awoo's blog

Educational Codeforces Round 162 [Rated for Div. 2]
By awoo, history, 2 months ago, translation, In English

Hello Codeforces!

On Feb/23/2024 17:35 (Moscow time) Educational Codeforces Round 162 (Rated for Div. 2) will start.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

UPD: Editorial is out

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2 months ago, # |
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Good luck to everyone!

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    2 months ago, # ^ |
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    HERESY! Weak CPers are not allowed to use a picture of the Lord as their pfp.

    I advise you to quickly change your pfp and repent for your sins before the Holy Church Of Bateman is forced to take punitive action.

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      2 months ago, # ^ |
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      Oh I'm sorry :/

      Spoiler
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2 months ago, # |
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Such a short and clear announcement I hope problems statements be like this too!

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2 months ago, # |
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Hope able to solve ABC this round

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2 months ago, # |
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Hope able to AK this round

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2 months ago, # |
  Vote: I like it -41 Vote: I do not like it

Good luck to everyone! (please upvote i need contribution)

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2 months ago, # |
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Good luck to all , i hope to solve ABC

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2 months ago, # |
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Excited!

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

Excited to get to Expert

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2 months ago, # |
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Hope to solve ABCDEFGHIJKLMNOPQRSTUVWXYZ in this round (i'm a divinity i can do things you dont think are possible)

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2 months ago, # |
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Hope to reach 2024

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2 months ago, # |
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I think the title should say [Rated upto div2] instead of [Rated for div2]

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    2 months ago, # ^ |
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    Div. 2 includes all ratings below 2100 (if there's no Div. 1 contest)

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      2 months ago, # ^ |
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      Oh then why we have div3 and div4 contests.. if no such division exists?

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        2 months ago, # ^ |
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        Div. 3 and Div. 4 are not exclusive to Div. 2. Usually Div. 3 includes Div. 4 and Div. 2 includes Div. 3 and Div. 4. The only exception so far, to my knowledge, was https://codeforces.com/blog/entry/121579 this Div. 1 & 2 & 3 contest.

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        2 months ago, # ^ |
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        Historically, there were no Div. 3 and Div. 4 long time ago. They were added later as a part of Div. 2 for new type of contests that are newbie-friendly.

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2 months ago, # |
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My last contest in this winter vacation

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2 months ago, # |
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Score distribution?

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    2 months ago, # ^ |
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    It's an educational round.

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    2 months ago, # ^ |
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    Damn, it seems to me or they specifically ask this question only under every Div3, 4 and Eductional rounds

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2 months ago, # |
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Who tested?

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2 months ago, # |
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good luck :>

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2 months ago, # |
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Educational rounds are mathforces af, thats a skip

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2 months ago, # |
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Just to let you guys know vimdhayak ji is here

Sabko dekh lenge
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2 months ago, # |
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Enjoy the contest!

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2 months ago, # |
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stay up to late again:）

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2 months ago, # |
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Small and clear announcement. Hopefully we will enjoy this contest. </>

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2 months ago, # |
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Good luck everybody. Hope to get positive delta :)

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

Ah Shit, Here We Go Again.

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

Missed opportunity to name problem C. as "Find C"

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2 months ago, # |
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Do we write outputs for each test case as it comes or all the way at the end?

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    2 months ago, # ^ |
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    Either way works, however it's much simpler and more efficient to output as it comes.

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2 months ago, # |
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Finally, Goodbye 2023 has a worthy competitor.

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2 months ago, # |
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It's like Div 1.5

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2 months ago, # |
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B>>>C

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    2 months ago, # ^ |
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    wtf? just kill the nearest mob

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      2 months ago, # ^ |
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      i couldnt get the implementation right for B , C was just prefix sum

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        2 months ago, # ^ |
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        priority_queue or even std::sort should work

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          2 months ago, # ^ |
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          hi can you please check with my code once i know i made it too much complex but i still used the same approach you mentioned. here it is

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            2 months ago, # ^ |
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            you cannot sort only distances without hp, after sorting your monster[1][i] doesnt correspond to monster[0][i], use vector<pair<int, int>>

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              2 months ago, # ^ |
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              I am sorry. Thanks i blindly believed gfg editorial without reading furtherly.

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          2 months ago, # ^ |
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          yup sorted it in the end

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    2 months ago, # ^ |
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    Same lol

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2 months ago, # |
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B < A

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problem D :(

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    2 months ago, # ^ |
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    it was just about binary search.

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      2 months ago, # ^ |
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      What's your approach?

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        2 months ago, # ^ |
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        for each index,i checked on both sides and if any side's sum exceed the number at current index than can compress the range.Also note that if any side's range have all number same then will not consider that side.

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          2 months ago, # ^ |
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          how did you find if all numbers are same in a range?

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            2 months ago, # ^ |
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            For example, arr=[1,1,2,3,3,3] i stored the ranges (0,1) , (2,2) , (3,5) in a vector 'vec' in sorted manner. It can be done in O(n).

            now suppose i want to check if (4,5) has all number same,what will i do? i will find the largest index in vec such that vec[i][0]<=4, now i will check if vec[i][1]>=5 than (4,5) lies under the range (vec[i][0],vec[i][1]) otherwise some numbers are diff.

            Just binary search things...

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              2 months ago, # ^ |
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              you can check it easy if all elements in the segments are the same or not you can use prefix sum and check if the summation in the segment is equal to r-l

              vector<ll>id(n+1);
for(int x=1;x<=n;x++){
    if(v[x]==v[x-1]) id[x] =id[x-1]+1;
    else id[x] = id[x-1];
}
if(id[r]-id[l]==r-l) all elemnts are the same
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              2 months ago, # ^ |
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              Understood it! Thanks

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            2 months ago, # ^ |
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            I did it with sparse table.

            Spoiler
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I couldn't get the implementation for B right for over an hour... fml

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E too easy to be an E, swap D and E

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cheaters ruined the contest again :(

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    2 months ago, # ^ |
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    just train more, yours perfomance is bad not because of cheaters

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2 months ago, # |
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Nice problem F! Thanks!

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    2 months ago, # ^ |
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    How to solve it? My intuition is that we will perform 2nd operation at most once. Since number of 1s stays constant , we will try to merge 1s as close as possible.

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      2 months ago, # ^ |
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      That's it! (I didn't realize we needn't do the 2nd operation twice in the contest QWQ)

      When we perform the 2nd operation only once, we can consider on the reversed string. We can try all positions for the highest bit, and binary search on the final length of the string ignoring leading zeroes.

      If the positions of the highest bit and the lowest bit are determined, the greedy solution is to put all 1 outside to the last few 0s in the interval, so the prefix of some length doesn't change. So we can use hash or suffix array to compare the prefixes, and then the best starting position can be determined.

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    2 months ago, # ^ |
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    But I think it's hard to write it.

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2 months ago, # |
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I finally joined the dark side, selling my soul to Atcoder!

If you want to build intuition and upsolve problem D without looking at the editorial, try out 1901D : Yet Another Monster Fight. I've added hints as well.

A major hint for this problem:

Spoiler
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2 months ago, # |
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after getting 15 WA on problem C, I wished contest end earlier.

screwed up, going gray again

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    2 months ago, # ^ |
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    can you share the approach for C? thanks in advance

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      2 months ago, # ^ |
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      it's called brute force approach, guess the solution and submit until AC

      so basically what I did was it must has something to do with prefix sum, so I computed that and may involve prefix cnt(0) so I used that, but couldn't figure out there is length involved so failed to AC.

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2 months ago, # |
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in B is it not optimal to kill the nearest monster?

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hint for D pls

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    2 months ago, # ^ |
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    think of binary search.

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    2 months ago, # ^ |
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    If you have an range minima data structure and a range maxima data structure, can you figure out if all elements in a range are identical?

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      2 months ago, # ^ |
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      it can also be done through binary search by storing the ranges of identical numbers in sorted manner.See my code.

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    2 months ago, # ^ |
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    2 times binary search,front and back

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      2 months ago, # ^ |
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      whats value to search

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        2 months ago, # ^ |
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        Prefix to all and to all the a(i) search 1~(i-1) and (i+1)~(n)

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          2 months ago, # ^ |
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          so as you can see it need two time search

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    2 months ago, # ^ |
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    hint
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    2 months ago, # ^ |
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    think about binary search + preffix sum

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Enjoyable contest. At least I will not come back to pupil ^-^

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Why so tough time limit in E?

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    2 months ago, # ^ |
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    The intended solution is most likely small-to-large merging. It is not uncommon for the centroid decomposition to have a large constant factor.

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2 months ago, # |
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how to approach B?

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    2 months ago, # ^ |
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    Loop from the closest monster to the farthest (you need to take only the absolute value of all monster positions and then sort them), but you need to know each monster's health after the sorting, so you need to use something like a vector<pair<int,int>> where the first value is for position and the second value is for the index of that monster corresponding for its health . Now just loop from the closest to the farthest and sum their health and check how many rounds you need to get rid of the current sum, Rounds_required = ceil(current_sum/k).

    If at any time the rounds required are bigger than the current monster position, then we can't kill that monster so break and print no

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      2 months ago, # ^ |
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      "it's better to use ( current_sum + k — 1 ) / k for ceiling ". said by LGM

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        2 months ago, # ^ |
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        is there a blog post for that? i wanna read more about why that's the case

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    2 months ago, # ^ |
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    think binary search on prefix sum

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2 months ago, # |
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.

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for Problem C why i'm getting an WA : submission

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    2 months ago, # ^ |
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    To determine if the answer for a query is yes, you need to the number of non-ones from L to R.

    Instead of sum-cnt*2 > 0, try sum-cnt*2-num_nons (where num_nons is the number of values in the range that is not 1).

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      2 months ago, # ^ |
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      After number of non-ones,then what ?

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        2 months ago, # ^ |
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        That's it. Just create another prefix array that allows constant time counting of the number of "non-ones" from [L,R].

        It is optimal to set b[i] to 1 when c[i] is not 1, and then set b[i] to 2 when c[i] is 1. If b's sum is too small, then it suffices to increase one b[i] to match c's sum. If b's sum is too large, the answer is "NO".

        What amoad forgot to do was consider all indices where c[i] is not 1, as b[i] would be 1 and therefore part of b's count.

        I hope this explanation makes sense

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    2 months ago, # ^ |
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    try this test 1 1 1 1 3 2

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    2 months ago, # ^ |
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    for the array 1 1 1 2 2 with query from 1 to 5 ,your code will say yes , while the actual answer is no. got -3 before figured this out finally and had to take a whole different approach

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Clutched problem C

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Need More Practice for Educational Rounds. Btw Can anyone tell me the approach for C.(B>C)

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    2 months ago, # ^ |
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    we construct "good" array just using 1 (but if there is already 1 in original array we can write 2 instead) then just compare sums between l and r (can be done with prefix sum) and if sum[l:r] of original array is less than "good" array its ans is "NO" else "YES"

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2 months ago, # |
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Very nice problems

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How to hack others?

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any hint for problem c????

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    2 months ago, # ^ |
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    Basically, if you have $$$1$$$'s in array $$$a$$$, they will be at least $$$2$$$ in array $$$b$$$ and all other elements will be at least $$$1$$$. So, for some $$$l$$$, $$$r$$$ you just need to check if sum on subarray from $$$l$$$ to $$$r$$$ is at least it's length $$$+$$$ amount of $$$1$$$'s in it. $$$l=r$$$ is an edge-case.

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Typo in problem D's tags: https://imgur.com/TvwOJcP

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Did anyone recognize C was 1856B

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    2 months ago, # ^ |
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    Yes, I solved the question today, but I'm still unable to solve Question C. I attempted to use a segment tree but failed.

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2 months ago, # |
← Rev. 4   Vote: I like it +16 Vote: I do not like it

A: Let L=the minimum position of i such that a[i]==1, R=the maximum position of i such that a[i]==1, the until occurences of 1 become a single block, R-L will decrease 1 if you do operation on position R, otherwise it will not decrease. So the answer is (R-L)-(cnt-1) where cnt is the count of occurences of 1.

B: For each 1<=i<=n we need sum(j:abs(x[j])<=i)(a[i])<=k*i to kill monsters with distance <=i in i turns.

C: If L==R answer is no. Otherwise, for each a[i]==1 we need some j such that a[j]>1 and let a[i]+=1, a[j]-=1. Then we need sum(i:a[i]==1)(1)<=sum(j:a[j]>1)(a[j]-1). We can check this condition in O(1) by prefix sum.

D: Assume slime i is eaten by some slime to the left (the other case is similar), and all slimes used to eat i will form an interval [j, i-1]. So we need to find the maximum j such that sum(j, i-1)>a[i]. If j==i-1, slime i-1 can eat slime i directly. If value of a[j...i-1] is not all the same, then there will be a largest slime which can eat all other slimes in range [j, i-1] then eat slime i. Otherwise, all slimes in [j, i-1] have the same size and cannot eat each other, and slime i-1 is not larger than slime i. So we need to extend this interval to the left to find some k such that a[k]!=a[i-1] (if such k exist).

E: Small-to-large merge. Let dp[u][c]= (the number of nodes v in subtree of u, such that v is color c, and nodes on the path from u to v (except v) are not colot c). Then the number of valid paths with lca=u is sum(v1)(dp[v1][color[u]]) + sum(v1,v2,c)(dp[v1][c]*dp[v2][c]), where v1,v2 iterates over childs of u, c iterates over all colors on subtree of u, and v1<v2, c!=color[u]. The first term means valid paths start from u, and second term means valid paths start and end in subtree of u. To calculate the value we have dp[u][c]=sum(v: child of u)(color[v]==c) (if c!=color[u]), dp[u][color[u]]=1, we can maintain imformation in std::map and merge them small-to-large.

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    2 months ago, # ^ |
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    Can you explain what merging small to large means? I had the same dp relation but got TLE on testcase 10.

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      2 months ago, # ^ |
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      using tmii = map<int, int>;

tmii& dfs(int node,int prev){
	tmii& mp=*new tmii();
	for(int next:adj[node]){
		if(next==prev) continue;
		tmii& mp2=dfs(next,node);
		if(mp.size()<mp2.size()) swap(mp,mp2);
		for(auto [color,cnt]:mp2){
			if(color!=c[node]) ans+=(ll)mp[color]*cnt;
			mp[color]+=cnt;
		}
		mp2.clear();
	}
	ans+=mp[c[node]];
	mp[c[node]]=1;
	return mp;
}
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    2 months ago, # ^ |
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    Problem E can also be solved with DP on auxiliary trees.

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    2 months ago, # ^ |
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    In problem C Why a[i]=1 ?

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2 months ago, # |
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Can anyone please tell me why does this code won't work for problem C :(

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    2 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it 
    3 1
1 1 3
1 3

    Your answer is $$$NO$$$, but should be $$$YES$$$ ($$$2, 2, 1$$$).

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      2 months ago, # ^ |
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      thanks man, got it

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      2 months ago, # ^ |
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      Will you help me in solution of D also?

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        2 months ago, # ^ |
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        Maybe your idea is wrong? In bs you did you should check if there exist more, then $$$1$$$ different value on a segment. Or i didn't get you code.

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          2 months ago, # ^ |
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          My idea was right, just making a small mistake. ~~~~~ Anyway thanks man ~~~~~

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2 months ago, # |
  Vote: I like it -39 Vote: I do not like it

B >>> D > E > A > C

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    2 months ago, # ^ |
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    lol

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

    I felt B was a standard implementation problem

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      2 months ago, # ^ |
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      I know it is, but I got frustrated by not being able to find out a slick way to implement it.

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        2 months ago, # ^ |
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        Ugm... It is just $$$\forall x \ge 1 \; x \cdot k \ge \sum_{i \in [1, x]}F(i)$$$, where $$$F(i)$$$ is sum of hp of mosters on point $$$i$$$ and point $$$-i$$$.

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2 months ago, # |
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Anyone who got WA on test 2 at problem D, then managed to solve it? What was missing? I did a binary search on prefix sums while checking for the existence of different elements using two segment trees (minimum != maximum). I got WA on test 2, and I couldn't find one where it was not working.

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    2 months ago, # ^ |
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    Same here . I had the same approach but WA on test case 2.

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    2 months ago, # ^ |
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    You don't just have to check the existence of different elements. If your segment found by binary search contains all equal elements, you will need to extend it until a different element is found.

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    2 months ago, # ^ |
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    I found the test I got incorrect on:

    3 
2 2 1

    My program gives 2 -1 -1, it should have been 2, -1, 1. I guess I didn't pay enough attention when implementing.

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

    Let's say we have 3 9 9 9 9 9. We're checking how far right we need to go to eat 3. Simple binary search will result in -1 (no answer for it), because it will first check 3 9s but they're the same so then it will try more than 3 9s, etc. You need to process 1-length case separately.

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      2 months ago, # ^ |
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      This is the answer. Took me a while to realize why my program was not working. Thanks!

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2 months ago, # |
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In problem D sample input and ouput it is given in the third testcase that 2 2 3 1 1 will give an output of 2 1 -1 1 2 but cant slime 3 be eaten when 2 eats 2 and then eats 3. Making the expected output 2 1 2 1 2 or am I going wrong somewhere

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    2 months ago, # ^ |
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    "A slime can eat its neighbor only if it is strictly bigger than this neighbor. " So 2 can't eat 2.

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      2 months ago, # ^ |
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      ah sorry, my bad, should have double checked. also, Thanks

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    2 months ago, # ^ |
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    slime1 can not eat slime2.

    A slime can eat its neighbor only if it is strictly bigger than this neighbor

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2 months ago, # |
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F is beautiful! Too bad I only managed to solve it a few minutes late lol

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2 months ago, # |
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B — Can anyone find out the my mistake Thanks

int main(){

int t; cin>>t;
while(t--){
   ll n,k; cin>>n>>k;
   ll hel[n];
   for(int i=0;i<n;i++){
     cin>>hel[i];
   }

   map<ll,ll> m;
   int mon[n];

   for(int i=0;i<n;i++){
     ll x; cin>>x;
     if(x<0) x=x*-1;
     mon[i]=x;
     m[x]=m[x]+hel[i];
   }

   ll cnt=0;      
   ll p=0;
   int flag=0;
   for(auto &ele:m){
      ll extra=cnt+((ele.first-p)*k);
      if(extra<ele.second){
        flag=1;
        break;
      }
      ll remain=extra-ele.second;
      cnt+=remain;
      p=ele.first;
   }

   if(flag==1) cout<<"NO"<<endl;
   else cout<<"YES"<<endl;
}

}

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    2 months ago, # ^ |
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    cnt+=remain; here is your mistake! you shouldn't increase the value of cnt everytime. Rather you have to update the value, as cnt is used here to calculate the remaining time. so it should be: cnt=remain;

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2 months ago, # |
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Can someone hack it? 247947284

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

Edit --> My question was dumb asf. I misunderstood the entire question so resolving now

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

247976858 Can you give me test case wrong ? My program WA on test 2 (Problem D)

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    2 months ago, # ^ |
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    test:
	1
	7
	1 5 10 4 4 1 1 
	
correct output:
	1 1 -1 1 2 1 2 
	
your output:
	1 1 -1 1 2 3 2
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      2 months ago, # ^ |
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      Thanks! when i revised my code but it's TLE (my code use O(nlog(n)^2)) 247988327

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        2 months ago, # ^ |
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        You can use a sparse table instead of segment tree to reduce the time complexity to $$$O(n*logn)$$$.

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      2 months ago, # ^ |
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      Your test case really helps, thanks! I realized that with the special case that the neighbor is greater than the current item, it doesn't satisfy the condition for a single binary search anymore.

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2 months ago, # |
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Can anyone find out the my mistake Thanks[submission:247947411]

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    2 months ago, # ^ |
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    void solve() {

    ll n, q;
cin >> n >> q;
vector<ll>a(n + 1);
for (int i = 1; i <= n; i++) {
    cin >> a[i];
}
vector<ll>b(n + 1, 0);
for (int i = 1; i <= n; i++) {
    b[i] = a[i] + b[i - 1];
}
while (q--) {
    ll l, r;
    cin >> l >> r;
    if (l == r) {
        cout << no << endl;
        continue;
    }
    ll tem = b[r] - b[l - 1];
    ll h = r - l + 1;
    if (tem >= h / 2 + (h - h / 2) * 2) {
        cout << yes << endl;
    }
    else {
        cout << no << endl;
    }
}

    }

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      2 months ago, # ^ |
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      6 1
1 1 1 1 1 4
1 6

      Your answer is $$$YES$$$, but should be $$$NO$$$.

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      2 months ago, # ^ |
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      Wrote exact same code and got so frustrated thinking what is wrong in this and left the contest.247975227

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    2 months ago, # ^ |
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    Test case
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2 months ago, # |
  Vote: I like it -18 Vote: I do not like it

COULD YOU MAKE SAMPLES IN TASK C BETTER?

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2 months ago, # |
← Rev. 2   Vote: I like it +77 Vote: I do not like it

I think problem D can be solved using binary search

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2 months ago, # |
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Do anyone know how can my O(n * log(n)^2) solution for problem D got TLE, since n * log(n)^2 is only around 10^8

247977945

Thank you so much

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    2 months ago, # ^ |
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    I think you don't consider Segment Tree constant per query, which is very big.

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      2 months ago, # ^ |
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      I've got TLE because of that too. Is the constant really that big? Why it isn't just log(n)?

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        2 months ago, # ^ |
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        Well, when you already have $$$10^8$$$ time complexity, it is big enough to get TLE. I don't remember the exact value.

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    2 months ago, # ^ |
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    .

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2 months ago, # |
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Hello ,

why i am getting wrong answer on problem B?

https://codeforces.com/contest/1923/submission/247969707

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    2 months ago, # ^ |
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    maybe because you sorted cordinate before filling it, not sure about the other things

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2 months ago, # |
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Why does this submission 247963822 fail problem B test 2? Thanks!

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

Why does C have so many hacks?

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    2 months ago, # ^ |
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    even tourist got hacked? tf

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      2 months ago, # ^ |
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      does anyone know that missing edge case?

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        2 months ago, # ^ |
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        I don't think there's an edge case possible for that task? (except L==R which his submission covers). My code is exactly the same as tourists so I'll probably get hacked as well smh

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          2 months ago, # ^ |
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          Maybe my view of problem will help you:

          Let's notice, that we can fill "one" to indexes with other elements, and then fit our sum with big numbers to indexes with one.

          a[1 1 1 100 100] -> b[67 67 67 1 1]

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2 months ago, # |
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What is an "easy" solution of problem E? I solved it exactly as this problem.

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    2 months ago, # ^ |
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    I also solved it using this idea, some people have solved it using small-to-large merging.

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    2 months ago, # ^ |
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    I just saw your comment. I already posted my solution elsewhere but here it is again.

    It consists in one dfs. The idea is to keep track of which colors were accessible before on the dfs tree, with multiplicity. When branching, we set the current color to multiplicity = $$$1$$$ because it then loses its multiplicity (the path needs to be beautiful). But then it needs to regain its old multiplicity $$$+1$$$ when we end the dfs.

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2 months ago, # |
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tourist C hacked

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2 months ago, # |
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D: For each index ,I gonna binary search on the answer. So consider position i, we have to check if after x seconds, we can eat this slime or not.

Im computing the left side first, that mean we considering only from 1 to i-1.

In the position i and after x second, there are two case:

  • All x-1 slimes to the left of it are the same , so the process can make the biggest slime is one of them.
  • Otherwise , the process after x seconds always make one slime that equal to the sum of all x-1 slime to the left of i-th one (this pretty obvious because at any time, you always exist a biggest slimes that near to another), then we just need to check that sum is bigger than size of i-th slimes or not

After finishing the left side, we do the same on the right side and compare the answer 247990076

E : I saw a lot of simpler solution for problem E (merge set/map on tree, etc...). Anyway, I have another technique for E that is "Auxiliary Tree", which was very useful and generic in some specific type of problems. You can find it here : https://codeforces.com/blog/entry/76955. This problem is similar to E, which you can try to solve it first : 613D - Королевство и его города. You can read my solution for E here : 247980215.

Bonus

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    2 months ago, # ^ |
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    Thanks! This make sense for me.

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    2 months ago, # ^ |
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    Thanks for the bonus, helped me fully understand Auxiliary Tree.

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

What is the hack on C? Asking for a friend..

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    2 months ago, # ^ |
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    There seems to have been some issue with the system (probably related to Polygon), they're rejudged and most of the WA hacks are gone now.

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

Unfortunately, hacks for C were judged incorrectly due to a very peculiar case of UB :(

We are sorry for that, all hacks have been rejudged. The solutions during the contest were judged correctly, so this affected only the hack phase.

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

    Yeah, I was so confused. I literally went through my code a dozen times to see what was wrong. Thanks for the prompt fix.

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

This comment has been deleted.

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2 months ago, # |
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Im not able to understand why im getting wrong answer 2 for problem D,247990642

Could anyone please help!!

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2 months ago, # |
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I wasted 30 minutes in Q2 just because I didn't used int64_t

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2 months ago, # |
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247954847 How this solution is getting TLE?

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2 months ago, # |
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it may not appropriate to write such comment but

There is a case where my code for C is failing, i tried a lot to debug even stress testing but still couldn't found anything! it's failing on token 59831 on testcase 3

i don't know what's minor mistake i'm making 247924659

can anyone help me ?

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

    Here's a counter test

    1
5 1
1 1 1 1 4
1 5

    Your code prints YES instead of the expected NO.

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

    What about this test!

    1
6 1
1 1 1 1 2 3
1 6
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      2 months ago, # ^ |
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      Ahmed_Mostafa, bro you have saved my life by telling me about this TestCase i wish for your long life

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2 months ago, # |
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Can someone explain why I got TLE 247937101

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

Approach for solving Problem E using auxillary tree and DP.

Suppose for each color $$$\textbf{c}$$$ we perform a $$$\texttt{dfs}$$$ on the tree and calculate number of nodes in subtree of node-$$$i$$$ of color $$$c$$$ such that none of it's ancestor has color $$$\textbf{c}$$$. Now we divide our answer in two cases:

Case-1: Node-R has color c
Case-2: Node R has color c

But this will take $$$O(n \cdot n)$$$ time. To optimise this, we compress the tree in a way such that we only take nodes of color $$$\textbf{c}$$$ (and some extra nodes less than equal to number of nodes of color $$$c$$$ in number) without losing any information that we could obtain for any node of color $$$\textbf{c}$$$. Now we can use the same logic without any problem.

For each color our complexity would then reduce to $$$O(\text{number of nodes of color c})$$$ (ignoring the extra $$$\log$$$ factor that comes because of $$$\text{lca}$$$ and $$$\textbf{sorting}$$$. Here is my submission for the problem.

Problems
Blogs and Videos
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    2 months ago, # ^ |
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    It can also be done with one dfs and an array. The code is really short.

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      2 months ago, # ^ |
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      Holy shit 🤯 I feel so dumb now.

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        2 months ago, # ^ |
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        But thank you for sharing your solution as well and explaining it. Edu rounds are meant to learn new techniques.

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      2 months ago, # ^ |
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      Hi, can you explain what the above code does?

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      2 months ago, # ^ |
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      Can you explain your code ??

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2 months ago, # |
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is problem D prefix/suffix sum problem? then using binary search on pref(left), suff(right) then take lower..? but how to handle case where having same size adjacent?

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    2 months ago, # ^ |
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    I took care of this case by first going through the whole array and extracting the adjacent subarrays of same values (in the form of a vector of pairs (left index, right index)). Then, when binary searching, I would call the condition not met whenever the currently considered subarray lied in one of the precomputed "adjacent subarrays of same value".

    To handle this with a good complexity, I used another binary search to find the good (left index, right index) candidate with which I should check if I am inside or not. It's like geometry, so maybe there are easier ways to do so.

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      2 months ago, # ^ |
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      To make the implementation easier, you could use DP. Define $$$dp[i]$$$ to be the index of the leftmost consecutive identical element that ends at $$$i$$$.

      • If $$$a[i] == a[i - 1]$$$, then $$$dp[i] = dp[i - 1]$$$.
      • If $$$a[i] \neq a[i - 1]$$$, then $$$dp[i] = i$$$.

      Using this DP array, how to check if the range $$$[l, r]$$$ has identical elements? Just check if $$$dp[r] \leq l$$$. If yes, then the range has identical elements.

      Now, you can do a binary search on $$$[0, i - 1]$$$ to figure out the answer.

      Submission

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

Any small counter testcase for my C 1923C - Find B; submission 247973573 as well? I have used all the TCs posted by AVdovin and 29logN

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

I guess we can solve problem F using greedy algorithm with string suffix structures :D

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

https://codeforces.com/contest/1923/problem/C

can anyone tell why my code is wrong...

update: Solved

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2 months ago, # |
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I got a time limit on C even though it passes now :( (barely though) Python always does me dirty like that

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    2 months ago, # ^ |
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    why not switch to java or cpp ?

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      2 months ago, # ^ |
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      Just laziness, I like some of the abstractions that let me focus on the solutions only (and then get screwed)

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

Hello Everyone, Hope Everybody is doing great. I am stuck at problem C, can someone help me, what's wrong with my code

248047400 this code is after i have take the input array

for(int i=0;i<n;i++) { prefix[i+1]=prefix[i]+a[i]; } while(q--) { ll l,r; cin>>l>>r; ll k=r-l+1; if(k==1) cout<<"NO"<<endl; else { ll sum=k/2+2*((k+1)/2); if(sum<=prefix[r]-prefix[l-1]) cout<<"YES"<<endl; else cout<<"NO"<<endl; } }

According to me it should be correct,but failed at some 58k token.

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

    Can you explain what's the meaning of sum in your code?

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      2 months ago, # ^ |
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      Basically sum is the minsum of the array for which good array is possible Suppose pattern {1 1 1 2 2 2} -->3 one and 3 two if we decrease any of the two then it wouldn't be good array and this pattern would be min sum possible for length 6 good array possible

      {1 1 1 2 2} max sum for which good array of length 5 is not possible. {1 1 2 2 2} min sum for which good array of length 5 is possible.

      let say length of array is k then we need min sum of k/2 one's and (k-k/2) two's for array of length k to be good array

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

        An example : 1 1 1 3. It should be "NO" but accroding to your solution your sum = 2 + 2 * 2 is exactly 1 + 1 + 1 + 3. Then you output "YES".

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

    Find possible b's for 1 1 1 1 4 and 1 1 1 2 3 on paper and with your solution

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      2 months ago, # ^ |
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      You cannot find right?

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

        They have same sum, but for 1st you cannot, while for second it is 2 2 2 1 1. I've just fallen into the same trap.

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          2 months ago, # ^ |
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          Oh yes, I meant the first. I thought as you said "find" there will be a solution.

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

    Video Editorial for problem C

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2 months ago, # |
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247904944 Look at this , my implemetation for 1st problem

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2 months ago, # |
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Why are red coders visible as rated contestants in the final standings?

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2 months ago, # |
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is it rating ? my rating stayed the same

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    2 months ago, # ^ |
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    It is system testing now. The rating is not calculated yet.

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      2 months ago, # ^ |
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      yeah i see

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      2 months ago, # ^ |
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      could you tell how much time does it generally take after the completion of the contest? (new comer/ first contest)

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        2 months ago, # ^ |
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        For edu rounds, it may last for 2-3 hours(without the 12 hour hacking phase and the preparation). For the normal ones(div.1/div.2), it will be done within about 1 hour.

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          2 months ago, # ^ |
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          so once the contest is completed first there will be this hacking phase for about 12 hours and upon completion of that it will require 2-3 hours more right?

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            2 months ago, # ^ |
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            Yes. And then you will wait for another 1-2 hours to get your rating changed.

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2 months ago, # |
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I have been a Pupil for ages now. Ya it hurts

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2 months ago, # |
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Video Editorial for problem C : YOUTUBE EDITORIAL LINK Audio : English

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2 months ago, # |
  Vote: I like it -41 Vote: I do not like it

unrated contest pleaseeeeee!!!!

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2 months ago, # |
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Can anyone give the solution of problem E with simpler explanation ??

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2 months ago, # |
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Where is editorial? The round is already over and even contestants' ratings changes have been published. Round was great but it's cool to see the solutions when you haven't forgotten the tasks yet. Thanks :)

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2 months ago, # |
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Hey folks, can someone please help me figure out why my solution for C fails in the second test case.

My code
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    2 months ago, # ^ |
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    use long long int as sum of elements of the array will be greater than the maximum capacity of regular int

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2 months ago, # |
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Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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2 months ago, # |
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Who else failed the system test for problem C because they used 2e5 instead of 3e5 for the array size? (And somehow passed the pretests?)

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2 months ago, # |
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Some printing mistakes in 1923 C ci > 1 is printed as ci > i and last line should be I this sum is less than

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2 months ago, # |
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lovely contest became specialist after a lot of struggle

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

+

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2 months ago, # |
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Hello CodeForces,

I have been marked because my solutions are similar to other people. This is not true and i am expecting that you are gonna change this because that is not true.

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2 months ago, # |
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In problem C, could you please check, why this segment tree approach is failing? code

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