AGrigorii's blog

By AGrigorii, 10 days ago, translation, In English,

Hello, Codeforces!

I'd like to invite you to Codeforces Round #484 (Div. 2). It'll be held on May/17/2018 19:35 (Moscow time) and Div. 1 participants can join it out of competition. Please pay attention that round starts in the unusual time!

Me and Andrey API Pikas prepared the tasks for this round.

Great thanks to Nikolay KAN Kalinin for helping us preparing the contest, to Mikhail MikeMirzayanov Mirzayanov for the great Polygon platform and Codeforces and also to Kirill KittyLover Samelyuk, Nikolay Apollon76 Permyakov, Dmitry Krainov_Dmitry Krainov, Andrey GreenGrape Rayskiy and Alexey Um_nik Danilyuk for writing solutions.

You'll have to solve six tasks in two hours. The scoring distribution will be announced later

Good luck!

UPD Score distribution 500-1000-1500-2000-2500-3000

UPD2 Congratulations to the winners of the division 2:

  1. Kotori_Is_My_Wife

  2. vosptu

  3. ez_zkj

  4. relativity

  5. iordache.bogdan

And also our congratulations to the winners of both divisions:

  1. Claris

  2. Benq

  3. Kotori_Is_My_Wife

  4. kmjp

  5. vosptu

UPD3 The editorial will be published tomorrow.

UPD4 Editorial

 
 
 
 
  • Vote: I like it  
  • +181
  • Vote: I do not like it  

»
10 days ago, # |
  Vote: I like it -40 Vote: I do not like it

Is there div 3 and is div 3 rated?

  • »
    »
    10 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Div2 rounds are rated for people on div3 too

  • »
    »
    10 days ago, # ^ |
    Rev. 2   Vote: I like it +9 Vote: I do not like it

    Questions about div3 is getting more annoying than "Is it rated?". Seriously, stop.

    "Div 3" means there is the special contests rated only for people under 1600. Nothing more. "Div 3" does not exists anywhere except for the special "div 3" rounds.

»
10 days ago, # |
  Vote: I like it +24 Vote: I do not like it

Great :). You could fix timeanddate.com link to make it display time in user's timezone. Take a look on Mike's comment: http://codeforces.com/blog/entry/59322?#comment-429178

Displayed time screenshots of #483 and #484:

»
10 days ago, # |
  Vote: I like it +16 Vote: I do not like it

Yeaaa! Round by my favourite problemsetter! I waited it so long. <3 <3 <3

I think, I can become Candidate Master tomorrow.

  • »
    »
    10 days ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Hi, bro! It seems that I can upgrade my color too!

»
10 days ago, # |
Rev. 2   Vote: I like it +21 Vote: I do not like it

Third palindrome round in 2018 !!

  • »
    »
    10 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    What do you mean?

  • »
    »
    10 days ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    The first square round in 2018!

    • »
      »
      »
      10 days ago, # ^ |
      Rev. 2   Vote: I like it +50 Vote: I do not like it

      The first square, whose each digit is a power of two, the second digit is the first time two, the third is the second divided by two, for division two, and each digit of the square root of the round number is two, and dividing the square root of the round number by two gives a number whose digits sum to two.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +278 Vote: I do not like it

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +19 Vote: I do not like it

        Damn, I forgot that the sum of the digits is a power of two, more precisely 2^2^2.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +21 Vote: I do not like it

        In case you didn't have enough...

        484 = 222

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
          Rev. 2   Vote: I like it +42 Vote: I do not like it

          never enough ...

          22 = 17 + 5 : date of contest ² = div 2

»
10 days ago, # |
Rev. 4   Vote: I like it -45 Vote: I do not like it

13, 14, 15, 17

»
10 days ago, # |
  Vote: I like it -25 Vote: I do not like it

I smell a lot of hacks

  • »
    »
    10 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I smell easy C and ot hard D

»
10 days ago, # |
  Vote: I like it +21 Vote: I do not like it

Q. Why do you start blushing when a new Codeforces round is announced?

A. It turns me Red. ^_^

  • »
    »
    10 days ago, # ^ |
      Vote: I like it -19 Vote: I do not like it

    How can div. 2 only contest make specialist a gradmaster? Does this make any sence to you?

    • »
      »
      »
      10 days ago, # ^ |
        Vote: I like it +9 Vote: I do not like it

      It is a joke. I am not talking about myself.

»
10 days ago, # |
  Vote: I like it +102 Vote: I do not like it


Reading comments be like...

»
10 days ago, # |
  Vote: I like it +4 Vote: I do not like it

is it rated for purples ?

»
10 days ago, # |
  Vote: I like it -36 Vote: I do not like it

Should have a contest on yesterday. 13,14,15,17....why missed 16??

»
10 days ago, # |
  Vote: I like it -17 Vote: I do not like it

What about for Div 3 members?

  • »
    »
    10 days ago, # ^ |
      Vote: I like it -26 Vote: I do not like it

    Most contests in Codeforces:

    1) Div.1 + Div.2: Candidate masters and higher (1900 and higher) will be in the first division while experts and lower (lower than 1900) will be in the second division

    2) Div.2 Only: Candidate master and lower (lower than 2100)

    3) Div.3: Specialist and lower (lower than 1600)

    So please stop commenting the same question every contest.

»
10 days ago, # |
  Vote: I like it -21 Vote: I do not like it

Hii

»
10 days ago, # |
  Vote: I like it -35 Vote: I do not like it

Wish high ratings to all.

»
10 days ago, # |
  Vote: I like it -17 Vote: I do not like it

If i remember correctly, now the contest is delayed by 1 hour compared to original announcement, which is not good for people waiting for this contest.

»
10 days ago, # |
  Vote: I like it -50 Vote: I do not like it

Bad timing for the user all over the world

»
10 days ago, # |
  Vote: I like it +130 Vote: I do not like it

I hope this round won't be like that 1vldm8

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Guess I'm a faithful guy...

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Hacking is such a ridiculous thing, maybe good for some guys but unfair... But not always bad, the stress and excitement is cool, agree?

»
10 days ago, # |
  Vote: I like it -20 Vote: I do not like it

no a good time but i like it!:)

»
9 days ago, # |
  Vote: I like it -11 Vote: I do not like it

better starting time for Bangladeshi contestant....

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Can't open problem C and D. Showing Error !!

  • »
    »
    9 days ago, # ^ |
      Vote: I like it -23 Vote: I do not like it

    Have you tried restarting your computer or opening Chodeforces in a different browser?

»
9 days ago, # |
  Vote: I like it -12 Vote: I do not like it
The comment removed because of Codeforces rules violation
»
9 days ago, # |
  Vote: I like it +66 Vote: I do not like it

How to solve F?

  • »
    »
    9 days ago, # ^ |
    Rev. 4   Vote: I like it +26 Vote: I do not like it

    If we look at the dfs tree of the graph:

    • If we got any backward edge (an edge connecting a node to one of its ancestors) then the answer must be one of the nodes between the two endpoints. If there is not such edge then answer is -1.
    • If we got another backward edge then we need to find the intersection (or remove nodes from solution which are ancestor of the other node or are not ancestor of the current node). For example, 1 -> 2 -> 3 -> 4 is a cycle, and 2 -> 3 -> 5 is another cycle. Both 4 -> 1 and 5 -> 2 are backward edges.

    • If we got a cross edge (edge connecting a node to another that is neither its ancestor or descendant) then if the other node is in cycle and topmost node of the cycle is a parent of the current node, we need to erase the nodes that are neither parent to the current node nor descendant of the other node. For example, 7->4 is a cross edge.

    Submission as a reference: 38382900

»
9 days ago, # |
  Vote: I like it +8 Vote: I do not like it

How to solve problem D?

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    You can brute force for all posible k using sets in O(nlogn).

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +9 Vote: I do not like it

      what is the test 5 ?

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        I used co-ordinate compression. I got WA TC #5 when I was printing K in terms of the compressed value instead of the real value.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
          Rev. 3   Vote: I like it +3 Vote: I do not like it

          I got WA on test 5 for a different reason.

          My answer: 57209037

          Judge: 45567032

          My algo iterated a values as candidates for k. When a is sorted by size, these values are consecutive in the input:

          57209037 (found best solution at this value)

          45567031 (found worse solution at this value)

          Obviously, the best solution can also be found at 45567031+1.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        If anyone needs complete test 5 input, I've put it here: https://pastebin.com/vHuv6LE4

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      What is stored in the sets? Also... Brute force requires to iterate through the whole array right?

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 3   Vote: I like it +3 Vote: I do not like it

        We can store lengths of locations and when they are reached, with a array or bst like std::set or std::map. To do this we can sort the sequence and enumerate k from 1 to n, using disjoint sets.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Can you elaborate please?

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      based on this we can simply use an array instead of std::set, thus we can do this step in O(n).

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    And of course k can only be equal to a[i] + 1

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I got the k=ai+1 part during the contest but can you tell me about how to brute force ??

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 4   Vote: I like it +9 Vote: I do not like it

        I have one set to store segments of elements less than k and one multiset to store lengths of all segments in first set. Then i add all a[i] from the smallest to the biggest, where current k = a[i] + 1. When you add one there're three scenarios: it forms a new segment, it unites with one segment, it merges two segments together. I think it's clear how to do updates and how to keep multiset of length.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Oh thanks I get it.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
            Vote: I like it +8 Vote: I do not like it

          Alternately, start with the segment [1...n], and add a[i] from largest to smallest. This way, it's guaranteed that the only operation happening is breaking up an existing segment, which makes implementation easier.

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 2   Vote: I like it +3 Vote: I do not like it

        So when at some step segment set size is begger than it's sizes at all previous steps and when all elements in multiset are equal I update the answer with a[i] + 1.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I thought of this, forgot to implement it because the samples didn't need it, and managed to implement it with about 10 seconds to spare in the contest.

      ...Needless to say, I wasn't able to submit in time :(

      Ah Codeforces, thou art a cruel mistress...

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    Maintain a dsu of active indices and process the indices with increasing order of values. At each step, let cur_k be value + 1. Check if neighboring indices are active and accordingly merge them. Also maintain the count of total active indices for a given size of set. Update the answer if the current size is equal to total active indices.

»
9 days ago, # |
  Vote: I like it -22 Vote: I do not like it

Problems are very hackable !

»
9 days ago, # |
  Vote: I like it +61 Vote: I do not like it
  • »
    »
    9 days ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    I took 30mins to solve it. Authors need to be more cautious before preparing a problem. :(

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    striver_79 try to solve problems rather than searching for problems

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      haha, if I was to search for the problem, I would have got all the pretests passed in 5 minutes. I tried for an hour, but when I did fail for it, then I was searching for the DFS part when I came across it. Learn to observe before saying something. :)

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 2   Vote: I like it +7 Vote: I do not like it

        Looking for the dfs part and then came across that nice explanation btw what are you looking in DFS. "WHAT TO OBSERVE". I also type DFS in google but I didn't came across that question maybe your browser is a legendary one.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
            Vote: I like it +7 Vote: I do not like it

          I do not owe an explanation to you that is for sure buddy. And stop blaming someone for pointing out a link which seems similar. :) I guess you learn everything without googling. You got so much of inner talent :)

»
9 days ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Problem E can be solved by the fact that this rectangle can be mirrored adjacently on the coordinate plane.

No, I didn't solve it.

I've done a similar problem in Google Foobar. If I had been given 40 mins I could've solved it. :(

»
9 days ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

What is the 6th pretest in C?

»
9 days ago, # |
  Vote: I like it +38 Vote: I do not like it

Problem C is available at Hacker Rank: Your text to link here... . Check this out..

»
9 days ago, # |
  Vote: I like it +11 Vote: I do not like it

I read "connected components with equal size" instead of "even size" in C. Yikes

»
9 days ago, # |
  Vote: I like it +23 Vote: I do not like it

Feedback:

C: nice

D: A bit ambiguous, I needed two clarifications to understand it.

E: The mirroring trick is very standard. I think I've seen a very similar (maybe the same) problem in uva.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    C was a direct copy from geeks for geeks

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Mirror trick? Can you say sth? I solve it using Chinese remainder theorem.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Do you know if there's a name for the mirroring trick? I've seen a solution that used it in the past, but I wasn't able to understand it then. I'd like to find more information about it and learn it now.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      just consider each side of the table as a mirror and copy the table based on its picture in the mirror. So anytime the ball hit the side it enters the new table and continues its direction.

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 3   Vote: I like it 0 Vote: I do not like it

        I don't understand what algorithm follows this. You calculate the intersetcion point of rectangle and a straight line. If you hit the origin or a corner, you are done. If you hit the wall, you reflect the whole rectangle and keep going.

        You repeat the above 10^9 and tle.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
          Rev. 2   Vote: I like it +20 Vote: I do not like it

          The point of the reflection isn't to simulate the process: that can be done without the reflection at all. The point of the reflection is:

          1. You can now travel in a straight line

          2. You get generalized coordinates for the corners: for example, using the labeling in erdos' picture in the other comment, point A will have coordinates (2mk, 2nl) for integers k, l.

          You can get similar generalized coordinates for the other 3 corners, and then it's simply checking whether integers k and l exist such that the point given by them lies on the line you have, in the direction you want.

          • »
            »
            »
            »
            »
            »
            9 days ago, # ^ |
              Vote: I like it +1 Vote: I do not like it

            Can you please provide a resource link to learn it? Thanks in advance.

            • »
              »
              »
              »
              »
              »
              »
              9 days ago, # ^ |
              Rev. 2   Vote: I like it +1 Vote: I do not like it

              I'm sorry, but I don't have an online resource for this sort of thing.

              Given the context, I'm pretty sure that there'll be a few problems of this type discussed on AOPS, so look there maybe?

              To be honest, this is my first time seeing the reflection being used in a coding problem. I've only used it in maths problems before, but it seems like a nice Olympiad problem trick, so AOPS seems like the place to look.

              I'll look around a bit and see if I can find something more concrete.

            • »
              »
              »
              »
              »
              »
              »
              6 days ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              This link might be helpful for the proof and solution of diophantine equations.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +78 Vote: I do not like it

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I agree with you and saw a couple of problems like E. (e.g. in Google foobar) I'm wondering why did I read it at the end of contest! :-/

»
9 days ago, # |
  Vote: I like it +8 Vote: I do not like it

I used tree dp to solve C but now i find out that there's some simple solution -_-

»
9 days ago, # |
  Vote: I like it +24 Vote: I do not like it

So are you sure what you want us to find in D is what D's statement says?

Thus, in the sequence of n days we can find consecutive nonempty segments when the shark traveled the distance less than k in each of the days: each such segment corresponds to one location.

Shit in bold's true but it it's not true that each location has a corresponding nonempty set.

Now getting back to what the problem asks for:

Find such integer k , that the number of locations is as large as possible.

And at the same time, what you want us to do is find the k which maximizes the number of nonempty sets, not the number of locations.

Even after the announcement, which was too late anyway, nobody bothered to actually update the statement.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This is because if a shark travels 2 days in a row, he will still be at a new location on the day in between, correct?

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    After reading the problem statement I thought the answer to this test case 8 1 2 7 3 4 8 5 6 should be 0 as it satisfies all the conditions asked i.e 1) the shark was in each location the same number of days, 2) the number of locations is maximum possible satisfying the first condition, 3) k is smallest possible satisfying the first and second conditions

    • »
      »
      »
      9 days ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      I have the same confusion regarding this problem (but not anymore). Let me help clarify this. Assume the answer is k, one location is counted as non-empty consecutive segment which every number in it is strictly lower than k.

      For first sample case, if the answer is 7, the number of location is 3, that is {1,2}, {3,4}, {5, 6} and valid because each segment size is two. If you choose 0 as the answer, the number of location is zero. So there's a better answer than 0.

»
9 days ago, # |
  Vote: I like it +23 Vote: I do not like it

Happy Ramadan holiday to all. I wish happiness and health to everyone.

»
9 days ago, # |
Rev. 4   Vote: I like it +5 Vote: I do not like it

This statement didn't exist at the beginning of the contest. My solution was hacked because of it. and there was no clarification about it!!!!!!! AGrigorii [ ](Ca)

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    it did exist

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +2 Vote: I do not like it
  • »
    »
    9 days ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    Actually we missed initially that the note was incorrect for n = 2. The mistake was fixed a few minutes before the round, as soon as MikeMirzayanov spotted it, and the update was pushed from Polygon to CF. It takes some time, but for me the statement opened a minute before the contest already contained the fix, so I decided that everything was ok and I did't have to make an announcement. It turned out that for some people the old, cached statement opened without the fix. Luckily almost everyone understood what we meant correctly.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it -13 Vote: I do not like it

      Till I submitted my solution ,the fix didn't exist. I wrote my solution based on what I read in the statement. It wasn't my mistake.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      You can`t re-test only A problem without test with size = 2? I think it would be fair

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it -22 Vote: I do not like it

      So this is over and my rate decreased beacause of a mistake which is not mine.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        Stop crying over dumb shit. Mistakes happen and this contest is not the end of the world.Also its just -20 which can be recovered very easily.

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          As for me, this worsened my result by 45, but I totally agree that this is just a rating, but it's not very pleasant either

»
9 days ago, # |
  Vote: I like it -9 Vote: I do not like it

A question copied directly from geeksforgeeks, wtf ???? Contest should be unrated.Its a big disgrace.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    The practice shows that it won't be unrated. There was a case when absolutely the same problem appeared twice even on codeforces itself but the round was rated.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      If I am not wrong it was in an educational round, in which the ideas can be common. And it did not affect many div 2 coders. I might be saying this since I performed bad in the contest. But still C was just google and first link and it shouldn't be the case.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        I think there were plenty of "just google and first link" cases too. But still I think we should let coordinators decide whether it's fair or not.

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 2   Vote: I like it +3 Vote: I do not like it

        I have seen many times(I guess at least 3) where exactly same problem was in other contest, but mostly all of such contests were rated.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it -17 Vote: I do not like it

      Seriously Mike, people copying code from other sites getting more credit than people working their ass off for more than half an hour. Fantastic work Mike.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it -6 Vote: I do not like it

    Yes. I carefully thought and coded it for 47 minutes only to realize after contest that it was googleable. Shame !!! -_-

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Was not able to open all problem statements in one window :( Bug

»
9 days ago, # |
  Vote: I like it +17 Vote: I do not like it

o.O

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    I think only his last solution got hacked, and all previous solutions were marked as hacked.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    did he cheated using clone account?

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

What topic should i learn to solve C div2?

»
9 days ago, # |
Rev. 2   Vote: I like it +5 Vote: I do not like it

Almost same as today's Problem C Is there any difference conceptually ???? May be it is not what the contestant expect from the problem setter..

  • »
    »
    9 days ago, # ^ |
    Rev. 2   Vote: I like it +16 Vote: I do not like it

    In the yesterday contest, solution for C was also in Wikipedia.

»
9 days ago, # |
Rev. 3   Vote: I like it -25 Vote: I do not like it
»
9 days ago, # |
  Vote: I like it -61 Vote: I do not like it

I will leave codeforces if the contest is not made unrated, or semirated for the sake of it. But it is not fair for people to copy codes and press submit button.

»
9 days ago, # |
  Vote: I like it +24 Vote: I do not like it

How to solve C if it would have been a graph with cycles ?

  • »
    »
    9 days ago, # ^ |
      Vote: I like it -18 Vote: I do not like it

    tree hasn`t got cycles

    • »
      »
      »
      9 days ago, # ^ |
      Rev. 2   Vote: I like it +3 Vote: I do not like it

      I know that. I am just asking how would one go about solving the problem if it would have been a graph with cycles as the approach used for solving the tree one will not work here.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it -11 Vote: I do not like it

        Sorry, I did not understand correctly what you mean

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 5   Vote: I like it -8 Vote: I do not like it

        I may be wrong but in case of undirected connected graph we have to consider only those edges which if removed results in more than one connected component otherwise clearly if removal of an edge does not divide the graph then answer depends on n is even or odd. Now for the previous one you can have a look at bridge in a graph --> https://www.geeksforgeeks.org/bridge-in-a-graph/ Removing such edges and doing the same what we did in actual C question should do the thing.

        But Edges might be of order which does not satisfy the constraints if n is 1e5.

        By doing the same thing (which i wrote above) i mean that edge should be removed only if it divides two parts in even number (each).

        • »
          »
          »
          »
          »
          9 days ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          The approach seems fine but I think finding out Bridges and removing would give "minimum" number of edges required for deletion and not "maximum".

          • »
            »
            »
            »
            »
            »
            9 days ago, # ^ |
              Vote: I like it -8 Vote: I do not like it

            No removing bridges does not govern whether we get minimum or maximum edges. When i say that you remove edge only if it disconnects the graph and both sides nodes should be even. So automatically all the edges which satisfy this property will get counted and that's what we did it in given C we count all those edges which divides the nodes into two even halves. Please correct if i m wrong !!

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      in case of a connected graph ???

»
9 days ago, # |
Rev. 2   Vote: I like it +204 Vote: I do not like it

About the choice of wording in condition 1 of A

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

In Question B why doesn't 3*nlogn pass the test cases. I tried using maps in c++ but it is showing time limit exceeded in test 45 . Can someone point out mistake in my code : Link

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    while(used[cnt2]==1)
    cnt2--;
    

    This part doesn't looks like O(1) or O(log n). I think, it's O(n) at worst case (because of "cnt2 = cnt1;" six lines higher).

»
9 days ago, # |
  Vote: I like it +15 Vote: I do not like it

Yay, I became master, but sadly only because of div2 rounds.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    What is the difference? Master is master. If you don't want to participate in div 2, then wait for div1...

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      well actually its opposite, there are some purples that skip div 1, because they don't want to be beaten up by reds and yellows and join div 2 to compete with blues instead.

      • »
        »
        »
        »
        8 days ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        They can be beaten up by blues then.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    I don't understand why everybody is complaining and thinks that it would be easier to get to 2100. The only true thing is that there will be more opportunities to get there. It doesn't mean it would be easier. If you don't deserve 2100 you won't get it, one way or another.

    For example I already managed to lose a lot of points in div 2 contest on Monday. And today I was 122. If there were no violet coders, I would have been 25'th which would have given me much more rating.

    • »
      »
      »
      9 days ago, # ^ |
      Rev. 2   Vote: I like it -31 Vote: I do not like it

      Lol You don't understand? Before this shit revolution, to get Master(old) you need to most likely solve 3 probs in Div 1 which is equivalent to all div 2 problems. Now get just good ranks which in most case is possible with just 4 div 2 problems. As you can see in this round, 370 officially solved 4th hardest, you think thats good enough for someone Master?

      "I don't understand why everybody is complaining and thinks that it would be easier to get to 2100" Because it actually is faggot. If you don't understand, then compare how many users have got there maximum rating in past div 2 only rounds.

      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        "to get Div 1 you need to most likely solve 3 probs in Div 1" — makes sense. Sorry about my previous post.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    You've became at least, not like me :D (9 points)

    Congradulations!

    • »
      »
      »
      9 days ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      At least you have a chance to participate to next Div2 round, not like me :D (passed to master with 11 points).

      Congratulations!

»
9 days ago, # |
  Vote: I like it +14 Vote: I do not like it

It looks very strange that O(nlogn) solution for D works without any used memory: 38374872.

»
9 days ago, # |
Rev. 2   Vote: I like it +5 Vote: I do not like it
  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    That claim about F is totally ridiculous

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +1 Vote: I do not like it
      • »
        »
        »
        »
        9 days ago, # ^ |
          Vote: I like it +20 Vote: I do not like it

        The Quora answer is wrong and comments don't seem to give any hint to actual solution.

        As for the Polish problem, the translation shows that it is a harder version of Problem F. However, it isn't possible for problem setters to verify that their problems don't exist even on non-English sites.

      • »
        »
        »
        »
        9 days ago, # ^ |
        Rev. 2   Vote: I like it +3 Vote: I do not like it

        Yes, it's the 4th time that I came across this problem.

    • »
      »
      »
      9 days ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      It is not. I didn't realize this and now I checked that my incorrect solution to kon (9/10) is accepted after 2 small modifications (for 2 or more disjoint cycles and for empty list of cities).

      Though I am not sure if I would have been able to figure out the first one during the contest...

      Anyway, top 5 has gone...

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Sorry but I have to say that CF won't care about such a thing!

    Sometimes CF copies exact problems from itself though.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      It's not copying if the second setter didn't realize that the idea for the problem was already used before.

      • »
        »
        »
        »
        6 days ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I mean that they won't care about it after realizing copying.

        CF be like: do we have issues with the tests? if yes, then it's unrated else it's rated for sure. I've never seen a round to be unrated for a more logical reason.

»
9 days ago, # |
  Vote: I like it +39 Vote: I do not like it

speed of light coder !! ****

»
9 days ago, # |
  Vote: I like it +20 Vote: I do not like it

I think systests on D are very weak. I found a big mistake in my code after submit but surprisingly gave me an AC. My AC code: http://codeforces.com/contest/982/submission/38384511 A simple test case when my output differs from another AC submissions: 11 7 9 8 100 1 2 3 200 4 5 6 There something invalid in this input?

  • »
    »
    9 days ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    n = 1488 in systests

    Problemsetter is the best ❤️.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    My AC submission give output 10 and your code output 7. Maybe your code do any undefined behavior?

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

In Problem B What if the number of extroverts(1) at any point becomes greater than number of introverts(0)? Many solutions will fail then ![problem:982B]

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It is mentioned in the problem statement itself that there is always a suitable seat for an extrovert, So no such case will occur.

    • »
      »
      »
      9 days ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      My bad ! I didn't understood the problem statement properly and wasted much time :/. Anyway Thanks for help !

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row

»
9 days ago, # |
  Vote: I like it +8 Vote: I do not like it

Can someone help me in understanding D ...??

  • »
    »
    8 days ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    For each different k[i] = a[i]+1, you check the size of all the contiguous days with a distance covered strictly less than k[i].

    You go through all the a[i] in ascending order.

    if a[i] does not have a neighbour lower than him, it creates a new contiguous day sequence of length 1. Then, you merge it with his neighbour(s) (if they are lower than a[i]), which destroys their sequence and creates a new sequence of length 1 + total length of the sequences of the neighbours. It's important because you need to keep track of how many different lengths of sequences you have at all time, cause you will need it at each step.

    To keep track of how many different lengths of sequences you have after each a[i], you can create a table that tracks how many sequences each length has, and a variable that tracks how many lengths you have (that you increase by one each time you insert a sequence whose length is not represented yet, and decrease whenever you remove the last sequence of a given size). If there is only one length in the table, then you know that the number of sequences for k[i] is (number of distances <=a[i]) /(only length of sequence represented).

    The answer is the minimum k[i] that leads to the maximum value of this ratio. You can do the merge with an union/find, making the merging a O(n*log(n)), and computing the ratio each time can be done in constant time, leading to (I think) a solution in O(n*log(n)).

»
9 days ago, # |
  Vote: I like it 0 Vote: I do not like it

AGrigorii When is the editorial gonna be published? If it's already published in Russian, please do provide the link. Thanks

»
8 days ago, # |
  Vote: I like it +21 Vote: I do not like it

I found a surprisingly simple solution for F. The key observation is that if you have a node with an outdegree of 1, you can delete that node and push all incoming nodes to the successor (with a special case for self-loops). This process can reduce the number of edges if an incoming node was already pointing to the successor node.

As an example, node 3 in the image below can be deleted since it has only one outgoing edge:

Which results in:

Note that this reduced the outdegree of node 1, making it now eligible for deletion.

If you have more than one node remaining after this reduction, then the answer is impossible. I haven't tried to prove this, but it resulted in an AC submission. :)

For each node, it is essential to maintain an unordered_set for incoming and outgoing nodes to detect duplicate edges. This makes it slightly slower (due to the set operations), but it's fast enough to pass in C++ (albeit not in Java, from my experience). My accepted submission: 38409568.

  • »
    »
    8 days ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    I thought about a proof of your solution, and I think it is correct. If anyone find any wrong with it, just point out the wrong point.

    The first part is about the validity of deleting the point with one out-degree. We can think about a point u that satisfies the requirement and we call v is a point that have a direct edge to u. We can easily find that every circle that pass the edge u-v must pass the point that u can only arrive at. So we can think u is useless and delete it from the graph (of course it contains the process of changing some edges).

    The second part is about why the answer is impossible when there are more than one point remained. Now, the graph satisfies every point has at least two out-edge, of course we can remove arbitrary out-edge to make every point's out-edge is two. Then we build the dfs tree of the graph, and it is a binary tree. We only need to discuss about the out-edge of the leaves. We assume that this graph is valid for answer. Then we need to construct the out-edge of the leaves by a optimal way to make the graph is valid. So we can first make every leaf link to the root, and the other out-edge link to the next leaf in a way you like. But we will find that one leaf can't make another link because it wouldn't be valid no matter where it connect to. So we can prove this part.

    This solution has a drawback is that it can't output all the points that satisfy the problem. But it realy easy to type.

    (If there are many grammar errors, please don't blame me. Because my English is realy not good.)

»
8 days ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I did not manage to solve div2-C using Python 2. My submission 38369140 got runtime error. I think the issue is recursive calls. Is there a trick in Python to get this code AC?

»
7 days ago, # |
  Vote: I like it 0 Vote: I do not like it

can someone point out the mistake in my code for 2B? 38450232. This is nlgn for sorting and then a map to get the index. Failing on test 11.

»
7 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Who can tell me what's the meaning of 'even size' in Problem C. The 'even' means equal? or %2==0??

»
32 hours ago, # |
  Vote: I like it +3 Vote: I do not like it

Request: Please post the editorial link with problems for this contest. Had a hard time to find this post to find editorials.