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Codeforces Round #844 (Div. 1 + Div. 2)
By tourist, 22 months ago, translation, In English

Hello!

Welcome to the Codeforces Round 844 (Div. 1 + Div. 2, based on VK Cup 2022 - Elimination Round) that will start on Jan/15/2023 15:05 (Moscow time). It will be a combined rated round for both divisions and open to everyone.

This round is a mirror of VK Cup 2022 Elimination — annual programming championship for Russian-speaking competitors organized by VK. VK Cup started in 2012 and has grown to be a five-track competition in competitive programming, Mobile, ML, Go, and JavaScript.

All the problems are authored and prepared by me. Thanks to KAN, errorgorn, lperovskaya, dario2994, Monogon, Arpa for making this round better.

You will be given 8 problems and 3 hours to solve them.

UPD: Editorial

Congratulations to the winners:

  1. orzdevinwang
  2. noimi
  3. Radewoosh
  4. gamegame
  5. QAQAutoMaton

and to maroonrk for getting the only accepted solution to problem H2.

  • Vote: I like it
  • +1287
  • Vote: I do not like it

Comments Comments (352)
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22 months ago, # |
  Vote: I like it +144 Vote: I do not like it

omg tourist round

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

Damn, 3 hours for only 8 problems. Scary

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

omg tourist round

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

    omg tourist round

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

      I am so interested in your graph, how did you maintain a slope of 45 degrees ?

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

        وانتا مالك؟؟

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          22 months ago, # ^ |
          ← Rev. 2   Vote: I like it -31 Vote: I do not like it

          What language is this, looks Arabic I don't know it. Speak in English. Also, I expressed my interest in speedy_boy graph NOT yours. Actually, your graph isn't interesting at all mr.newbie! The comment directed to you is below : Newbie detected, opinion rejected XD

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

            Respect to newbie

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              22 months ago, # ^ |
              ← Rev. 2   Vote: I like it +17 Vote: I do not like it

              yes sir. all due respect! but he seems talking in a rude manner to me by a foreign language tho I didnt talk to him and replied to ur comment only :-( However, your graph with slope 45 is so cool ✪ω✪ speedy_boy

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

        He didn't maintain it after this contest though, if he did he should be orange by now.

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

    newbie detected.. opinion rejected xD

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

Shortest announcement ever *_*

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22 months ago, # |
  Vote: I like it +192 Vote: I do not like it

Tourist can't compete against benq in this round, but he can make some crazy geometry problems to force benq to lose rating and get back to rank #1.

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

3 hours? How difficult the problems will be

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

Maybe my "master experience card" will be expired in this contest.

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

I don't care about the difficulty lvl of the problems if it's Tourist Round. All i know that it is going to be fun.

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

Note the unusual timing

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

omg tourist round ...^,^

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22 months ago, # |
  Vote: I like it +458 Vote: I do not like it

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22 months ago, # |
  Vote: I like it -120 Vote: I do not like it

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

what you guys think is Benq gonna or not gonna participate this contest?

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

    Benq will not participate, because tourist has intentionally made problems with topics where Benq is weak. This is to make Benq fall to second place in top rating chart, so that tourist can become #1 again.

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

      Oh,man !! Is there any topic on earth in which these legendary guys are weak?

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

        I don't know, maybe some optimization of matrix multiplication algorithm from $$$O(n^{2.43})$$$ to $$$O(n^{2.42})$$$.

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

          To make benq lose rating you have to specifically make it hard for him. If it’s hard for everyone he won’t lose rating.

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    22 months ago, # ^ |
    ← Rev. 2   Vote: I like it -18 Vote: I do not like it

    I think Benq only wanna compete with tourist who is well-known as the No.1 in competitive programming. In my opinion, he will not participate contests without tourist such as this contest

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

Omg 3hr Round ><

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22 months ago, # |
← Rev. 3   Vote: I like it +4 Vote: I do not like it

I think, Benq will not perticipate in this contest. But I will participate and face to the world number one Programme's problems...☝️

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

Score distribution.......???

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22 months ago, # |
  Vote: I like it -44 Vote: I do not like it
tourist is a great person
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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

omg tourist round

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

omg tourist round

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

Omg clash with ABC.

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22 months ago, # |
← Rev. 2   Vote: I like it -43 Vote: I do not like it

.

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    22 months ago, # ^ |
      Vote: I like it +45 Vote: I do not like it

    My IQ has decreased by 20 points while trying to read this. Thanks.

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

This is the shortest announcement for a round that I have ever seen :))

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

    Legend don't need to say much. Their presence is enough alone sir!

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

omg tourist round . Probme 1 will be 1200+ Rated now

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

It is rated ？(⊙﹏⊙)

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

    well, obviously, sorry bother……

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

TFW you expect tourist to claim back first place then see that tourist himself is the author

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22 months ago, # |
← Rev. 3   Vote: I like it -39 Vote: I do not like it

.

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    22 months ago, # ^ |
    ← Rev. 2   Vote: I like it -16 Vote: I do not like it

    It was good one . Don't know what's wrong ? Maybe too rude or offensive for some people ?? Atleast you should have blurred the authors above. It's disrespectful for them.

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

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

Contest Clashing with atcoder Beginner Contest 285.

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

    Tourist round >>> Any Other round on this planet. As Simple as that .

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22 months ago, # |
  Vote: I like it -19 Vote: I do not like it

I might just end up top1 because problems are being designed by tourist.

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

score distribution when?

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

omg tourist round

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

ok

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

omg tourist round

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

Thanks to MikeMirzayanov for amazing platforms Codeforces and Polygon.

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

Contest Collision (CF & ABC):

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

    CF is based off VK cup. It can't ve postponed

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

i do bad in tourist rounds, hopefully this will change tomorrow

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

why problem in this year contests are didn't given any rating till now?

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

conflict with ABC285... what a pity that i can't participate in both contests...

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

tourist gang

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

The One Piece is real!!!

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

Originally I am not planned to participate this contest because I have a date then. But when I see the author, I just postponed the date and register for this contest. Wish I can turn cyan once again.

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

omg tourist

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

I wish the next round will be written by Benq

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

Obviously tourist round will be full of fantastic problems. Good luck, yeah! ^=^

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

I will always be a fan of tourist!

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

It is briefly before the beginning of the round now, and score distribution hasnt been announced yet...

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22 months ago, # |
  Vote: I like it +79 Vote: I do not like it

Where is score distribution?

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22 months ago, # |
  Vote: I like it -55 Vote: I do not like it

I think I may not be able to participate because I feel like going to the toilet right now. What a wrong timing. :(

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

glhf

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

OMG tourist round

I regret I couldnt participate in it as I got stuck with some personal work

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

why the rating of H2 is higher than the rating of H1

it said that H2 is the hard version,doesn't it?

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

    its not H1 vs H2 but rather H1 vs H1+H2

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

    A solution for H2 always works for H1. So if you solve H2 you get the points for both.

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

F is an amazing problem! couldn't code it in time though, but mindsolved it

Can somebody please tell the solution?

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

    $$$dp[i][j]$$$ = $$$P$$$(Minimum Prefix $$$\geq$$$ $$$ -j $$$ )

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

In E, finding maximum total area is not that hard but finding the exact subrectangles seems quite difficult. Stuck on it's implementation. Any approach regarding the same?

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

    Maintain a stack for the rightmost intervals for u, d, ud rectangles. It was very painful implementing this, especially when the contest started at 4am for my local time.

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

      Hey, thanks. I just implemented the logic using your stack idea. Yeah, felt like handling a lot of cases!

      My solution

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

Pretty hard contest.

No idea for D. Have idea for E but got WA. Now I'll return to CM.

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

    how's C done ....?

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

      It's pretty annoying. You need to consider all j that n is multiple of j, calculate how many char will be changed if you make the string to j kinds of different chars. You need consider 2 cases: j<j0 , j>=j0, where j0=the number of different chars in the initial string. If j<j0 you need change some chars with low frequency.

      After you decided the optimal j, you need to add all chars you'll put to the final string into a "char pool" (implemented as int pool[26]), first try to make every chars remain the same(if(pool[s[i]]>0) pool[s[i]]--; flag[i]=true;) the assign i where flag[i]=false any char in the pool.

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

    Well consider two indices i and j. now calculate all such x that make both a[i] + x and a[j] + x a perfect square. the rest should be easy

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

    In E we can first shrink rectangles with same type (row1, row2, row1+row2) and then consider for each 2-height rectangle, shrink it if it's penetrated by any 1-height rectangle, and shrink every 1-height rectangles who doesn't penetrate it but cross with it.

    However I got WA at pretest 15. Also C was very annoying. I spent about an hour for it.

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

    What I did is calculate the answer firstly for two numbers, let's say a and b. We want to transform a into x^2 and b into (x+k)^2 (because b > a), and that means that $$$b-a = (x+k)^2 - x^2$$$, so $$$x = {(b-a-k^2)}/{(2k)}$$$. You can just try every k from 1 to sqrt(10^9) to see whether that k satisfy this condition, then, for each valid k, calculate the number of perfect squares you'd also obtain from the other numbers in the array.

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

    Let's choose some x. If ai & aj becomes perfect square after adding x, then: $$$a_{i}$$$+x = $$$u^{2}$$$ & $$$a_{j}$$$+x = $$$v^{2}$$$ Substracting , we get $$$a_{i}$$$ — $$$a_{j}$$$ = $$$u^{2}$$$ — $$$v^{2}$$$

    => $$$a_{i}$$$ — $$$a_{j}$$$ = (u-v)*(u+v)

    Now split ($$$a_{i}$$$ — $$$a_{j}$$$) into 2 factors f1 & f2 such that f1*f2 = $$$a_{i}$$$ — $$$a_{j}$$$.

    So, f1 = u-v & f2=u+v.

    Find, u from here. Substitute in $$$a_{i}$$$+x = $$$u^{2}$$$ & you will get x.

    Find the count over all possible x in this way & take the minimum one.

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

      I thought trying for every factor of a 10^9 number for 50*(50*49/2) times would cause TLE…

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

        my thinking was similar to you though instead of tle i got wa

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

          Now I upsolved D. My submission:189361440

          The proof of that the submission is available:

          In the algorithm we consider for all divisor k of (aj-ai) where j>i. The maximun number of divisors we check in one case is:

          sum(1<=i<j<=n, sqrt(aj-ai)/2)

          The "/2" is because if (aj-ai)%4==0 k must be even, if (aj-ai)%4==1 or 3 then k must be odd.

          Notice that sum(a[i+1]-a[i])<=A (A=max(a[i])=10^9). We can see for each d, sum(1<=i<=i+d<=n, sqrt(a[i+d]-a[i])/2) is maximized when all a[i+d]-a[i] are same (can be proved by the mean value inequality) and the maximun value is about (n-d)*sqrt(A*d/n)/2=n*(1-t)*sqrt(A*t), where t=d/n. We sum up it for d=1...n-1 and the sum is about O(n^2*sqrt(A)), which fits the time limit.

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

        It doesn't TLE, because you're factoring the differences of $$$a_i$$$ and $$$a_j$$$, all of which can't be around $$$10^9$$$ simultaneously.

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

        it's around 4e7 operations (which should be doable even with 1s time limit), plus the time limit is 4s.

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

          But for case a[i]=1+20408160*(i-1), it runs for 10622587 operations.

          My testing code
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          22 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          OHHHHH I forgoted there's this line in the statement:

          It is guaranteed that the sum of n over all test cases does not exceed 50.

          I missed in the contest.... I've thought sum(n) could be 2500 and missed the intended solution.

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

      I just upsolved D with a similar solution. I iterated over all n^2 pairs of numbers. For each number, i iterated over all values of f1 (so sqrt(max(a)) time) and did some math to see if it worked. Then I took all the possible values of x I got, and I simulated the process for each value of x.

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

Problem A was kind of ugly. C also required (at least in my case) a quite hefty implementation.

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

problem C is tough :((( Can anybody show me some hint? :((

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

    Definitely, not an easier one!!

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

      I spend almost 3 hours but still couldn't solve it :((( But i see so many people could solve it :((( Pardon me for my bad english

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

        Just brute every i from 1 to 26(i is how many letters will be left in string -> n % i == 0) and then you should calculate how many ops(call it cnti) you needed to make string balance(i used greedy in this part). So your first answer is min among cnti.

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

      Implementation is nightmarish.

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

    Suppose you want to modify the string to have exactly $$$x$$$ distinct characters. How many times will each character appear? Which $$$x$$$ characters should you pick? Can you calculate the minimum number of changes to reach $$$x$$$ distinct characters?

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

ofc the first question immediately got hit by annoying geometry question lol(don't worry imo it's also interesting)

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

how's problem C solved ...... nothing struck my brain for like 1.5 hrs i took multiple examples but i couldn't generalize it .

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

    You can keep only 1 <= i <= 26 characters with n/i frequency for each one

    Try them all and minimize the answer

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

Can't believe I clutched C like that.

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

Seeing the drawing for problem A before reading the problem statement was intimidating lol

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

Anyone know what pretest 11 was about in problem E ?

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

    I got WA11 and 2 1 3 2 3 2 3 2 4 helped me find the mistake. When I was using only 1 row out of 2-row rectangle I forgot to clear the other row in the segment tree

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

How to solve D? :(

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

    See here

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    22 months ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    The answer can always be 1. Now let's see what will happen if the answer is $$$>1$$$.
    So suppose the answer is 2. This means that there are any two indexes(let's assume i and j) such $$$a[i]+x=c*c$$$ and $$$a[j]+x=d*d$$$. Here c and d can be any valid integer. So now $$$d^2-c^2=(d-c)*(d+c) $$$.This is equal to $$$(a[j]-a[i])=(d-c)*(d+c).$$$
    Now if we factorize $$$a[j]-a[i]$$$. We can find the value of $$$d. and .c$$$. Using those values we can find the value of $$$x$$$. Now we will store all the eligible values of x and use the best possible option.

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

      I didn't get how you solved to get, d=(A+B)/2 and c=B−d

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

        since $$$d+c=B$$$ and $$$d-c=A$$$. If we add both the equations we get $$$2*d=A+B$$$ which is equal to $$$d=(A+B)/2$$$. For the value of $$$c$$$ it comes from equation 1 by shifting $$$d$$$ from left side to right side.

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

I swear F is easier than E...

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

any hint on d?

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

D is simple but the constraints are way to big. Any idea how to solve it ?

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

    Suppose that $$$b_i^2 = a_i + x, b_j^2 = a_j + x$$$, then we have $$$(b_j - b_i)(b_j + b_i) = a_j - a_i$$$. By enumerating $$$b_j - b_i$$$ we know what $$$b_i$$$ and $$$b_j$$$ are, and also the $$$x$$$.

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    22 months ago, # ^ |
      Vote: I like it +130 Vote: I do not like it

    So ... it is not simple, then?

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

D was very good

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

    Yeah, it was nice. I initially thought it was some sort of meet-in-the-middle trick, but was pleasantly surprised to see it was a much nicer brute-force.

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

    How did u solve D ?

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      22 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it
      Spoiler
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        22 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        As x can be of the order 1e18 wouldn't O(√n) will be of the order 1e9? which should TLE?

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

          We only need to find the factors of a[j]-a[i] to deduce all the x's and a[j]-a[i] can at max be 1e9.

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

            Can you please Explain About How to Find X using (a[j]-a[i]) what two factors represents of difference and how this term come (factor1 + factor2)/2
            i am not getting it :(

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

              see for $$$n_{i}$$$ and $$$n_{j}$$$ to exist $$$n_{j}-n_{i}$$$ should exist and be a factor $$$a[j]-a[i]$$$ because this equation $$$a[j]-a[i] = (n_{j}-n_{i})*(n_{j}+n_{i})$$$ holds.

              therefore if we iterate on all the factors of $$$a[j]-a[i]$$$ , (p,q) such that p*q = $$$a[j]-a[i]$$$ then we can treat smaller of p,q & p<q as $$$n_{j}-n_{i}$$$ and the bigger one as $$$n_{j}+n_{i}$$$ and solve these two equations

              $$$n_{j}-n_{i}=p$$$

              $$$n_{j}+n_{i}=q$$$

              find a feasible $$$n_{i}$$$ , $$$n_{j}$$$ if it exists we can get the respective x from equation

              Unable to parse markup [type=CF_MATHJAX]

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

    That's what she said

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22 months ago, # |
  Vote: I like it +111 Vote: I do not like it

trash problem E, only boring implementation.

I will never think a strong competitor to be a certainly good author any longer.

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

    I thought it has to be obvious after first stage

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

    Implementation is an important aspect of problem solving too; in my opinion, there's nothing wrong with problem E. You need to find a way to systematically reduce the rectangles, and figuring out how to structure your code to do this as efficiently as possible is part of the challenge.

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

    I agree. Compaired with implementation, observation is much too easier in this problem. That makes me feel really bad.

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    22 months ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    I can't implement it, too. But I think it's not all the author's fault. E is not very good but also not trash.

    We should consider to improve ourselves. I have already suffered a lot for my poor implementation ability.

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

    Problem E is a good problem. Please do not send such stupid comment only because you can't solve it. This will make you look like a loser.

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

      I don't solve FGH, but I don't criticize them. You never get to know why I am saying it's trash and others are upvoting me. So you are a fool.

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

        That's because you can solve E in usual, while you have never been solved problem FGH in the past. You just arrogantly make silly comment on problems you can't solve, and try to infect others with your trash emotion. You never get to know why I am saying your comment so brainless, you will just like a mouse hide in the hole and won't listen to others opinions.

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

          OK, next time at div4 A I will provide you with a math problem that has 10,000 corner cases and requires 100KB code and you cannot solve it and I say to you that you should improve your coding skills and not complain about the problem itself

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

            Actually, E was not that difficult to implement.

            For me, implementation is like an art: if you do it the right way, it feels godly.

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

            Yes, of Course he will not complain.

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

              I really hate a homeless dog barking at me when I'm walking in the street. It's really annoying. But I can't have too many requirements to a dog, right? They just want a bone.

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

                The same to you, fucked bitch

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

                  Angry fool, LOL. I don't mention anyone. If you are angry, that's only because you are shame on yourself<3

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                  21 month(s) ago, # ^ |
                    Vote: I like it 0 Vote: I do not like it

                  I know.

                  But why are you grey?

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

          More often than not I cannot solve problem E even D in div1+2 but never have I criticized them if it's my fault that causes me to fail to solve it

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

        You are being offensive. You are not offering any explanation for your words. You are attacking when attacked. I really can't stand this behavior by entitled contestants. Many comments by you in the past have a similar attitude, that makes you a bad member of the community. Please be more respectful.

        Problem E was rather easy to mind-solve in quadratic time (and still interesting). Rather hard to implement properly in pseudo-linear time. That is one of the main points of the problem. The implementation itself (at least mine) is short, but I wasted more than one hour because of wrong assumptions, wrong ideas, not having the whole picture clear in mind, and blah blah. That does not make the problem trash, it makes me weak. I was very satisfied when I finally solved it.

        You are the fool if you assume that codeforces upvotes are a valid support for the content of your messages.

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

      pls shup up until you can solve E

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

        What make you mention that i can't solve E? Please shut up until you can prove it.

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

          so how can you know he can't solve E

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

            He agree it himself bro, please don't ask such stupid question next time.

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            22 months ago, # ^ |
            ← Rev. 2   Vote: I like it +32 Vote: I do not like it

            they ACed A-D 1:17 into the contest, then proceeded to WA on pretest 1 on E for two hours.

            Now, perhaps I'm too low rated, too unexperienced to judge a master, so take my word with a grain of salt all you like, but that doesn't sound like they can solve "only boring implementation". If they only had a hour or so sure it might have been a time issue, but two hours for ~150 lines of implementation that only involves greedy and sortings? Please.

            When I get stuck on implementation for two hours in a contest, I don't say the problem is trash. I say that I'm too bad at programming and that this problem exposes an issue in me. I should be grateful to this problem even though it gave me -delta because it shows me how to improve my ability.

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

    That's very offensive to say Sir. Either problem is Good or it is not meant for you. Tourist has put his hardWork and time into that problem, atleast he deserve a valid constructive criticism sir. Codeforces upvotes are merely a number to me. You are still wrong to me. Also Someone hardwork isn't trash at all.

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

thanks for this great round

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

How to solve problem C?
My idea was to enumerate the number of occurrences of each character.
It took me about 2 hours. But I got TLE on pretest2.

(Forgive my poor English.)

Edit:
Why many people downvote this comment?

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

    My idea was to first, implement a function that optimally changes the string to be balanced with $$$x$$$ different characters ($$$O(n\log(n)$$$) in my implementation). And then go to all the divisors of $$$n$$$ less than 27, because the number of characters cannot be greater than 26. Then in total is something like $$$O(26 n\log(n))$$$.

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22 months ago, # |
← Rev. 2   Vote: I like it +19 Vote: I do not like it

How to solve F?

I turn the problem into this:

There's an array S indexed from -n to n, the number on index 0 is 1, rest is 0.

Do n operations, on each operation there's S[i]/sum(S) possibility to choose index i, add 1 to S[i], then P possibility to add S[i+1] 1, 1-P possibility to add S[i-1] 1.

Output the possibility to make S[-n...-1] both 0 after n operations.

But I have no idea how to solve this :(

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

If D doesn't use long long type to save $$$x$$$, will it be WA on #5? I got WA on #5 and I think it was because I didn't use long long type to save $$$x$$$, but I didn't have enough time to change. :(

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

F was a lovely problem, absolutely loved it !

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

How to solve Problem B

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

    More generally, how to approach these questions based on sorting?

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

    Check if you can take k people. You can take k people iff there are exactly k people for whom a<k and there is no person for whom a=k

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

In problem D, suppose any two consecutive square numbers are p^2 and (p+1)^2 . Then their difference will be (p+1)^2 — p^2 = 2p+1. Now given the maximum element is 1e9. so after choosing some x and changes a[i] to a[i]+x. The maximum number of all a[i]+x will be at most 2e9 or even less than 2e9. Isn't it? Or I am wrong somewhere? If this is the final solution we can brute force all the perfect squares of a[i]. But this solution gives me WA at test 5. Where is this solution wrong then? My solution . TIA :)

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

    Hint: 1e9*1e9=1e18

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

    no, difference is at most 5e8, but actual squares can be up to 3e17,and number of squares is still 1e9

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

    "so after choosing some x and changes a[i] to a[i]+x. The maximum number of all a[i]+x will be at most 2e9 or even less than 2e9"

    How you came to this fact?, like x can be upto 10^18.

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

    Got it. Found the mistake. Thanks a lot.

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

8

6 0 3 3 6 7 2 7

For the third testcase in problem B (reproduced above), why couldn't 2 people (persons 2 and 7) come? They'd both be happy (2 because at least 0 come and 7 because at least 2 come) and the rest because they're each happy if and only if 3 or more come; less than 3 come, so the rest should be happy as well. Why isn't this a valid result?

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

    a person is happy if at least ai persons OTHER THAN i are coming

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

    The problem statement says a[i] excludes the i-th person themselves. So 7 would not be happy, since there aren't two other people who went (just one).

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

    How can you participate in 234 contests consistently for 4 years and still be a newbie?

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

When will we be able to upsolve?

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

Wrote solution for 1782D - Many Perfect Squares few minutes after the contest and got WA on test 6. Can anybody tell me what's wrong? (189360636) It passed more than 50000 random tests with $$$n\leq{7}$$$.

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

    You shouldn't stop iterating over $$$k$$$ when you found an appropriate one. If you try out all valid $$$k$$$'s, your solution will pass the tests (check out this submission: link)

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

A pretty hard contest tbh especially C, so hefty implementation

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

    I liked C. Good problem where you think how to implement rationally.

    ...Cofeforcer codefofced coresorces!...

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

The ratings are updated preliminarily. Tomorrow I will remove cheaters and update the ratings again!

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

    Thank for kicking me back to div2. Maybe I need more practice before being a genuine master.

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22 months ago, # |
← Rev. 5   Vote: I like it +26 Vote: I do not like it

Now I've upsolved E. My submission:189365023

Basic idea:

First the maximum total area is the initial area. Proof: for any 2 overlapping rectangles, if they are same type (row1, row2 or row1+row2; we'll call them type1, type2 and type3 later), we can shrink one of them to remove the overlapped area and remain the same total area,like this:

[-------[+++++]||||||] -> [-------] [|||||||||||]

('-': left '|':right '+':overlap)

So we can remove all overlaps caused by same type of rectangles. If they are still 2 rectangles who are overlapping, then one of them is type3, the other is type1 or type2 (since type1 cannot overlap with type2, and we've removed all overlaps of same type in previous step). WLOG assume they are type 3 and 1. If type3 is "penetrated" by type1, which means type1.L<=type3.L && type1.R>=type3.R, we need to shrink type3 to row2 (and it becomes type2), like this:

[---[+++]---] --> [---------]

.....[| | |]...............[| | |]

('-': type1 '|':type3 '+':overlap)

Otherwise, we can shrink type1 to remain the area (since type1 can only get out of type3 from at most 1 side). Now we could solve the whole problem.

The naive approach is O(n^2), but by classifing rectangles by different types and sort them by the left borders, we can get an O(n*log(n)) solution. In the first step, for each type of rectangles, we keep prevR=the largest right border of rectangles we've checked, and for every rectangles with left border <= prevR, we move it’s left border to prevR+1 (and if it's left border is greater than the right border, we mark it as removed). In the second step, we maintain 3 pointers p1, p2, p3. For each type3 pointed by p3, we check for every type1 and type2 until there's no more that type of rectangles, or it's left border is greater than type3.R. If there's any type1 penetrates the type3, we mark flag1 as true, similar for flag2. After checked type1 and type2, if(flag1 && flag2) we mark type3 as removed, elif (flag1) we change it's type to 2, elif(flag2) we change it's type to 1. (However, the code implementation is very very very annoying)

Update: Now I've kicked back to div2. Maybe I need more practice to make sure that I can solve all solvable problems in contest.

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

Worst problem C ever

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

Cheating case Codeforces Round #844 (Div. 1 + Div. 2, based on VK Cup 2022 — Elimination Round) problem C

Zaoldyeck code :- 189324513

HAKUNAA_MATATA code :- 189349431

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

Thanks for the fast editorial

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22 months ago, # |
  Vote: I like it +54 Vote: I do not like it

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

I suddenly know why 19000+ people registered but there were only 9000+ people submit.

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

    People either got scared of problem A, or they accidentally slept in. (I started the contest 30 minutes late bc I slept in XD)

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

      Unusual timing also didnt helped the cause. I couldnt participate due to contest getting started ealier than usual

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

This is the hardest race I've ever done

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

Can someone plz tell where this code goes TLE. Solution code for problem C 189395773

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

omg tourist round

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

Hi there!

If you have any doubts in problem B and C you can checkout the tutorials here- www.youtube.com/@grindcoding . Happy coding!

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

Anyone tried to solve the problem D using Binary sesrch?

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22 months ago, # |
← Rev. 2   Vote: I like it +43 Vote: I do not like it

H is a problem designed cleverly (although most haven't ever clicked the link of it). For case k=0 it's a trivial div2C problem (consider the smallest "bounding-box" of a pattern and use inclusive-exclusion principle to count for each bounding-box). For case k=1 it's a normal div2E problem (consider every possible position of the broken light relative to the bounding-box, they'll form a rectangle, if it's contained in the bounding-box, subtract the number of patterns where all lights in this rectangle are turned on). For case k=2 we get 2 rectangles, and we need to guarantee that there must be at least one pair of corresponding positions of rectangles where lights are both off. We consider how many patterns will violate this condition: If for every pair of corresponding positions, there must be some position with light on, we construct a graph whose vertexs are positions in the 2 rectangles, and add an edge between each pair of corresponding vertexs, we need to find the number of ways to color these vertexs where every edge has a vertex with a light on. Since the graph can be disposed to several chains, and the number of such coloring for a chain is Fibbonaci numbers, we need to find the length of each chain (if any chain has 2 adjacent vertexs out of the bounding-box, the answer is 0, because they can't be light on).

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

    I'd rather say it is too classical and easy (to come up with a solution, implementation not included) for most who did click the link of it, and totally not worth higher points than G. And I bet many solved G but could not solve H1 only because of a lack of time. (For me, an additional 5 minutes would suffice.)

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

      It's a waste that H is not a problem with case of k=0,1,2 as easy, medium, hard subtasks, where much more people would try to think about it instead of ignore it completely.

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

Can anybody teach me that how to solve F? many thanks.

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22 months ago, # |
← Rev. 3   Vote: I like it -11 Vote: I do not like it

tround

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

When editorial will be published?

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

when will we get editorial of this contest?

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

I feel like D was a lot more easier than C, I still am not sure how to do it. I guess D and C should have swapped places

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22 months ago, # |
← Rev. 2   Vote: I like it +253 Vote: I do not like it

Solution sketch for F:

Let us say that two characters are paired if they were inserted in the same operation. Note that there are always an even number of characters between any character and its pair. Let $$$g(n, k)$$$ be the probability that after $$$n$$$ operations, there are $$$2k$$$ characters between the leftmost character and its pair. Note that $$$g$$$ does not consider the type of pairs inserted (() vs )().

Compute $$$g$$$ by DP. $$$g(1, 0) = 1$$$ and $$$g(1, k>0) = 0$$$. From any state there are three possible transitions. A new pair may be inserted left of the leftmost character, producing state $$$(n+1, 0)$$$. A new pair may be inserted between the leftmost character and its pair, producing $$$(n+1, k+1)$$$. Finally, a new pair may be inserted entirely to the right of the leftmost character's pair, producing $$$(n+1, k)$$$. $$$g$$$ has a quadratic number of states each with a constant number of transitions.

Suppose that ( has value $$$1$$$ and ) has value $$$-1$$$. In a regular bracket sequence, all prefix sums must be non-negative. Let $$$f(n, s)$$$ be the probability that in a string produced from $$$n$$$ operations on an empty string, all prefix sums are $$$\geq s$$$.

Compute $$$f$$$ by DP. $$$f(0, s \leq 0) = 1$$$ and $$$f(0, s > 0) = 0$$$. To compute $$$f(n, s)$$$, suppose there are $$$2k$$$ characters between the leftmost character in the resulting string and its pair, and suppose that pair is of type )(. Inside the pair, we have a sub-problem $$$f(k, s+1)$$$. To the right of the pair, we have a sub-problem $$$f(n-1-k, s)$$$. Each $$$k$$$ contributes a term $$$(1 - \frac{q}{10^4}) \cdot g(n, k) \cdot f(k, s+1) \cdot f(n-1-k, s)$$$ to $$$f(n, s)$$$. Pairs of type () are handled similarly. $$$f$$$ has a quadratic number of states each with a linear number of transitions.

Finally, output $$$f(N, 0)$$$ for $$$N$$$ in the input.

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    22 months ago, # ^ |
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    Thank you!! I've been waiting for someone to make an editorial on F the whole day.

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    22 months ago, # ^ |
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    In computing of f, if all prefix sums at the left are not less than s, then at the right we should make all prefix sums not less than zero? Because if all prefix sums on the right are not less than x, then by adding to each of them s from the left we get that they are not less than x+s?

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      22 months ago, # ^ |
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      The condition that prefix sums of the right part are not less than zero is sufficient but not necessary. Note that $$$s$$$ is non-positive.

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      22 months ago, # ^ |
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      Because we always insert () or )( pairs, the prefix ending at the leftmost character's pair has sum 0.

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    22 months ago, # ^ |
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    Some corner cases in the DP

    Also, this problem is quite tight on time, and $$$\text{mod}$$$ operation is quite costly, so try to minimize the use of it by using long long and only modding a state after the third loop has finished.

    AC submission with a few comments: 189490185

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    22 months ago, # ^ |
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    Compute f by DP. f(1,s<=0)=1 and f(1,s>0)=0...

    I think f(1,s<=0) is not 1.

    There is p possibility that f(1,0) is 1 and (1-p) posssibility that f(1,0) is 0.

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      22 months ago, # ^ |
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      Thanks, it should be $$$f(0, s \leq 0) = 1$$$ and $$$f(0, s>0) = 0$$$.

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22 months ago, # |
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First time seen that there is no thanks to MikeMirzayanov for Codeforces and Polygon platforms.

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22 months ago, # |
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When will editorial be released

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22 months ago, # |
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Can someone share a solution to C?

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    22 months ago, # ^ |
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    Here is the code : 189565190

    Basically what i have done here, is start iterating from 1 to 26, and for every integer when n%i == 0, called a function solve which gives the most optimal string when the no of distinct characters present in the string are i.

    Now, In solve function, first I calculated the no of times a character should be present in the string, that is n/i, then replaced all the characters whose frequency in the string is greater than n/i to of those whose frequency is less than n/i, till all the i characters have n/i frequency.

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22 months ago, # |
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So I have a solution for F which requires O(N^3) memory and time but the memory is way too much it is something like this : dp[i][j][k] = the probablity of it being a balanced string after i moves and having exactly j sub balanced strings (any balanced string is a number of balanced strings next to each other) having k positions in the whole string were if string )( is added the number of balanced substrings go up by one (in other words if you write the prefix sums of +1 and -1 the prefix sum is +1 at that point). now updating each cell of this dp is O(1) because all you have to consider is where the string () or )( is added. in case of string () if it is added between 2 balanced substrings it adds one to the number of them and also adds one to k value. if it is added in the k positions where the prefix sum is +1 it doesnt add the number of balanced substrings but adds one to k and in other positions it doesnt change k or j. for string )( , it cant be added between the balanced substrings, in the chosen k positions it adds one to the number of balanced substrings and also adds one to k and in other positions it doesnt change anything.

is this approach wrong ? and if it is not can i somehow iterate on something to change memory to O(n^2) ?

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    22 months ago, # ^ |
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    I've thought this kind method would work, but consider for (.(*).) where k=2 (represented by '.'), if we add )( at *, it would become (.().().) where k=3.

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22 months ago, # |
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No editorial?

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22 months ago, # |
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Please Publish Editorial.

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22 months ago, # |
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where is the tutorial？？？

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22 months ago, # |
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omg tourist round

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22 months ago, # |
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in C after fixing the number of distinct characters in the final string how to decide which characters i have to keep and which to delete so as to minimise operations ... mathematical proof is appreciated .

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    22 months ago, # ^ |
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    Sort the frequency of the initial string and draw a histogram, then draw a rectangle represent the target frequency, see what's the difference between them.

    For example, if the frequency of the initial string is 6,4,4,1 ans you need to change all frequency to 5 (and there will be 3 different chars), the histogram will be like this:

    -----!

    ----*

    ----*

    !

    ('-':the character will remain the same '*':the character will be added '!':the character will be removed) in this case the number of chars will be changed is 2.

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22 months ago, # |
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internet explorer tutorial

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22 months ago, # |
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where is tutorial ? we want to upsolve..

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22 months ago, # |
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Please give the editorial before the next contest starts !!

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22 months ago, # |
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The slowest editorial ever. Oh, wait, there's no editorial yet!

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22 months ago, # |
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Spent a day looking at B. Still can't solve. tourist Please make tutorial easy and if possible early.

Thanks.

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22 months ago, # |
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It is the first announcement to have those triples of no:
1- No editorial
2- No score distribution
3- No thanks to mike mirzayanov

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22 months ago, # |
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There are thousands of problems in Codeforces, you guys could solve them instead of waiting for editorial and you guys can upsolve the problems anytime later after getting the editorial, Mike won't delete any problems. So instead of worrying about it, solve some other problems:)

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22 months ago, # |
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Can someone share a solution for problem H?

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22 months ago, # |
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lets go

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22 months ago, # |
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TUTORIAL IS OUT!!!

TUTORIAL IS OUT!!!

TUTORIAL IS OUT!!!

TUTORIAL IS OUT!!!

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