vovuh's blog

Codeforces Round #656 (Div. 3)
By vovuh, history, 4 years ago, translation, In English

<too-old-joke-about-copy-paste>

Hello! Codeforces Round 656 (Div. 3) will start at Jul/17/2020 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) 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 rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) 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 participants of the third division, you must:

  • take part in at least two 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 my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

</too-old-joke-about-copy-paste>

UPD: Also thanks to infinitepro for help with statements and testing the round!

UPD2: Editorial is published!

  • Vote: I like it
  • +511
  • Vote: I do not like it

Comments Comments (201)
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4 years ago, # |
  Vote: I like it -64 Vote: I do not like it

will it be rated??

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4 years ago, # |
  Vote: I like it +76 Vote: I do not like it

Third unrated round for me in a row :(

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    4 years ago, # ^ |
      Vote: I like it +207 Vote: I do not like it

    codeforces nowadays .

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      4 years ago, # ^ |
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      xD

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

        Hi, i'm new, how do i know if a round is rated or not?

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

          It is written in the contest announcement, and if the round becomes unrated there will also be an announcement saying that. But most rounds held are rated.

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4 years ago, # |
  Vote: I like it +28 Vote: I do not like it

Monogon contest again got rescheduled :(

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    4 years ago, # ^ |
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    It's good that it got rescheduled. Would have been sad if it became unrated like the previous two rounds.

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

I think they are running this round because they think there will not be a crash and an infinite queue. looking forward to it

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4 years ago, # |
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Codeforces is my new School ... teaching me and hosting Contest :XD Thanks a lot CodeForces!!

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4 years ago, # |
  Vote: I like it +223 Vote: I do not like it

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it -30 Vote: I do not like it

    meme.png

    In English : Mike says "do not spoil my idea with tell everyone"

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4 years ago, # |
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"the round will be rated for you." Hope this time,,, this line will be true for me..Btw Thanks for your hard work..

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4 years ago, # |
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Deleted

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

Hopefully this wont be unrated !!

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4 years ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

I was eagerly waiting for the announcement :)

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4 years ago, # |
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Let's all pray for smooth conduction of this round.

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4 years ago, # |
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It was better make a test round before real contest :D

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

    There is one testing round arranged, it will start in ~45 min. check here

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4 years ago, # |
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This will be my first ever time competing.. any pointers?

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

Situation or system in last 2 rounds.

hope all will be fine this time :)

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4 years ago, # |
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Why there is unusual time for round 657 ?..

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    4 years ago, # ^ |
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    May be authors comfortable with that time.

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

      Codeforces contest time are generally xx:05 or xx:35 according to GMT. But that rounds scheduled at 09:00 GMT(not 09:05). So that i assume that round is tied with some other contest(may be some onsite contest).

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        4 years ago, # ^ |
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        on july 19 there is the cookoff on codechef. maybe to avoid a clash

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4 years ago, # |
  Vote: I like it -8 Vote: I do not like it

</too-old-joke-about-copy-paste> not too old for me :D

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4 years ago, # |
← Rev. 3   Vote: I like it +6 Vote: I do not like it

To qualify as a trusted participants of the third division, you must:

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

I believe this is incorrect :)

Edit: Nevermind, just noticed it's the same in previous rounds. There's a different limit for being a trusted participant and being ranked. Sorry for the noise.

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4 years ago, # |
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Hoping that this contest goes on smoothly _____

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4 years ago, # |
  Vote: I like it +20 Vote: I do not like it

Who could tell me how to remove or report that comment?so dirty..

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

i'm coming back to this site after a break of 3+ years. so i'm new to this "12-hour open hacks" system.

firstly, is it something new? or has it been a regular thing in contests recently?
also, does it mean there will be a kind of "hacking phase" which lasts for 12hrs after the 2-hour "coding phase" has ended?

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    4 years ago, # ^ |
      Vote: I like it +21 Vote: I do not like it

    1. It's a regular thing for div.3 and Educational Rounds.

    2. Yes.

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      4 years ago, # ^ |
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      oh this is new for me ... i guess we would need to solve the problem (or atleast, pass the Pretests) in order to hack other participants' solutions?

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        4 years ago, # ^ |
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        Nope. You can hack any problem, even if you didn't solve it.

        What's more, sources can be copied for local test.

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    4 years ago, # ^ |
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    dolphins returning to Italy and footballers returning to CodeForces, Cool!

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4 years ago, # |
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Hoping positive delta change for all :p

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

    That literally can't happen afaik.

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      4 years ago, # ^ |
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      What If everyone has exactly the same rank(same score with same penalty).

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        4 years ago, # ^ |
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        even then the higher rated participants will have a negative delta while lower rated ones will have positive

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        4 years ago, # ^ |
          Vote: I like it -7 Vote: I do not like it

        Everyone will get 0 rating change.

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

I believe there should be a half-hour unrated short contest some time before the actual contest, to test the system's current load handling capacity.

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

As a tester, I won't say anything, since you already know how good vovuh contests are!

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

Hope atleast first problem do not take time.

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4 years ago, # |
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really excited for this contest. I just hope it doesn't get ruined unlike past few contests. Looking forward to learn a lot from vovuh's tasks.

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4 years ago, # |
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why is the time extended 15 min?

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4 years ago, # |
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Good luck and high rating to everyone! :)

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4 years ago, # |
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Hey, I don't think this is right:

  • do not have a point of >>1900<< or higher in the rating.

Although you can find this sentence after it:

if your rating is less than >>1600<<, then the round will be rated for you.

I think this may puzzle those people who are new and without any experiences before in Codeforces.

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

Seems more like a Div2 than Div3

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4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

Great contest, thanks a lot for this, now I see what vovuh hype was about.

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

Deleted
Just realized 2^26 is more than 200000 lol the meme is pointless

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

    living in a country with a different number of alphabets surely make you confuse

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4 years ago, # |
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how to solve question D???????????????

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    4 years ago, # ^ |
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    Just brute force with recursion.

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      4 years ago, # ^ |
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      what will be the time complexiy in that case

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        4 years ago, # ^ |
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        O(2^logn) i guess

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          4 years ago, # ^ |
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          Simple bruteforce . set l=0 , r=n-1 , mid=(l+r)/2 , at each moment you can either change range [l,mid] to current character or change [mid,r] ; Time complexity : O(2^log2(n)+26n) which is O(n) Ps : O(26n) to compute prefix sums of occurences of each character .

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

          O(2^logn)=O(n)

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          4 years ago, # ^ |
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          With every time using "count", it should be O(n*2^logn) which is equal to O(n^2) right?

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            4 years ago, # ^ |
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            Actually length of string is also halved each time. It could have been precalculated,it thinked this way also, but realized that count within solve() is better in term of time.

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    4 years ago, # ^ |
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    Divide and Conquer, complexity should be O(n), because each character would be counted at most twice. (I am not sure about the complexity part tho)

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

      yes, it´s $$$O(n)$$$. Look: let $$$T(n)$$$ be the amount of operations for an array of length $$$n$$$.

      $$$T(n) = 2\cdot T(\frac n 2) + O(1)$$$, for $$$n > 1$$$.

      $$$T(1) = O(1)$$$.

      if you solve this: you´ll get that $$$T(n) = O(n)$$$

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4 years ago, # |
  Vote: I like it +47 Vote: I do not like it

My solutions (untested, but overall ideas should be right):

A
B
C
D
E
F
G
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    4 years ago, # ^ |
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    in D do you need the prefix sum? since it will iterate in ln(n) anyway?

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

      I did D without prefix sum

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      4 years ago, # ^ |
      ← Rev. 3   Vote: I like it +1 Vote: I do not like it

      no, it's not needed (although I used it), look:


      int cost(int i, int j, char c){
    int res = 0;
    while(i <= j)
        res += (s[i++] != c);
    return res;
}

int solution(int i, int j, char c){
    if(i == j)
       return s[i] != c;
    int mid = (l+r)/2;
    int s1 = solution(i, mid, c+1) + cost(mid+1, j, c);
    int s2 = solution(mid+1, j, c+1) + cost(i, mid, c);
    return min(s1, s2);
}
/// inside main()

cout << solution(1,n);

      This has complexity $$$O(n\cdot \log n)$$$, so it is fine. If you want to do it linearly, use a prefix sum to compute the cost.

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    4 years ago, # ^ |
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    is DP was necessary in question D or normal recursion will work

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      4 years ago, # ^ |
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      Bruteforce recursion works in just 30ms

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      4 years ago, # ^ |
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      Actually, simple recursion does work. Oops.

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        4 years ago, # ^ |
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        I am not understanding why my solution to D is giving TLE when I don't use s.substr() and instead pass the limits in the function(Link to submission) and gets accepted when I pass the substring to the function(Link to submission). Can anyone tell me the reason for it?

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

          Passing a string to a function passes it as a copy if you don't take the reference as the attribute and hence it leads to a O(n^2*log(n)) solution which is TLE. Instead, when you pass the substring, it is of the same order as the number of operations in your for loop and the complexity is not affected. You can instead do:

          int func(string &s,int st,int ed,char c)

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            4 years ago, # ^ |
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            Thanks!

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            4 years ago, # ^ |
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            Thanks sayam! I had the same doubt!I even used prefix sum still I got TLE, was unable to understand the issue.

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            4 years ago, # ^ |
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            Can you please elaborate?

            I got TLE on my first submission which i thought was having O(2*logn) as time complexity, after reading your comment i just changed (string s) to (string &s) and boooom the sol got accepted :(

            Doesn't that change only affect the space complexity? How time complexity is getting affected?

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              4 years ago, # ^ |
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              creating a copy of a string of length n takes O(n) time and O(n) extra space. (this happens if you call by value) which when done over n times gives time complexity O(n^2).

              & operator in parameter name helps it to pass by reference, (as if using the same string but with a different name) so no time or space required.

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      4 years ago, # ^ |
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      Normal recursion will work

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

    For D you can also try every possibility of the good array and find the minimum answer from all possibility. The total possible good array will be n

    Solution

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      4 years ago, # ^ |
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      How did you come up with the number of good arrays for a given n? I thought there is only n good arrays for a given n...

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        4 years ago, # ^ |
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        Yeah! Thank you for pointing out. I just recalculated, It should be n. My math sucks.

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    4 years ago, # ^ |
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    I used the local minimum approach on C but getting WA on test 876 pls help

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

https://www.geeksforgeeks.org/assign-directions-to-edges-so-that-the-directed-graph-remains-acyclic/ Question E

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

Was G component wise 2-SAT or there is a easier approach to it?

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    4 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    The easier approach is just bipartite coloring. You just need to build a bit special graph to apply it.

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

That was an amazing contest! Thanks Vovuh

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

Problem E explanation can be found here : https://www.geeksforgeeks.org/assign-directions-to-edges-so-that-the-directed-graph-remains-acyclic/

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    4 years ago, # ^ |
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    Its Problem E I think. Also it shattered me when I cam to know that a good problem can be googled easily.

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    4 years ago, # ^ |
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    Ohhkk.. So whosoever read that article just had to implement it.

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4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Is it me or the contest was at a bit higher level than usual Div3?

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

This can help you with first 4 problems. https://docs.google.com/document/d/1Fq4UAJBh9Tn48hiF6bLzwvl7ce5XikqWeFmFfw1Q0OE/edit?usp=sharing

Help me with the last 3 problems.

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4 years ago, # |
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How to solve E ?

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4 years ago, # |
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G was implementation heavy i believe.

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4 years ago, # |
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Okay , my rating got fucked up again anc I need a quick editorial now.

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4 years ago, # |
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My Solution to C , i try to provide a proof of the solution too . Hope You like It.

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

in D I forgot to call the string by reference and got one TLE omg

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    4 years ago, # ^ |
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    Just realized what I was doing wrong after reading your comment. Big facepalm! I usually always pass things by reference.... F

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      4 years ago, # ^ |
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      And here I am trying to use DP, thinking my solution is too slow instead of passing string by reference :(

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4 years ago, # |
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Did anyone solve F with dp on trees + rerooting technique ?

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

    I solved it by using the same technique . You can check out my code.

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

I think there was typo while naming the round It should be Codeforces Round#656(Div 2)

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4 years ago, # |
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https://codeforces.me/contest/1385/submission/87160668

Can anyone tell me what's wrong here? I guess the complexity O(n),still giving TLE.

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    4 years ago, # ^ |
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    What is your expected time complexity for memset?

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    4 years ago, # ^ |
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    First thing is you should pass string by reference otherwise even if it is right you will get TLE, but even after that your code gives TLE, so there must be some other problem.

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    4 years ago, # ^ |
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    Here's your AC, don't use memset in this case as there are many test cases and you are resetting the whole prefix array. Also pass string by reference. https://codeforces.me/contest/1385/submission/87164492

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4 years ago, # |
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Is there anyone who solved E without topological sort? I tried with dfs but getting WA on test 2.

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

    I did. First check for cycle with directed edges. If not found you can always direct the undirected edges and still have an acyclic graph. I had two adjacency lists, one for directed edges and the other for undirected edges. Have 3 states each for visited, completed and not visited separately. In a dfs run from a vertex, if any of the undirected edge is directed to a completed vertex, then this can never be a part of the cycle. So remove this vertex from the adjacency list of the completed vertex. Code : 87141053

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4 years ago, # |
  Vote: I like it -72 Vote: I do not like it

D is annoying complecated statement. Once understood the defintions it is "implement what statement says".

Not sure why such problem is in the set. From my point of view the most idiotic kind. It is not more than decypher obfuscated problem statement.

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

Tasks were interesting, more div 2 than div 3 in my opinion.

Only thing I do not like, that you are allowing more than two occurrences of same element in $$$G$$$. That is not funny joke.

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    4 years ago, # ^ |
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    Yep, I thought it could be stupid because it is the only case when $$$-1$$$ appears, but now it's too late to change anything.

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      4 years ago, # ^ |
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      Problem G test case:

      5
5 3 5 1 4
1 2 3 2 4

      ans:

      2
1 2

      why it is wrong? help please :)

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        4 years ago, # ^ |
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        If you swap columns $$$1$$$ and $$$2$$$, you get $$$a_1 = [1, 2, 5, 1, 4]$$$ and $$$a_2 = [5, 3, 3, 2, 4]$$$. Both rows are not permutations.

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          4 years ago, # ^ |
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          see the other way, I am swapping 5 from 1st with 2 from 2nd i.e [2,3,5,1,4] n [1,5,3,2,4]

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            4 years ago, # ^ |
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            You are not allowed to perform such operations. The element can't change its initial column.

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    4 years ago, # ^ |
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    Wasn't that included in the samples?

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      4 years ago, # ^ |
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      Yeah, you are right. I have noticed it after contest. I had another case when I expected ans to be $$$-1$$$ (and that situation is impossible, except case where more than two numbers are equal). I need to check samples more careful.

      Anyway, still it is not funny :)

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

What is the proof of E sol.?

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    4 years ago, # ^ |
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    Consider an undirected edge $$$(u,v)$$$. Suppose making this edge directed, gives rise to a cyclic graph, then both the directed edges $$$(u,v)$$$ and $$$(v,u)$$$ must belong to some cycle with the directed edges. But in this case, even after removing the undirected edge, the graph is still cyclic. Otherwise, either $$$(u,v)$$$ or $$$(v,u)$$$ (or both) could have maintained the acyclicity. Thus when the graph is acyclic with the directed edges, we can always direct the undirected edges and still get a $$$DAG$$$.

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4 years ago, # |
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can someone till me the difference between this submission 87132940 and that 87161171 I thought that they are the same idea, but I don't know why the first submission failed

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4 years ago, # |
← Rev. 4   Vote: I like it +32 Vote: I do not like it

My solutions :

A. Iterate on all possible triplets from set of numbers 1, x, y, z, 1e9 for each a,b,c, and check if a given triplet exists that satisfies the given condition

B. Take only first occurrences of each number while iterating make the permutation

C. One obvious way to generate array c is to always take minimum end element of the array and generate the final array. So as we move right, we get a smaller array and better odds to generate a valid array c. So binary search on the answer from range(0,n-1) and check if array C formed by above method is correct.

D. To build the answer for (1,n,c), we obviously need the answers for the range (1,n/2,c+1) as well as for (n/2+1,n,c+1), so either use memoization dp or just build the answer for each character for a range recursively, ie, for (1,n) for 26 characters, build the answer for (1,n/2) and (n/2+1,n) for 26 characters each. Then generate answer for each character for (1,n) using conditions described in the question.

E. An observation was that if there is no cycle in directed graph using only directed edges, then all other edges can be directed to avoid a cycle. One such orientation is to direct edges on the basis of topological sort. So check if graph built using only directed edges contains cycle, and if it doesn't ,generate a topological sort from that, and order edge nodes on the basis of topological order.

F. An observation was that if at some time we have k leaves for a node, or k leaves for many nodes, it won't affect the answer's maximum if we decide to do operation on any such valid node because if leaves of that node stay, then that node is as good as unusable node at any point in the future. So a possible algorithm is to take the node with highest leaf count, remove (cnt — cnt mod k) leaves and hence do floor(cnt/k) operations, where cnt is the count of leaves attached to that node. We can do this by maintaining the adjacency list, adjacent leaf nodes list, and leaf frequency list and a set to store the node with highest leaf count and updating these as we pop elements from the set.

G. I did this using bipartite graphs. First we check if each value from 1 to n has exact frequency of 2 for making 2 valid permutations. If so, build a bipartite graph where nodes are pair of (row,column) for each value. 2 nodes are in disjoint parts(of bipartite graph) or have edge in between if they are either in the same column or have the same value. Build this graph, and check if it is bipartite. If it is, then an answer exists. To find the optimal answer , let s just find the cost for a connected component. It is the minimum of the cost of moving one disjoint part of the component to row 1 and the adjacent part to row 2, and vice versa. That is where implementation might become a bit heavy.The total cost is the sum of cost for each component.

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    4 years ago, # ^ |
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    master_noobie How do we prove this observation for E?

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      4 years ago, # ^ |
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      Consider the graph with only the directed edges. If there is a cycle, then obviously we can't remove it, so the answer is NO.

      If there isn't a cycle, then there exists a topological ordering of the nodes, that is an array $$$pos$$$ such that if there is a path from node $$$u$$$ to node $$$v$$$, $$$pos[u] < pos[v]$$$.

      Now consider an undirected edge $$$(x, y)$$$ and assume WLOG that $$$pos[x] < pos[y]$$$. Then if we direct the edge from $$$x$$$ to $$$y$$$ and add it to the graph, the topological order of the new graph remains the same as the topological order of the old graph (there are a few cases here I guess, but nothing difficult). Thus, since the order still exists, there cannot be a cycle in the new graph. Thus after directing and adding all the undirected edges in this way, the resulting graph will have no cycles.

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4 years ago, # |
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First result of google search adding directed edges without creating a cycle

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4 years ago, # |
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Thanks for such a nice problemset vovuh :)

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4 years ago, # |
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Here are my solution ideas for all problems if you are waiting for the editorial. (video published after the contest ended of course). It is possible that my idea for F isn't actually right; I don't have a proof for it yet.

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4 years ago, # |
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Why is this submission wrong? 87125487 It passes the tests if I clear the dp array, but it should not affect the answer, if I understand resize() correctly.

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    4 years ago, # ^ |
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    Resize does not reset the existing values in a vector. It only expands/shrinks the tail to match the new size.

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      4 years ago, # ^ |
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      Thank you, I found the bug. I kinda knew that, but for some reason thought that it only concerns values (not data structures, and new values are initialized, but old values stay the same), but vector is also a value. This bug has ruined several contests for me, and I am very glad I finally tracked it down.

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4 years ago, # |
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I will be adding detailed editorials for CF problems with appropriate proofs and implementations on my YT channel

Editorial for problem C: https://youtu.be/vjRHaZb16bU
Editorial for problem D: https://youtu.be/9gVpnosPKec
Editorial for problem E: https://youtu.be/yn6ZPaqwlso

Enjoy watching!!

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4 years ago, # |
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Thanks for the contest, in Problem D — I did learn that passing strings to recursive function can lead to TLE, so pass reference :(

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4 years ago, # |
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Wait is it unrated??? when did it change to unrated???

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4 years ago, # |
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Very interesting tasks!! Almost every task required something new. Now after seeing solutions of E, i am awestruck with the use of topological sort in it.

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4 years ago, # |
← Rev. 3   Vote: I like it -16 Vote: I do not like it

can anyone tell me whats wrong with my code? Problem C https://codeforces.me/contest/1385/submission/87169051

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4 years ago, # |
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can someone explain why i am getting TLE in problem D here is link to my code https://codeforces.me/contest/1385/submission/87163991)

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

vovuh nooinenoojno awoo

d_rushabh is hacking the solutions that he submitted using his fake accounts. Remove him from the contest.

+adding similar contestant: mihirarora

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    4 years ago, # ^ |
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    I think he doesen't know that in div. 3 rounds you don't get points by hacking. If he hack all his alts, I guess there is no problem then

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      4 years ago, # ^ |
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      i_say_what_i_want Where is that mentioned?

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        4 years ago, # ^ |
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        In div. 3 rounds you don't get points if you hack someone. Also, if all his alts gets 0 problems solved, then they won't change the rank of contestants, then, it wont change other contestants performance or rating change.

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4 years ago, # |
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I have tried to explain my approach for E , basically explaining the code for topological sort .

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4 years ago, # |
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Can anyone tell what is this code wrong for solution of Problem B? I take all integers in permutation1 until next integer of permutation2 appears. But it throws wrong ans.

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    4 years ago, # ^ |
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    There is a possibility that when you increase the value of j, it might become more than the last index of B and the value at that memory location might be equal to the next A[i], in this situation your code will skip that value and might give you the wrong permutation.

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4 years ago, # |
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8 bbaaddcc explain this pls

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4 years ago, # |
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Isn't it strange today's contest E question is my doubt post on codeforces only that has been answered ? https://codeforces.me/blog/entry/68765

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    4 years ago, # ^ |
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    It is imposible to check your problems against all other problems in the world, especially if they weren't actually published in an OJ. Of course this shouldn't happen, but in general it is common for easy problems to collide with each other.

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4 years ago, # |
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Can somebody help me in question E .My code is giving error on test case 2 and the error says jury has an answer but your code doesn't .Pls help link to my code https://codeforces.me/contest/1385/my

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    4 years ago, # ^ |
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    Please use the correct link to your submission! I'm not sure what the issue is yet, but judging from my experience with this question, have you cleared the adjacent matrix for the vertices after each test case? That might have caused your code to give a wrong answer on testcase 2.

    My code is here 87179787 (I hope it helps)!

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4 years ago, # |
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Can anyone please help me on this in C problem? Getting RTE on Test 2.

https://codeforces.me/contest/1385/submission/87169468

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    4 years ago, # ^ |
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    I looked at your submission. For starters, I think you have overcomplicated your code. However, here is why you're getting your run time error.

    Try running this as a test case.

    1
2
4 3

    If you dry run your code, in the second iteration of your while loop, you'll see that, you have popped all elements from your deque, but you're still running the loop and comparing the elements, which throws an exception.

    I have not corrected your code, but a simple approach that will help you is this. The solution idea is to look for a suffix of the array such that it's one of the three types. 1. All numbers are non-decreasing 2. All numbers are non-increasing 3. At first the numbers are non-decreasing, then they are non-increasing.

    Among all such suffixes, you needed to find the one that had largest length (ensuring smallest prefix to be removed). I did it in a simpler way. See it here 87122310

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4 years ago, # |
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Again div3 with the same difficulty as div2 bruh.

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    4 years ago, # ^ |
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    Really? I can't think of making the problems easier than this without compromising the quality of contest. If anything, its the div 2 rounds that seem more like div 3 nowadays.

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4 years ago, # |
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Div2.5

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4 years ago, # |
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Can someone help me debug this https://codeforces.me/contest/1385/submission/87198122

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4 years ago, # |
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Rated.

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4 years ago, # |
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Hello i'm newbie, can someone explain why is my contest doesn't get rated?

This is not my first contest, I've participate more than 2 contest. I solved atleast 1 problems for each contest.

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    4 years ago, # ^ |
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    The past contest you attended became unrated, due to server problems. The rating for this contest will get updated in some time.

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4 years ago, # |
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This was my first RATED contest. I'm a new user. I participated in last two contest Codeforces Round #655 (Div. 2) and Educational Codeforces Round 91 (Rated for Div. 2), but somehow both the contest got unrated. So in this #656 Contest I solved 3 question. Few moments ago I got my rating, which surprised me.....only 460. How can this be possible? Please help me understanding this.

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4 years ago, # |
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A and B should have been swapped :P

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4 years ago, # |
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Edit : Wrong Post

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4 years ago, # |
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edit: commented on wrong post

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4 years ago, # |
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Hello, I got quetion, in the problem D the test case include el case asdfghjk and the answer is 7, but the answer shouldn't be 6?. The string of the answer would be ddddgfee

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