a very good problem, but for me, writing about this kind of problem makes it a not that enjoyable experience :(

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14 months ago, # |
← Rev. 2   Vote: I like it +15 Vote: I do not like it

B1 and B2 are interesting problems, but I solved them for too long time.

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14 months ago, # |
  Vote: I like it +40 Vote: I do not like it

Congrats to null_awe, finally GM almost half a year after 2399.

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

My screencast

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

I wrote something wrong in B1,B2 and I only fst B1. Maybe B2 test case too weak? I found a hack of my submission 222782134. But it shows Unexpected verdict in hacks.

The test:

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

    Sorry, the tests for B2 are a bit too weak. You got the Unexpected verdict because one of the testers' solutions (which is marked with "Accepted") got hacked in your challenge. The test case will be added soon.

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

It's a really good contest and let me return to violet, thanks to problems setter~

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

I posted a video editorial of problem C from Div. 2, I hope you enjoy it and find it interesting.

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

tho it seems that this round isn't that good, but it's still not so bad I think. But ig it's more like guessforces in div.1 A&B1B2.

A: I just observed the pattern of the sample. Once you noticed that, it's trivial. Not the best problem in my opinion, but still not bad.

B1&B2: I just did it the same as hos.lyric's. Guess the claim mentioned in the editorial, and don't prove it at all. then just pass it.

C: Quite educational in my opinion. Though it may be quite standard for some high-rated geniuses?

Overall this round isn't that bad. Sure, the problems aren't that nice, but it's not a reason to downvote.

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

Why is there a popup notification "Can't read or parse problem descriptor" every time I try to open any of problems of these rounds? For example, 1869F - Flower-like Pseudotree won't open.

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    14 months ago, # ^ |
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    It looks like it's been a temporary glitch. Now everything is working.

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

I usually like chinese rounds. Problems will be observation based.

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

Has the editorial blog been deleted? I couldn't open it

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