Educational Codeforces Round 45 [Rated for Div. 2]
By awoo, history, 7 years ago, In English

Hello Codeforces!

On Jun/10/2018 13:05 (Moscow time) Educational Codeforces Round 45 will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extented ACM ICPC rules. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

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

The problems were invented and prepared by Adilbek adedalic Dalabaev, Roman Roms Glazov, Ivan BledDest Androsov, Mike MikeMirzayanov Mirzayanov and me.

Good luck to all participants!

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 KrK 7 225
2 isaf27 7 231
3 BigBag 7 325
4 Motarack 7 327
5 TangentDay 7 331

Congratulations to the best hackers:

Rank Competitor Hack Count
1 halyavin 202:-52
2 2014CAIS01 26:-2
3 djm03178 20
4 bitcoin 19
5 antguz 25:-17
549 successful hacks and 525 unsuccessful hacks were made in total!

And finally people who were the first to solve each problem:

Problem Competitor Penalty
A tzuyu_chou 0:01
B DoomzGay 0:05
C 562225807 0:08
D teja349 0:12
E eddy1021 0:18
F nhho 0:45
G AChen142857 0:14

UPD: Editorial is out

  • Vote: I like it
  • +213
  • Vote: I do not like it

Comments Comments (190)
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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Good luck to all!

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

Starts at unusual time :)

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +39 Vote: I do not like it

    I can finally participate in a contest without staying up late as a Chinese.

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

Welcome back BledDest!

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

During 20:00-20:05(BJS), I have my left hand on a computer that open Codeforces, and my right hand on a computer that open Atcoder.

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    7 years ago, # ^ |
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    There is a hacking phase too in this round so we will be very busy around 1205(UTC) lol

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

    Also if there something happens, they will have to increase contest time. Which is unfortunate from every angle.

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7 years ago, # |
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Sometimes only for unusual time many regular participants can not participate...!!!

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

Thank MikeMirzayanov for codeforces and polygon platform. :)

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

Tester halyavin haven't registered yet)

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7 years ago, # |
  Vote: I like it -46 Vote: I do not like it

Is it rated for Div. 3 then?

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Finally, a contest after more than a week of non-practicing :D

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7 years ago, # |
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Contests with hacking phase make me nervous before even participating.

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

The one in which solution can be hacked on some base cases rather than edge cases :: Educational round.
So, think 4 times before submitting

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    7 years ago, # ^ |
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    the "education" in educational rounds stands for that.

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    7 years ago, # ^ |
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    If I think 4 times, contest would already be over :p

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7 years ago, # |
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Done with semesters finally!! Wishing higher ratings to everyone including myself :P

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7 years ago, # |
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For me, it is first time contest starts in reasonable time zone. (19:05 KST) I was struggle on midnight-beginning contest heretofore :)

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7 years ago, # |
← Rev. 3   Vote: I like it +10 Vote: I do not like it

float ship; // must be a float otherwise ship sinks

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

I couldn't help but notice that people with rating >= 2100 are not marked as out of competition.

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

    "Educational rounds are rated for users with rating below 2100, but it doesn't mean other users are unofficial. I think, In those rounds all users are welcome to participate officialy."

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7 years ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

This time is friendly for Chinese!

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

    That's true. Also it is for all people living in the Far East)

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

    I agree with you, but there is a spelling mistake.. Think of "int mian" "inlcude" "isotream" and so on.

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      7 years ago, # ^ |
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      oops...I didn't notice that spelling mistake.

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7 years ago, # |
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Can someone explain what are the extended ACM ICPC rules? In what do they differ from the regular ACM ICPC?

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

    The differences are the hacking phase afterwards and full retest on the final set of tests with the hacks included.

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7 years ago, # |
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For problem D.. the only "NO" testcase is n > 1 and b > 1 .. is this true ?

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    7 years ago, # ^ |
    ← Rev. 4   Vote: I like it +36 Vote: I do not like it

    I think answer is NO for n = 2 and a=1, b=1, also n=3, a=1,b=1, and min(a,b)>1.

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    7 years ago, # ^ |
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    I think its NO when neither of them is 1 or if both of them are 1 (except when n=1) The rule here seems to be that either of G and H will be connected (either of a and b must be 1) ^Not sure if it's entirely true though

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      7 years ago, # ^ |
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      Both of them can be connected at the same time also. (Consider the graph 1--2--3--4 with 4 vertices and 3 edges)

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7 years ago, # |
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I wonder what's the matter with problem D?

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    7 years ago, # ^ |
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    I forgot when the a==1 and b==1... how foolish I was!

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7 years ago, # |
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Give me some strong test case for problem C

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7 years ago, # |
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What's the 3rd test case for problem D?

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7 years ago, # |
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Can anyone tell me how many points will I get for successful hack and how many I will lose for an unsuccessful one?

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

    none. in educational rounds there is no points for hacking

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

      Then I don't get what's the point of giving 12 hours for the hack, if there is no point for hacking, I mean at last they will check each code on their test code then why 12 hours for hacking. @Infinite_M can you please explain. I am actually a beginer here.

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

        There is a point. You can increase your rank if you hack solutions of people ranked above you ;)

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        7 years ago, # ^ |
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        All successful hacks will be added to system tests. existing test cases don't guarantee that program is correct.

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          7 years ago, # ^ |
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          [user:killer_good] Can you please clarify this point? I was in an impression that codeforces provide some basic test cases to check code during contest.But they do have a test file including all corner test cases(like other websites , codechef,etc) after the contest they run every code through these test files(including corner test cases). is it true?? Or They just prepare base test cases on which they check submissions during contest and let participants find the corner test cases during hack??I mean don't they have test files for possible corner cases?

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

            for normal contests what you said is true. but for educational round system test will consist of successful hacks from users that's why we have a separate hacking phase unlike other contests.

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7 years ago, # |
← Rev. 2   Vote: I like it +74 Vote: I do not like it

User Fortza_Gabi_Tulba submitted F(39097227) at 00:28, and submitted E(39097570) at 00:29.

Since it is (almost) impossible for human to solve and code problem F in only one minute,

does it mean he submitted problem F while coding problem E?

Or he coded E and F, and then submitted F and E?

Or, is he a genius?

same thing for problem B(39091953) and D(39092747)...

I doubt if Fortza_Gabi_Tulba coded with friends...

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

    Тhe style of these codes is also different.

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    7 years ago, # ^ |
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    Some coders such as tourist use this strategy sometimes.

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      7 years ago, # ^ |
        Vote: I like it +35 Vote: I do not like it

      tourist does not do this in codeforces. It only makes sense to do this in judges where the penalty is calculated by the time of the last submission, which is not the case of codeforces where the penalty is calculated as the sum of times of all accepted submissions.

      So here it is better to submit a problem as soon as you coded it, and not doing that is suspicious.

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

    Fortza_Gabi_Tulba is a joke account.... Gabi Tulba is purple on codeforces. Take the joke

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7 years ago, # |
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Bloody build stupid graph for a=1,b=1 and n>=1,damm i should have check more carefully for problem D :(.

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

Test number 31 on problem F is not correct. It has n = 1, while this is not possible. According to the statement m >= 1 and you can not have non-zero m for n = 1

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

    Yes, sorry, there was a mistake in the statement. Luckily, no one got problems on that test during the contest.

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7 years ago, # |
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test case 10 in C?

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    7 years ago, # ^ |
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    Just open any solution which has reached to test case 10.

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      7 years ago, # ^ |
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      It is almost impossible to understand from someone's solution what I am failing at as my logic can be different than to the referred code.

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        7 years ago, # ^ |
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        When you open a solution you can see test cases it has reached:

        in this case

        Perhaps that will help.

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

        If you open a solution, you can view all of the test cases that solution has reached. Just scroll down. That's what Emil_PhysMath meant.

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    7 years ago, # ^ |
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    You must not include cases such as )( or )((() in your answer

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7 years ago, # |
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First contest during travelling ^_^

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

I was suffered from the word in problem A "commentary delegation burle demolish blahblah" I should study english more :(

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

How to solve E?

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

    Just precalculate for each street index the index just lower or equal where you can place the lamp-post. Then iterate over all possible values of powers and compute for all values the mincost by directly jumping to the place where you place next lamp-post and take minimum over all these. 39119701

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      7 years ago, # ^ |
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      Why O(nlogn)? I can prove only O(nsqrt(n)).

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

        Proving that for power i it needs O(n/i) should be enough. Let us say that we are computing for power i and length n. Now, say there is a lamppost at position l and the next lamppost be placed at r now even next lamppost must be placed at least l+i, clearly except in the case that it is not possible to illuminate the street. So there should be atmost about ~(2*n)/i iterations.So, the complexity should be O(k+nlogn). Please correct me if something is wrong.

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

    First check the maximum count of consecutive blocked positions, if this count (let this count's name be maxcon) is >=k or if the first position (0) is blocked then the answer is -1. Then search for the cheapest power starting from maxcon+1 to k. For power i, you will calculate its cost by putting its lamps greedily starting from j=0 and moving with j+=i, but if you reached some blocked j you should change j to the value of the last unblocked position before j (as if you were putting this lamp in the last unblocked position before j not in j itself).

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      7 years ago, # ^ |
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      For the 1st test case,

      6 2 3

      1 3

      1 2 3

      Can't we use the lamp with power 1 to light the whole street?

      If we put it at position 0 , it will cover [0,1] . then at 2 covering [2,3] , then at 4 covering [4,5] and at 5 covering [5,6]

      That way the total cost will be 4, which is cheaper.

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

        You need also to cover the segments [1, 2] and [3, 4]. There was an announcement regarding this during the contest which was:

        "You have to illuminate the whole segment, not only its integer coordinates. For example, if n = 3, illuminating [0, 1] and [2, 3] is not enough."

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7 years ago, # |
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can I get the rating in this round?

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

Does anyone know solution for G? My solution keeps getting TLE on test case 95, and I think mine has O(n log^2 n) time complexity and I can't think of any better solution. I used centroid decomposition

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    7 years ago, # ^ |
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    This should work, because there are only log(ai) different values of gcd, so complexity will be n*log(ai)^2.

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      7 years ago, # ^ |
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      Thank you. I was thinking too complicated solution, you are very smart

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      7 years ago, # ^ |
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      can you please explain the idea a little bit.

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      7 years ago, # ^ |
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      Intuitively I felt that there should only be around log(ai) different values of gcd, but I can't think of a good formal argument for why that is true. Do you have any good proof for it?

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      7 years ago, # ^ |
      ← Rev. 2   Vote: I like it +8 Vote: I do not like it

      Hacked huh. There are actually O(divisors) gcd values for paths (think about a leaf for every divisor and the rest of the tree has the same value). What you were thinking about is another thing, there are at most O(log) CHANGES of gcd in a path.

      So your solution is actually O(N * maxDivisors(2e5)^2) along with bad constant from using map (mostly chache misses I guess).

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

        It works, just forgot to clear maps after using ;w;

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          7 years ago, # ^ |
          ← Rev. 5   Vote: I like it +16 Vote: I do not like it

          http://codeforces.net/contest/990/submission/39135974 I submited the exact same solution that got hacked and it passed in the same time as before the hack hmmm.... Looks like that hack didn't get into systests lol

          Edit: we can see the hack now http://codeforces.net/contest/990/hacks/458169/test. I tried to test it locally and the second code crashed in my computer (probably couldn't handle the memory lol). Maybe someone could add this test to the test set? This is a intuitive solution that has already appeared in codeforces but with downward paths only iirc, I wouldn't be surprised if a similar solution passed systests just because it didn't got hacked.

          Edit2: I tested it in a custom invocation and the second submission passes in a bit under 4 seconds (ignoring input) and without the clear it's MLE. Can someone prove a lower bound than what I said for this solution?

          Edit3: the hacker that did this did most hacks for G http://codeforces.net/contest/990/hacks?chosenProblemIndex=G and the test didn't get added. Hopefully he didn't miss anyone :P

          Edit4: this case was added!

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

        Hey, can you explain your approach?

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          7 years ago, # ^ |
            Vote: I like it +6 Vote: I do not like it

          For each i, compute h(i) = number of paths that have value multiple of i. For each i, dp[i] = number of paths that have value exactly i = h(i) — dp[i * 2] — dp[i * 3] — dp[i * 4] — dp[i * 5] — ...

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7 years ago, # |
← Rev. 6   Vote: I like it -34 Vote: I do not like it

I was getting wrong answer in D coz I printed
YES
0 1 0 0 0
1 0 1 0 0
0 1 0 1 0
0 0 1 0 1
0 0 0 1 0
instead of
YES
01000
10100
01010
00101
00010

basically the same result with spaces. Shouldn't that be allowed to pass?

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

what I did in C was ignored all cases of brackets which gave )( this sort of thing, and kept a count of regular expression brackets, then if suppose (())), then this one has 1 closing bracket more, hence it will require any bracket sequence which has 1 opening bracket (such as (, (() or many like this)

I iterated for every sequence if it was a regular one, I added the total number of regular ones in the list given. If it was something like (())), which means it requires one open one, hence I added number of sequences with one open bracket, and when I came to the bracket with one open required, I ignored, hence I only calculated for the regular and closed ones, as open would be covered while doing for closed ones.

Can anyone suggest me what am missing?

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    7 years ago, # ^ |
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    I don't understand your solution

    what I did in C was ignored all cases of brackets which gave )( this sort of thing Do you ignore "()( )( "? Then your program won't count ()( )( + )( )() as a regular bracket sequence.

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    7 years ago, # ^ |
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    At last I found it. Your program gives wrong answer to this test case. I hope it will help:)

    input
    right answer
    your programs output
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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I saw one of the test case of Problem D is 1 1 1, which means a graph with just one node.

The correct answer is "NO", can anyone explain why there is no answer when n = 1?

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7 years ago, # |
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Remember to use long long everywhere, even though you think that it can't be over 2^31. Especially when you can't find your mistake anywhere.

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    7 years ago, # ^ |
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    That is my strategy!!!

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    7 years ago, # ^ |
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    Once this slowed my solution down to make it get TLE though so it's not a strictly good idea

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      7 years ago, # ^ |
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      Yeah you are right. But once on a USACO contest I got WA. Then I changed only one variable(I thought only it could cause a WA) to long long and again got WA. After the contest I read the editorial and the solution was suspiciously similar to mine. I changed all variables and got AC:(

      After that changing int to long long is the first(or maybe the second) thing I do after getting WA.

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7 years ago, # |
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What is the solution for D?

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

    Connecting vertex to all others means that this vertex will make isolated component in graph's complement. So if b > 1, you need b - 1 vertex connected with other vertexes in main graph. If a > 1, you need complete graph on n - a + 1 vertexes, and a - 1 isolated ones.

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      7 years ago, # ^ |
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      For the second case I think you meant complete graph on n-a+1 and a-1 isolated ones.

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7 years ago, # |
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Can one hack this solution with making an anti-hash test? http://codeforces.net/contest/990/submission/39097287

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    7 years ago, # ^ |
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    Yes.

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    7 years ago, # ^ |
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    Probably not because the constraint on the input is a[i] <= 1e6 so there aren't many numbers to create collisions.

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    7 years ago, # ^ |
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    Unordered map doesn't use a fixed hash fucntion so it's "improbable" to find a hack.

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      7 years ago, # ^ |
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      I once got several WAs in a problem because of a test designed to break unordered_map hash. Isn't it the same for hacks? Or does testing behaves differently using a fixed seed?

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7 years ago, # |
  Vote: I like it -24 Vote: I do not like it

awoo For the Problem B, I submitted this in C++ 11, it gave me a wrong answer. Then I submitted the same code in C++14 this and got accepted.

Because of wrong submission I had to face a penalty. I think this is wrong and it should be removed.

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    7 years ago, # ^ |
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    The code has an uninitialized variable. In test 3, there's nothing assigned to the variable 'got' in the binary search code. You can initialize it with a specific value and see what happens.

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

    you're right since your code has a mistake 2nd submission should be ignored.

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

      I don't think any action would be taken by MikeMirzayanov and CF admins. I hope they take decision in my favour, if they do something. Well I enjoyed the contest, though I didn't perform well. I think next round would be favorable.

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

Are there any editorial now?

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7 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Why this 39120549 fails while this 39119719 passes for problem C?

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

    I don't think op[N]={} initializes an array with zeroes, but I think op[N]={0} do.

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

      Even that failed. 39127177 . I am unable to understand the Diagnostics. Can someone check the Diagonostics of the above solution and explain to me. Thanks.

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

      it actually does. Not an undefined behaviour.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +20 Vote: I do not like it

    You are writing in an empty string (st) without changing its size, so it's undefined behavior. When stuff like this happens it's usually because you allocated the memory badly.

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

How to solve F? i figured i would put in a super source and a super sink to reduce it to a max flow prob. But complexity for solving flow vector is too bad

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +24 Vote: I do not like it

    Notice that if we add a cyclic flow on a valid flow, it's still valid. So, if the graph has cycles, we can use cyclic flow to zero an edge on the cycle, effectively removing it. Repeatedly remove edges from cycles until the graph becomes a tree, and then the problem is pretty simple.(recursively solve it from leaves) Complexity is O(n).

    I didn't solve F at contest but shortly after the contest I figured this out. Hope that I am correct.

    Update: confirmed it works

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      7 years ago, # ^ |
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      Beautiful solution, thanks! I just submitted using your idea and can confirm that this indeed works

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

        why if sum !=0 test incorrect? This test looks correct.

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

          Look at the bottom right node, it has a netflow of 9, and not -9. As for why the sum should be zero, you are flowing from one node to the other, so the net inflow must be equal to the net outflow

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

Why My solution for D is wrong??

http://codeforces.net/contest/990/submission/39122954

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

    If you scroll down, you can literally see the test together with the message "wrong answer The number of components in the complement graph should be 84, but found 1"

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7 years ago, # |
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when will the ratings be assigned regarding the educational round.

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

Awesome educational as usual, but I think two hours might have been a bit too little for seven problems. Most people didn't have the chance to work on F or G, which is a pity considering they were good problems.

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

    Yes i worked out problem E as i found it was greedy and almost wrote the code but just needed 5 mins to do some case testing.

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

I can't figure out why my solution hits the time limit. It should be O(nlog(n)). http://codeforces.net/contest/990/submission/39097135

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    7 years ago, # ^ |
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    Brother i don't code in java but as i have seen several times in codeforces comments that arrays.sort() has a worst case complexity of O(n*n).

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

    Sorting Primitive types in Java uses quick sort so it takes O(N^2) in worst case. It's actually common for some reason to always include these anti quick sort tests in codeforces problems.

    To get around these you can do two things, declare your types as wrapper class (i.e Integer instead of int) so it becomes an object so runs in O(nlogn), or shuffle the input array.

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      7 years ago, # ^ |
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      Thank you, I will try that tomorrow, I have never had that kind of problem before.

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7 years ago, # |
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How to solve E if a lamppost of power l were to cover [x - l, x + l] ??

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

    So it turn out we can shift the lamppost to its left end in this segment to cover [x, x + 2 * l] only if [x + l] was not blocked and the problem just reduces to the same version in here.

    This idea was suggested by lrvideckis :)

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

    Fck myself. I was trying to solve this problem during contest. And couldn't create solution. So I decided to solve G instead.

    Only after your comment I reread the stamenent. (((

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      7 years ago, # ^ |
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      trust me you're not alone :'( I had the same misconception about this problem :|

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

The tragedy of every educational round.. waiting for the final test and the rating change.

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

How long do i have to wait to get my rating?

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

    Until the admins of the contest wake up! We need to wait for final tests.

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

    Well it's 10:10pm here so I will simply go sleep soon and check the next morning instead of worrying about it now. :P Good night!

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

when will rating change ?

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7 years ago, # |
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Well if you cannot wait to see the final result, you may submit your code in the archive and see how it works.

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7 years ago, # |
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Can anybody explain me the strategy used to start solving Problem C or any similar question question's hint or link.

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

    One bracket sequence can be:

    1. Good.

    2. Need some left brackets ( to make it good.

    3. Need some right brackets ) to make it good.

    4. Can never be good.

    Then you can do bracket matching algorithm to determine which type one sequence is and how many left or right brackets that can make it good. You can see my code at http://codeforces.net/contest/990/submission/39098111

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

    A bracket sequence will be regular if, starting count from first character, for any position of that sequence, number of opening brackets "(" is more or equal to the number of closig bracket ")" ,and finaly number of opening bracket is equal to the number of closing bracket.

    Now , if you want to make a regular sequence by concatenating two sequences , there will be some options

    1.choose two regular sequence

    2.choose one sequence that have n more opening brackets(here n=total number of opening brackets-total number of closing bracket) .and for any position of that sequence number of opening brackets are not less than number of closing brackets.and choose another sequence that have n more closing brackets than opening brackets (here n is similar to previous one).and for any position of this sequence number of closing brackets should not be more than n+number of opening brackets. now you can concatenate these two sequences in one way , 1st+2nd will be a regular sequence .

    [Don't know how much I could explain , sorry for my bad English :) ]

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

why my rating hasn't change until now? How long does the change take after the contest finished?

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7 years ago, # |
  Vote: I like it +64 Vote: I do not like it

Why couldn't system tests start automatically?

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

It’s really annoying to wait for rating changes after completing Hacking phase in every Educational round. please @MikeMirzayanov solve this.

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

    Do you know that there is a chrome extension for codeforces? You can check the expected rate change quite precisely.

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

      The issue, system tests have not happened yet so expected rating change is not final.

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7 years ago, # |
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after round 487 there will be no contest for 5-6 days and it's not good

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7 years ago, # |
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Can anybody explain how to use generators for hacking ?

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

    It requires a code that prints out the test case to stdout. i.e. Use printf / cout to make a test case.

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7 years ago, # |
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http://codeforces.net/contest/990/submission/39138201 can anyone help me finding whats wrong with my code for F?

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

Will the ratings beupdated after today's contest?

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

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7 years ago, # |
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@ MikeMirzayanov regarding submission for the question b.microworld my submission is http://codeforces.net/contest/990/submission/39094917 it is getting correct answer(in local g++ compiler) for the test case which is showing as wrong output in the system run for one test case. this is looking very odd.

can you please look into this, and also wanted to know if any one facing same.

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    7 years ago, # ^ |
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    adedalic Roms BledDest can anyone respond to this.

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    7 years ago, # ^ |
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    ar.size() - 2 can underflow (unsigned), which results in a large value and you access the array at that value, i.e. undefined behavior

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      7 years ago, # ^ |
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      i think that case wont be arrising. please check by running , i checked it.

      test case: 1 1 1 output: 1

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        7 years ago, # ^ |
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        @yassin_ reply please??

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          7 years ago, # ^ |
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          It's undefined behavior, results can vary for different runs or compiler versions

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        7 years ago, # ^ |
        ← Rev. 2   Vote: I like it +12 Vote: I do not like it

        Test 15 as you can see is: 1 1 1 and your solution gives: -1199 what is definitely UB. It seems like it is caused by ar.size() - 2.

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          7 years ago, # ^ |
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          @adedalic I checked it using local compiler and some other online as well which are working well for that input. can you once check it once by running the code. since i dont feel some thing undefined should be occuring for this code.

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            7 years ago, # ^ |
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            who are down voting this,..please explain me whats wrong in the code.(which is working fine in other compilers )

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          7 years ago, # ^ |
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          yassin_ adedalic

          so you guys are saying problem arise at ~~~~**~** while(n1>=0&&ar[n2]-ar[n1]<=k&&ar[n1]!=ar[n2]) ~~~~~ line 30 of code where compiler runs differently for this line. is it??

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7 years ago, # |
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when will ratings get updated???

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7 years ago, # |
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How is rating change determined here?? Pls help.

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7 years ago, # |
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I want to become purple...

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

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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

My following normal cin, cout solution for problem E timed out on test 6 as it was not able to read the input.

http://codeforces.net/contest/990/submission/39115276

whereas the following solution with ios sync passed.

http://codeforces.net/contest/990/submission/39145593

Is this justified?

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

Your crafting.oj.uz ratings are also updated!

We are seeking for a way to somehow differentiate the performance between the first and the 100th user, since currently both are the same because of the low $RATEDBOUND$s. If we just raise the bound, everybody's rating seems to be increased and this problem just occurs again.

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

Can anyone explain the logic behind the editorial of Problem C?

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