Aris's blog

Codeforces Round #764 (Div. 3)
By Aris, 3 years ago, In English

Hello! Codeforces Round 764 (Div. 3) will start at Jan/10/2022 17:35 (Moscow time). You will be offered 7-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks.

You will be given 7-8 problems and 2 hours and 15 minutes to solve them. One of the problems in this round is interactive. Don't forget to read the guide on interactive problems.

Note that the penalty for the wrong submission in this round is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them)
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of our work. Problems have been created and written by ITMO University team: MikeMirzayanov, MisterGu, myav, Gol_D, Aris, senjougaharin and Vladosiya.

We would like to thank: itohdak, Yogi79, smtcoder, Ra16bit, Tlatoani, nigus, MrDindows, leaf1415, Kniaz, Alireza, mini4141, Jostic11, BitHashTech, An_yujin, oversolver и sodafago for testing the contest and valuable feedback. List of testers will be updated.

Good luck!

UPD: Here is a photo of our team answering your questions during the round:

UPD 2: Editorial

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

Comments Comments (148)
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3 years ago, # |
  Vote: I like it +26 Vote: I do not like it

I wish every gray, become green! Good luck!

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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Excited to participate in the first Div. 3 round of $$$2022$$$ and my first unrated round, yay!

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3 years ago, # |
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Good luck to everyone!

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

Interactive problems in div.3 round are epic.

Look forward to this contest!

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3 years ago, # |
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Could we please have div3 rounds on the weekends? Most people in div3 probably have school Thanks

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    3 years ago, # ^ |
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    agree! I haven't had the contests on cf for a long time, although my score is low

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3 years ago, # |
← Rev. 5   Vote: I like it -30 Vote: I do not like it

I hate Div 3 so boring

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3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

I wish my cyan color become blue!

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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Another permutation or binary search on problem C

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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Will there be div 4 contests in the near future

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3 years ago, # |
  Vote: I like it -16 Vote: I do not like it

I tried solving a few interactive problems but got the idle limit exceeded! how to solve them?

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    3 years ago, # ^ |
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    I don't think there is any interactive problem that doesn't warn about flushing the output. Read the problem statement carefully from now on.

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    3 years ago, # ^ |
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    There is a link in the announcement of the round about interactive problems

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3 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Oh,no. I am very unhappy, because I will go to school on Tuesday. I want to have more rated, but I really won't have time. ):

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

The status of the problem I submitted 40 mins ago is still in queue. Is the same happening with all of you?

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3 years ago, # |
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So excited to see blogs on Codeforces( this the second one). I haven't seen them for a year. (just a joke)

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I wish my cyan color become blue!

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

I hope you done the best! Good luck for everyone!

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3 years ago, # |
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Here we go again...............

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3 years ago, # |
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First Div. 3 round in 2022!

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

All the best :) everyone, stay hydrated.

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3 years ago, # |
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I hope I do good ;)

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3 years ago, # |
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wow you guys are having fun (photo)

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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Thank you for this amazing round!

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

G has more submissions let's solve it.

can't solve G -> move to F

can't solve F -> move to E

can't solve E -> back to G ;_;

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In the photo are all of these guys also the setters of this round or just for answering the question asked by everyone?

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    3 years ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    All the people in the photo are also the authors of the round

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3 years ago, # |
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How to solve problem D,

I just thought of putting all the occurances of each letter into a set , & do the process optimally for k times , can someone help me, I know my solution will fail at cases like aaaaaaa, k = 2.. :

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    3 years ago, # ^ |
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    Hint — Binary search for the answer

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    3 years ago, # ^ |
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    Can be also solved greedily.

    Count occurences of letters in original string and count number of singles and doubles.

    (doubles — how many pairs of letters there are, singles — number of letters minus doubles * 2).

    Knowing singles and doubles values you can calculate solution.

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    3 years ago, # ^ |
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    There's no need for maintaining sets, you only need the counts of the possible number of pairs you can form for all the letters.
    This divided and then multiplied by k (integer division) will give the number of pairs that will be assigned to each color.
    If there are still other letters which have not been selected in any of the selected pairs yet, you can add 1 such char to the string of each color (eg: cac).
    My Submission.

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

DAAAMN, I had so much fun solving E (although I didn't quite finish it).

I thought that out of all solutions we were supposed to pick one that has least segments — which imo is even better problem (not sure if harder then original though).

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    3 years ago, # ^ |
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    Actually my solution to that problems is O(n * m * n). Does anybody know how to solve this variant in better time?

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I somehow frequently mess up the easier problems smh. Got 3 WAs on B when it was just basic implementation, and wasted a lot of time on D, when my first instinct was greedy with priority queue, but ended up trying to solve it with Binary Search unsuccessfully and finally solved it with my first instinct smh.

Great problems though.

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

It’s so great to solve all problems without a WA for the first time!

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3 years ago, # |
← Rev. 2   Vote: I like it +25 Vote: I do not like it

Solved C using Kuhn's algorithm... LMAO :D

If you are interested:
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    3 years ago, # ^ |
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    Same. The first thing that came to my mind after reading the problem is bipartite matching.

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    3 years ago, # ^ |
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    I solved it using priority queue :)

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

    Could you explain your approach, please? I just don't see how's this problem about bipartite matching...

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      3 years ago, # ^ |
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      For each number a_i in the array, consider the set of numbers from 1 to n that appear when you keep doing floor(a_i/2).

      For example, a_i=25 and n=5, then the set is {1,3}.

      Then we have a bipartite graph of positions and the numbers that could go there. As we can only have one number in each position, this implies a bipartite matching in the graph.

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3 years ago, # |
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how to solve F? Tried something close to a binary search but got WA.

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

I felt crippled at binary search in F :))

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

    F made me realise that if my usual binary search implementation isn't suitable for the problem, then I am screwed :)

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

This round is amazing! I really enjoy the problems. Thank you very much :)

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

Video Tutorial B:https://www.youtube.com/watch?v=W-y8gQFrFOY

Video Tutorial C:https://www.youtube.com/watch?v=6k8YudeK_04

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

NOice:))

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3 years ago, # |
  Vote: I like it +7 Vote: I do not like it

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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Why is G so simple, although I won't

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Interactive problems and BinarySearchForces.

Name a more iconic duo. I'll wait.

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3 years ago, # |
  Vote: I like it +23 Vote: I do not like it

Why? Just why??

Spoiler
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3 years ago, # |
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Got the binary search idea for D but did not know how to handle the odd case

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3 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Don't know why but That photo looks like a team of hackers from some Hollywood movie lol :/

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

What I learned from this contest . Take all inputs before even thinking of outputting an edge case

How foolish can I be at times

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

Great round!

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3 years ago, # |
  Vote: I like it +11 Vote: I do not like it 
Then you can swap any two symbols painted in the same color as many times as you want.

Imagine not noticing this line in problem D.

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3 years ago, # |
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Are exceeded queries in interactive problems always tagged WA. I kept on thinking about issues with the logic when my solution was actually asking for an extra query than needed.

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    3 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    You can have something like this in your code. Now you get MLE if you ask more than 10 queries.

    if(asked>10){
     vector<ll> check(MAX*MAX);  
 }
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      3 years ago, # ^ |
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      Thanks. Looks like a very useful trick.

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

      You can just keep an assert statement which will give you a runtime error if you exceed the query limit.

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3 years ago, # |
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Hi

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

Great round. I really enjoyed solving the problems, especially problem G.

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3 years ago, # |
  Vote: I like it +7 Vote: I do not like it

In F the condition that x < n was written on top, and I skip it so in 1h30m I can't reduce 11 queries to 10 queries lol :L rip everyone liked me :(

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    3 years ago, # ^ |
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    In the exact same boat as you lol. Hurts a lot since it's a rated contest for me.

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    3 years ago, # ^ |
      Vote: I like it -34 Vote: I do not like it

    yea but howcome the answer for the first testcase is 3, it should be less than n right Aris?

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

      How about you read the goddamned statement before trying to point fingers nowhere?

      Spoiler
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        3 years ago, # ^ |
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        well the statement wasn't clear for me, it hurts to be polite !?

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          3 years ago, # ^ |
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          Spoiler because i won't flood the page with useless writing
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            3 years ago, # ^ |
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            lool, did you even read whatever you pasted? I didn't offend anyone, I was asking a question, I didn't complain about the problem during/after the contest, I also didn't blame the problem writer, nothing happened like that.

            what I really see is you trying to mimic um_nik's attitude, and it just doesn't look good.

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              3 years ago, # ^ |
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              Yet again
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                3 years ago, # ^ |
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                Congratulations you're not illiterate, because you were able to read the problem statement.

                Try to have peace in your life, good luck ;)

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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Was problem G taken from somehwere? It has way too many submissions.

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3 years ago, # |
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Nice Div.3 contest. Thank you!

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3 years ago, # |
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Does "wrong answer Jury has answer, but participant doesn't (test case 3648)" means my answer is -1 but Jury is not or my answer is empty? Moreover, I just touch the length of segment must be 2 or 3 in problem E, but I implement it so complex, and finally get a wa2 without anymore clue... https://mirror.codeforces.com/contest/1624/submission/142294335

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    3 years ago, # ^ |
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    Yes, it means that the jury was able to find an answer, but you could not. For example, consider this testcase.

    1

2 3
000
000
000

    The required phone number is just a duplicate of the existing numbers. Hence, an answer definitely exists. However, your code produces $$$-1$$$.

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      3 years ago, # ^ |
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      I get why my algorithm is fake. Thanks a lot~

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3 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Video Solutions for anyone looking

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3 years ago, # |
  Vote: I like it +16 Vote: I do not like it

This round was very special for me because for the first time I have solved a problem in live contest. I loved it. I am happy and only I know how happy I am. We need more division 3 contest.

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3 years ago, # |
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Got a binary search solution for F, but it gets W/A on 6th test. Can anyone explain mistate or send a test that fails https://mirror.codeforces.com/contest/1624/submission/142288442 Thanks!

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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it
    Participant Jury Interaction
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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

Amazing contest!

The problems were very interesting, especially problem G!

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3 years ago, # |
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oops, I tried to do something more clever with F and figure out the bits one by one. was wondering why I was retarded for the past hour lol

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3 years ago, # |
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I'am gray. But it is showing under 'only unrated' in my profile's contest page.

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Upsolved E by matching a string using extended patterns (with code) within regexes (DFS simulation). Perl: 142302044.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Solution to Problem C using "Max Flow" :)142222070
Resource For Dinic's algorithm

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    3 years ago, # ^ |
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    If someday there will be an super-algorithm to solve all problems, everyone will be using that !

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      3 years ago, # ^ |
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      That will kill the very fun of problem solving

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3 years ago, # |
← Rev. 5   Vote: I like it -15 Vote: I do not like it

nice contest

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3 years ago, # |
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Please help, For which TC I am getting WA (Problem C) Your text to link here... thanks in advance

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    3 years ago, # ^ |
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    Input
    Expected Output
    Your Output
    Comments
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3 years ago, # |
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O(32(n+m)) was not acceptable in problem G?

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

for E consider the following test case

2 4

1278

6935

7878

output->

2

3 4 1

3 4 1

I wanted to ask whether the output is correct? that is can we repeat a particular segment?

UPD-> Got it, read the announcement now:)

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Editorial?

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3 years ago, # |
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Can somebody please explain me why I didn't get any rating for participation in this contest?

I don't want to create a new post, so desperately hoping my comment here will get some attention.

I'm a newbie in competitive programming and just recently joined Codeforces. I hoped to start acquiring rating and earning experience by participating in Div.3 contests. Even if this contest (#764) is the second contest I'm participating (the first one was Hello 2022 where I solved just one problem), I assumed I'm qualified to be rated in Div.3 contests. However, this one didn't affect my rating at all.

Please, help me understand the rules of the system!

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    3 years ago, # ^ |
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    Ratings haven't been updated yet. Normally in div3 rounds, there is a 12-hour hacking phase. When the hacking phase is done, ratings will be updated in a couple of hours.

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    3 years ago, # ^ |
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    you can always wait until the record of this contest appears in your contest history.

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3 years ago, # |
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I got wrong answer "wrong answer Jury has answer, but participant doesn't" in Problem E. I use greedy and kmp algorithm. I match target number as much as possible. Once match is one, i will backtrace one and rematch again. If it can't match target number or one of the segments is one, it will break and output "-1". My algorithm complexity would timed out. But is this algorithm is wrong? I try many tests and verify my answer is correct.

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    3 years ago, # ^ |
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    No, this is correct, you must have made a mistake somwhere.

    I've done something very similar and it fails only on big tests (tle).

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      3 years ago, # ^ |
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      Okay, I overestimated your solution. To make up for my mistake here is a test where your solution is failing :).

      1 1 10 abcxcdxdcx abcdcdcdcd

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3 years ago, # |
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Still rating is not updated

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    3 years ago, # ^ |
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    it's system testing. I guess too many successful hacks.

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3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Again! I turned into a zombie after solving 2 problems

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3 years ago, # |
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Are there any problems with the players ranked 212？Can anybody check it?

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3 years ago, # |
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I saw someone was gonna stream the solution of these but I could not find it now, do you have that link?

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3 years ago, # |
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Why this contest is unrated for me? Wasn't it rated for all participant under rating 1600?

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    3 years ago, # ^ |
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    Instead of killing your nerve cells in waiting you can always use CF predictor (it is quite accurate)

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3 years ago, # |
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rating changes when??

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3 years ago, # |
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I know this isn't the intended solution of C, but can someone explain why this wouldn't work?
My Submission (WA on 5).

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    3 years ago, # ^ |
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    Input
    Expected Output
    Your Output
    Comment
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3 years ago, # |
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Time limit of Problem G is not java friendly ;(

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3 years ago, # |
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This was my submission — 142369596 for 1624E - Masha-forgetful.

The checker log gives me the following message:
wrong answer Integer parameter [name=r] equals to 4, violates the range [1, 2] (test case 249)

Does this mean that my code prints r = 4 for some line while the length of the number is 2? (correct me if I'm wrong)

I can't seem to find my mistake. Any help would be appreciated!

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    3 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    No, it doesn't.

    You forgot to print the number of segments used if $$$m$$$ is 2 or 3.

    For example, one failing test case can be.

    1

1 2
00
00

    A possible answer is

    1
1 2 1

    but your code would produce

    1 2 1

    So, in the original testcase, the jury thinks that you're going to print $$$L$$$ segments, while you actually intended to print just one. Hence the WA.

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3 years ago, # |
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This is my submission for problem E 142402622. The checker log gives the message wrong answer Answer phone is not "s" (test case 532). I am not able to understand the meaning of this message and also not able to find the bug in my code. Could somebody please help?

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    3 years ago, # ^ |
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    Input
    A Valid Output
    Your Output
    Comments
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3 years ago, # |
  Vote: I like it +14 Vote: I do not like it

MikeMirzayanov my submission 142240970 coincides with 142233988 from bernardo_amorim, after looking at his code I think it was triggered because of the main, he have a similar style to mine and this problem is a classical 15 lines problem, I do not know this guy and I did not cheat in any way!!. Please remove the cheating accusation. I saw that he complained too here https://mirror.codeforces.com/blog/entry/8790?#comment-876676.

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The hardest problem of this round: Figuring out what the pun in problem F's name is.

I was like: "Interac-dive? But the problem has got nothing to do with diving"

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's probably 'Interac-div-e' where 'div' stands for the division operation.

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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The girl on the bottom-left looks cute!

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