Educational Codeforces Round 62 [Rated for Div. 2]
By awoo, history, 6 years ago, translation, In English

Hello Codeforces!

On Mar/22/2019 18:05 (Moscow time) Educational Codeforces Round 62 (Rated for Div. 2) 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 extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

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

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

Good luck to all participants!

Our friends at Harbour.Space also have a message for you:

Hello Codeforces!

We want to remind you about the two fully funded scholarships we currently have available:

Master’s in Data Science Scholarship & Master’s in Robotics Scholarship

Both scholarship opportunities include: - Full coverage of the Programme’s tuition fee (€23,000 value) - 3 hours of study a day at Harbour.Space University - 4 hours of internship a day with one of our industrial partners - €12,000 euros a year (living allowance)

If you’re interested in applying for the Robotics Scholarship, apply here.

If you want to apply for the Data Science Scholarship, fill out the form below and we will contact you about the next steps.

GO TO FORM

If you need more information about either, please contact us at [email protected]

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 QDEZ604 7 250
2 dotorya 7 254
3 300iq 7 586
4 dreamoon_love_AA 6 165
5 Hazyknight 6 169

Congratulations to the best hackers:

Rank Competitor Hack Count
1 Haunted_Cpp 8
2 Tqk 4
3 Jobaidul 3
4 Abu_Musa_99 3
5 yahia 2
37 successful hacks and 149 unsuccessful hacks were made in total!

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

Problem Competitor Penalty
A Sonechko 0:01
B usertab34 0:04
C KhaleD_ 0:04
D edisonhello 0:02
E Roundgod 0:23
F QDEZ604 0:38
G 300iq 0:44

UPD: Editorial is out

  • Vote: I like it
  • +167
  • Vote: I do not like it

Comments Comments (191)
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6 years ago, # |
  Vote: I like it +12 Vote: I do not like it

It's raining contests

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

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

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

Expecting this round to be full of awesome questions with strong pretests!!!

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6 years ago, # |
  Vote: I like it -39 Vote: I do not like it

Thank you, Harbour.Space University for the Educational Rounds. Best of luck everyone :)

Happy Coding

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

    I think Codeforces community doesn't like wishes and greets...

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      6 years ago, # ^ |
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      I think so :|

      There are some idiots also, who put down vote in every comment after a bad contest.

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6 years ago, # |
← Rev. 2   Vote: I like it -37 Vote: I do not like it

A contest by vovuh after such a long time.

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

ROBOTICS a.k.a. the biggest lie and way to tell yourself “i’m good i’m good” and live your whole life in a lie. listen kids, no one cares about your utter trash and useless robots because you’re not the next einstein and the world has enough smart vacuums, smart fridges, smart washing machines etc. (which are better in every way than your “creations” could ever cease to be.)
it’s literally “santa’s real” for big kids / grown adults.

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

May the force be with us! Happy Coding :)

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6 years ago, # |
  Vote: I like it -37 Vote: I do not like it

Is it rated?

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

    its rated for you

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    6 years ago, # ^ |
      Vote: I like it -9 Vote: I do not like it

    Educational Codeforces Round 62 [Rated for Div. 2], Isn't just the title enough to answer your question ?

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

.

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

there is no "*" mark before the handles of those who are not in DIV2! So is there a change in displaying people who are participating out of contest or only I am seeing this???

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

    There is no out of contest for Educational Rounds. They are for everyone. Just, it is rated for the division 2 participants.

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

Im running back home for this contest. Wish I would make it home before start...

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

Contest has started.

My mind:donkey mode has been activated

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

I want to add some scores.^_^

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

Hope for no 502 bad gateway like there was 3 minutes ago

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6 years ago, # |
  Vote: I like it -27 Vote: I do not like it

de ce coaie

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6 years ago, # |
  Vote: I like it -12 Vote: I do not like it

I think the order of Problem C and Problem D should change.

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

Problem statements for problems A and B could be written better. Took me a lot more time than I care to admit to understand the problems.

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    6 years ago, # ^ |
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    What is the test case 2 for Problem B?

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

      Probably <<>> Ans = 2

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

      I don't know what test 2 is, but the string is good if and only if it starts with a '>' or ends with a '<'. From here on, it should be clear how to solve it.

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

        How's the answer for <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> [2nd test case, 5th one] 50? I could choose the last < repeatedly until all characters before it are deleted, then choose the first > repeatedly until all characters after it are deleted. If you do that, the answer is 1. The problem statement doesn't place a restriction on the number of times you can choose a character, it only specifies that it must exist in the string. If repeated choices are not allowed, then how does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> evaluate to 0?

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          6 years ago, # ^ |
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          You're allowed to remove characters only before you start performing operations on the string, so you can't reduce it first and then remove.

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

    True. The statement for A ruined the contest for me.

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

      After 4 minutes I went to read and solve B, then I came back to A. I don't know why, but reading the problem later helped me understand it.

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

      Yes even I felt doomed

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      6 years ago, # ^ |
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      Lmao, felt the same D:

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

Again really nice C it's always fun solving problems that needs thinking

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

What is solution for E?? What is test 5?

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

How do we solve E?

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    6 years ago, # ^ |
    ← Rev. 3   Vote: I like it +55 Vote: I do not like it

    Note that there are no odd palindrome substrings of length > 1 if and only if there are no palindrome substrings of length 3. Then, note that a substring is a palindrome of length 3 if and only if the first character equals the last character in the substring. Thus, you can decompose the array into odd and even index arrays, and the problem becomes how many ways to fill in the array such that no two consecutive elements are equal. Then, we notice that we can further decompose each array into arrays of the form "# -1 -1 -1 .... -1 -1 -1 #", where # is either nothing or a positive element. From here, figuring out the number of constructions is an old HackerRank problem, with a solution on GeeksForGeeks.

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      6 years ago, # ^ |
        Vote: I like it +37 Vote: I do not like it

      Yeah, you are right. That problem in Hackerrank was mine. :)

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

      Can u please provide some example or proof for your statement "_Note that there are no odd palindrome substrings of length > 1 if and only if there are no palindrome substrings of length 3_".

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

        Assume there is an odd palindrome of length >1. Then the middle 3 characters are a palindrome of length 3.

        Assume there is a palindrome of length 3. Then obviously there is an odd palindrome.

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      6 years ago, # ^ |
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      My approach was all the elements at odd indexes must be unique and same for even indexes, what's wrong with this approach?

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

        Well, it's not that they must all be unique, it's that they must not have consecutive elements that are equal.

        For example, the array [1, 2, 2, 1, 1] is good, even though your approach would say it is not.

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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it +16 Vote: I do not like it

    Note that "no palindrome of odd length=no palindrome of length three", then solve for odd/even indices of the array separately.

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

    You can see my dp solution here 51717191

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6 years ago, # |
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How to solve Problem B and C.

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

    You have to find first '>' from left and '<' from end. From this points you can get delete rest of them.

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

      Why? Can you please explain the question and how did you get this. Because I think I am missing something in the question.

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

        From '>' you can delete everything left to it and from '<' — right ones.So, you have delete characters until '>' is first or '<' is last

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

      don't like to criticise individuals, but whoever wrote the statement for problem B is not deserving one. It says " for each operation you have to choose some character that still remains in the string", according to this, one may think that we can chose same character any number of times and then result will either be 1 or 0. But I don't know what he/she intended to write. Disappointed from this round! Also problem A was poorly written too. Took me 25 minutes to understand what the author wants.

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

        That's what I thought in Problem B. Got WA on testcase 2. Can't understand how the problem statement can mean anything else other than the logic which gives 1 and 0 as answer.

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

        You CAN choose the same character any number of times, but the answer is NOT always 1 or 0 :)

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          6 years ago, # ^ |
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          Give a counter case

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            6 years ago, # ^ |
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            <<>>>>>>>>>>>

            Answer here is 2: delete the fist 2 characters, then keep selecting the first character however many times you need :)

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

              For <<>> if i choose 2nd character('<') then the string will become <>> as choosing < will delete its preceding character. Now i can choose the first occurence of '>' which will delete the last character. The string becomes <>. Now delete any one of the characters will make it a good string. So there's only one deletion. How the ans is 2 and not 1? Please correct me if I am wrong

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

              [copied from above, since this thread seems to have more activity] How's the answer for <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> [2nd test case, 5th one] 50? I could choose the last < repeatedly until all characters before it are deleted, then choose the first > repeatedly until all characters after it are deleted. If you do that, the answer is 1. The problem statement doesn't place a restriction on the number of times you can choose a character, it only specifies that it must exist in the string. If repeated choices are not allowed, then how does >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> evaluate to 0?

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

It seems dotorya was hacked and 0ahmad0makki0 copied dotorya submissons.

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6 years ago, # |
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What is the TC #16 of Problem E?

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6 years ago, # |
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The problem C is interesting ! can't wait the Editorial... any hint ?

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    6 years ago, # ^ |
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    If you are told the minimum beauty value of your elements, could you easily find out what the optimal elements to pick would be?

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      6 years ago, # ^ |
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      Yes ! sort the Length and pick at most $$$M$$$ songs such that the beauty of the selected songs greater than or equal to the minimum beauty i told. right ?

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

        Yes exactly, you would just pick the greatest $$$k$$$ lengths such that the beauty is greater than or equal to the minimum beauty.

        If you sort the songs according to beauty, then you can preprocess these $$$k$$$ lengths as Ari said below using a heap, and then iterate over all the possible minimum beauties and take the maximum answer.

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

          could you please explain this part of your code for problem E ...

          if(v[0] == back) {
         dp[0][0] = 0;
         dp[0][1] = 1;
          } 
 else               {
         dp[0][0] = 1;
         dp[0][1] = 0;
       }
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            6 years ago, # ^ |
              Vote: I like it +1 Vote: I do not like it

            Sure. dp[i][0] is the number of constructions for the first i elements such that the i’th element is not the same as v.back, and dp[i][1] is the number such that it IS equal to v.back.

            That is the base case, so if the first element is equal to the last, dp[0][0] will be 0 and dp[0][1] will be 1, and vice versa if they are not equal.

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      6 years ago, # ^ |
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      You can dynamically keep the sum of the k largest elements in the range you care about with a heap

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    6 years ago, # ^ |
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    Store beauty and length in a vector of pairs. Sort by beauty in decreasing order. Now travel from index 0 to n-1. If index < k , then take sum of all lengths till index and multiply it by current beauty and keep max of the answer. Store the lengths in a min-heap or a priority queue with min on top. If index>=k then check where current length is greater than min length in heap or not. If it is greater than min length in heap, delete min from heap and add current length to heap and accordingly change the sum of k top lengths. Then take the required product and keep the max.

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

    sort according to beauty in descending order . Then use priority queue to keep track for k-1 largest length songs .

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

    Sort songs in decreasing beauty, and check for each of them what elements to pick such that length of k or less songs are biggest (and those elements have beauty greater or equal than picked song)

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

A is so poorly written I spent literally 20 minutes on figuring out what they want from me.. So I it was too late to compete.. Imho almost every educational has some type of really badly written tasks and that you can't understand anything (mostly D)..

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    6 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    I actually thought D was probably one of the most nicely written problems in the set. Their definition of triangulation was a bit different than what I'm used to, but triangulation of a polygon is a very widely used thing that you should know.

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

      I said mostly D. I was pointing out on previous contests.

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    6 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    In my opinion, none of the problem statements were badly written, they presented exactly what they wanted to. The matter of fact is that if somebody finds it difficult to understand the task, he marks it as badly written. Believe me, not only is solving a task a skill that comes from practice but understanding a task quickly also comes from practice and experience. So don't blame the statements. Sometimes statements are just complex and not "badly written". Just because you are taking time to analyse them and sort everything out in your head, doesn't mean they are badly written.

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6 years ago, # |
← Rev. 3   Vote: I like it -21 Vote: I do not like it

Statement of B was confusing, will give more attention to statements in future contests.

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

    You can choose same character any number of times. But that doesn't imply the answer will either be 1 or 0.

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

    don't trust how you understood the problem always read the Note that explain samples ... I did this mistake several times but now I will always read samples note

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6 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Somehow I found A much harder than D.

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

In problem C I thought that I need to pick adjacent songs only... I was implementing logic similar to KMP and completely wasted 30 minutes lol

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    6 years ago, # ^ |
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    weird how the examples show that's clearly not the case?

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      6 years ago, # ^ |
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      That shows I was not careful enough

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        6 years ago, # ^ |
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        well at least you got accepted, unlike me

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

          Don't compare yourself to other guys, you are just you. Have fun on studying algorithms

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

          Well Bro It is mid night now in East Asia, You are from Georgia Tech so by default I think you are in Atlanta USA, solving problems in mid night sometimes will be pain in ass cause your brain is not functioning well

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

            actually on Betelgeuse, there is no time here

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    6 years ago, # ^ |
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    made the same mistake :(

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

F was very nice.

Is the solution:

My idea ??
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    6 years ago, # ^ |
      Vote: I like it +24 Vote: I do not like it

    The easiest way to maintain all components is to create a separate vertex for each x-coordinate and y-coordinate, and treat each point as an edge that connects its x-coordinate with its y-coordinate. Then the answer for each component is the product of the number of x-coordinates and the number of y-coordinates in that component.

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      6 years ago, # ^ |
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      How do you erase?

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        6 years ago, # ^ |
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        I think this is for incremental version. For dynamic version you have use the idea of dynamic connectivity combined with this.

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        6 years ago, # ^ |
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        Just like teja349 has written, I get rid of erase operations by applying divide-and-conquer on time axis (just like in classic Dynamic Connectivity Offline problem).

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6 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Just for the sake of curiosity, can problem E be solved if we also ban palindromes of even length?

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6 years ago, # |
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If I look for something good in this competition, perhaps that will be that the test case wasn't weak.

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6 years ago, # |
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Problem E?!!!

I think I understood it incorrectly, can someone please explain it here, It has poor statements.

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    6 years ago, # ^ |
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    I think it's the way to fill all -1 in the array so that there isnt any value of i such that a[i] = a[i+2]

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

    We have to calculate the total number of ways to assign values (in range[1, k]) to the -1's in the array, such that no palindrome of odd length is present in the resulting array.

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    6 years ago, # ^ |
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    Ok thanks

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6 years ago, # |
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Can't deduce the Problem B logic Correctly!! Doesn't the problem statement means that every string we have, after some number of operations will either become "<>" or "><"? So the number of removal will either be 1(in case "<>") or 0(for this case"><").

This was the logic I applied. But it seems like I'm missing something. I built this logic because there is no limit to number of operations that i can apply on any given string. So why is this logic wrong? LINK TO MY SUBMISSION Please help!!

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

    We have to remove first, then apply operations

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

    If the original string is like "<<>>", then although remove the most out "<" (or ">"), there is another inner "<>". so the correct answer is 2, not 1.

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      6 years ago, # ^ |
        Vote: I like it -32 Vote: I do not like it

      I get this now. After complete period of 2 hours. Thanks Buddy!! Poorly Written Question played with my ratings :(

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        6 years ago, # ^ |
          Vote: I like it +43 Vote: I do not like it

        What do you mean "Poorly written question"?! It literally says Before applying the operations in BOLD in the problem text. It couldn't be more clear than that.

        It triggers me to no end when people don't read the statement carefully and then they say it's unclear. It's not unclear, you just didn't read it properly.

        (╯ರ ~ ರ）╯︵ ┻━┻

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

          it wasn't unclear. i analyzed it poorly. gonna read statements more carefully. ಥ╭╮ಥ

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    6 years ago, # ^ |
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    you need to check bold statement in the problem.

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      6 years ago, # ^ |
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      Exactly, it helped me to sort out my confusion. Thanks to the author :)

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

I know we all solved D by some trial and error, guessing, etc... then how can we prove the solution? Any proof?

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

    The edge i..(i+1) is a side of the triangle, in any triangulation. I.e: it would always be an edge. So the best way to connect it to 1 as the third node, and therefore create the minimum possible value.

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      6 years ago, # ^ |
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      This assumes separate edges are in separate triangles which need not be the case.

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    6 years ago, # ^ |
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    The key idea is that vertices $$$i$$$ and $$$i+1$$$ will always share a triangle.

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

    You are going to have to use all edges in the outer polygon in a triangulation, and for each edge you have to choose one additional label, and using 1 works. So the answer is simply

    $$$ \sum_{i=2}^{n-1} i (i+1) $$$
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      6 years ago, # ^ |
        Vote: I like it +31 Vote: I do not like it

      I don't think this proof is correct. This assumes all edges are part of separate triangles. But if there is a triangle with vertices $$$i$$$, $$$i+1$$$ and $$$i+2$$$, then two consecutive edges are accounted for in the same triangle.

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        6 years ago, # ^ |
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        That's true, some of the outer edges can be used together. The triangulation I'm saying is optimal is $$$(1,2,3), (1,3,4), (1,4,5), ..., (1,n-1,n)$$$. While intuitive that this is optimal, what I wrote was not a correct argument.

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

    Actually not all solved it by trial and error. I proved it during contest time :)

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    6 years ago, # ^ |
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    Actually I tried all possible states xD 51705295

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

    it seems i have proved this solution. Suppose there isn't any diagonal from 1 in optimal triangulation. Then we can move this triangulation in clockwise and decrease the cost. So there is some diagonal from 1 to i. Let's consider triangulation of "1 to i" polygon. If there isn't any diagonal from 1 we can do how described above and so on. So, to minimaze the triangulation cost it needs that all diagonals starts in 1.

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

    This comes in somewhat late, but I haven't seen a full proof in the comments yet.

    Let the value of a segment be the product of its endpoint. Then clearly the value of a triange is greater than or equal to the value of any of its sides. Moreover, unless one of the vertices is $$$1$$$, the value of the triangle is greater than or equal to the sum of any two sides. This is because $$$abc - ab - ac$$$ factors as $$$a((b - 1)(c - 1) - 1)$$$ which is non-negative unless $$$b$$$ or $$$c$$$ is $$$1$$$.

    Now note that every side of the polygon belongs to a triangle in the triangulation. If this triangle contains no other side then its value is at least the value of that side. If it does contain another side then the value of the triangle is greater than or equal to the sum of these two sides, unless one of the vertices is $$$1$$$. By adding up it follows that the sum of the values of all the triangles is at least the sum of the values of the sides, except for the two sides that include vertex $$$1$$$ (because they may be included in a triangle that has another side of the polygon).

    Triangulation $$$(1, 2, 3), (1, 3, 4), (1, 4, 5), \dots, (1, n - 1, n)$$$ gives exactly that sum of values, so that's our answer. Also note that the exact labels on the vertices don't matter, all that is needed is that exactly one label is equal to $$$1$$$.

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

banning mkisic_ from point buff

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

I couldn't find a reason why problems D had so many AC's. None the less I found that this was a direct answer to the problem. Actually, Problem D was easier than the problem here, but just a bit of modification to the code would do the logic.

Maybe this is the way, Problem D was supposed to be solved, using that DP approach with O(n^3).

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

    That's not really true. We just greatly underestimated contestants. Personally, I thought that majority will stuck in some sort of dp, so after some discussion we decided to leave a place to a straightforward solution. In the end community surprised me in the good way.

    PS: linear is not a limit: OEIS/interpolation gives us closed formula and $$$O(1)$$$ solution.

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      6 years ago, # ^ |
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      It’s unfortunate because I would’ve been stuck in a DP type solution if I wasn’t thrown off by the fact that many people were solving it within s few minutes, so I realized it must be some type of greedy solution. If it was a contest with a frozen scoreboard I think a lot fewer people would have solved it.

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    6 years ago, # ^ |
    ← Rev. 3   Vote: I like it +29 Vote: I do not like it

    Actually you can solve problem D in O(1) time complexity.

    Let S = 2 * 3 + 3 * 4 + ... + (n-1) * n

    3S = 2 * 3 * (4-1) + 3 * 4 * (5-2) + ... + (n-1) * n * (n+1-(n-2))

    3S = 2 * 3 * 4-1 * 2 * 3 + 3 * 4 * 5-2 * 3 * 4 + ... + (n-1) * n * (n + 1)-(n-2) * (n-1) * n

    3S = (n-1) * n * (n + 1)-6

    S = ((n-1) * n * (n + 1)-6) / 3

    P/s: I don't know how to use Latex :( So the signs and numbers may be ugly

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6 years ago, # |
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Can someone explain how to solve C and the reason behind it?

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

    Okay, the main intuition behind the answer can be said that, if the beauty is fixed, then you can easily find the answer, by taking the sum of all the largest k songs (sorted by length). Using this intuition, we iterate over through all the different beauties present in the array and dynamically maintain the sum of largest k elements, using a priority queue, and updating the sum while traversing. Here is the link to my code — https://codeforces.me/contest/1140/submission/51714832

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

Hello, Div.2, my old friend

I've come to solve your tasks again...

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

Can we solve C using DP by iterating on k?

If we can't use DP , can you explain why not?

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    6 years ago, # ^ |
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    Probably not since you would have (3*10^5)^2 states

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6 years ago, # |
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check this 0ahmad0makki0 is fake account of dotorya. Same solution and also same template.

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

    this is also pointed out above, but for some strange reason all the comments are heavily downvotes.

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    6 years ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    It's not my account, and I think that my account was hacked. (Password leak or something) Now I changed my password, and hope it won't happen again.

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

    I can't believe people are upvoting this. Why would dotorya make a fake account and then compete with both of them in the same contest submitting same solutions? I don't think a legendary grandmaster would be that stupid. It must be a hacker's account, should be banned soon.

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

How to solve G?

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6 years ago, # |
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I thought the answer could be only "1" or "0",and it took me a while to understand the meaning of Before applying the operations in problem B. ;P

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6 years ago, # |
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Any hint for G?

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

    Let's divide all edges into two types: edges of the first type connect vertices having different remainders modulo $$$2$$$, and edges of the second type connect vertices having same remainders.

    First of all, let's solve an easier version of the problem: we have to minimize the number of edges of the second type in our path, and minimizing the length of the path has lower priority. Then we can build one tree where each vertex is a union of vertices $$$2x + 1$$$ and $$$2x + 2$$$ in the original graph, and use something like binary lifting or centroid decomposition on this tree. For example, if applying binary lifting, we may maintain an array $$$up[N][LOGN][2][2]$$$, where $$$up[v][k][f1][f2]$$$ is the length of the shortest path that starts in vertex $$$2v + f1$$$ of the original graph, goes $$$2^k$$$ steps up the tree, and ends in the vertex having remainder $$$f2$$$ modulo $$$2$$$ in the original graph.

    Okay, but we can't use this approach in the problem from the contest, because sometimes it is better to traverse some additional edges of the second type, so we may choose cheaper edges of the first type. To handle this, let's preprocess all edges of the second type in such a way that cost of each such edge will be equal to the shortest path between its endpoints. This can be done with a Dijkstra-like algorithm.

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      6 years ago, # ^ |
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      Thanks for the solution！

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      6 years ago, # ^ |
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      How can we made this part "To handle this, let's preprocess all edges of the second type in such a way that cost of each such edge will be equal to the shortest path between its endpoints. This can be done with a Dijkstra-like algorithm." ?

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        6 years ago, # ^ |
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        Let's create a graph with $$$n + 1$$$ vertices. $$$i$$$-th vertex will represent an edge between vertices $$$2i - 1$$$ and $$$2i$$$, and in the end, we want the distance from vertex $$$0$$$ to vertex $$$i$$$ be equal to the length of the shortest path between $$$2i - 1$$$ and $$$2i$$$.

        How can we find the shortest path between $$$2i - 1$$$ and $$$2i$$$? We either pick the edge between them, or go to some other neighbor of $$$2i - 1$$$ (let it be $$$2j - 1$$$), take the shortest path between $$$2j - 1$$$ and $$$2j$$$, and take the edge from $$$2j$$$ to $$$2i$$$.

        So, we need to create the following edges in our graph: $$$n$$$ edges going from $$$0$$$ to $$$i$$$ and having weight equal to $$$w_{2i - 1, 2i}$$$ in the original graph, and for every pair of odd neighbors $$$2i - 1$$$ and $$$2j - 1$$$ an edge going from $$$i$$$ to $$$j$$$ and having weight $$$w_{2i - 1, 2j - 1} + w_{2i, 2j}$$$.

        All that's left is to run Dijkstra from vertex $$$0$$$ in this graph, and that's how we get preprocessed weights.

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

if The CF-Predictor broken ? ( Chrome Browser)

=> Use this Website

=> About the Issue

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

Second problem ruined the whole contest for me. The problem statement must be properly written.

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6 years ago, # |
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What's the approach for F?

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

    I have the following observation but I didn't solved it.

    Consider vertex $$$(x, y)$$$ as an edge connecting $$$a_x$$$ and $$$b_y$$$. Then the answer is the sum of (number of $$$a_i$$$ $$$\times$$$ number of $$$b_j$$$) in each connected component. If we can maintain the connectivity supporting deletion efficiently, the original problem can be solved.

    Edit: I upsolved it by doing divide-and-conquer on time axis. Distribute the lifetime of each vertex to $$$\text{O}(\log n)$$$ time intervals. Then traverse the segment-tree-like structure with disjoint set union operations and restore operations.

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      6 years ago, # ^ |
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      Can you give more explanations? It's hard for me to understand your code))

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        6 years ago, # ^ |
        ← Rev. 6   Vote: I like it +20 Vote: I do not like it

        Firstly, suppose a vertex exists during $$$[2, 9]$$$. Then you can split the interval similarly to range query in segment tree, i.e. $$$[2, 9] \rightarrow [2, 3] + [4, 7] + [8, 9]$$$ (it may vary due to different implenmentations of segment tree/divide-and-conquer).

        Call the following function recursively.

        f(left, right):
  Add vertices with interval [left, right] and record every changes
  if left = right
    Output the current answer
  else
    mid := (left + right) / 2
    f(left, mid)
    f(mid + 1, right)
  Undo the stored changes

        It is easy to see that:

        1. each answer is printed in correct order

        2. all vertices have the correct state (existing or not) when each answer is printed

        3. there are almost $$$\text{O}(n \log n)$$$ operations of adding vertices

        4. consequently, there are almost $$$\text{O}(n \log n)$$$ undo operations

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          6 years ago, # ^ |
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          It's amazing. I have understood the algorithm and the code!))

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6 years ago, # |
  Vote: I like it +32 Vote: I do not like it

please update the ratings.

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6 years ago, # |
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Can someone explain their approach in solving C, I dont know why I got hold of a very wrong track in solving it, and finally it became a mess.

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

    Let $$$x = [(\text{beauty}_{k_{1}}, \text{length}_{k_{1}}), (\text{beauty}_{k_{2}}, \text{length}_{k_{2}}), ...]$$$ This array is sorted by beauty.

    Then if you pick $$$(\text{beauty}_{k_{i}}, \text{length}_{k_{i}})$$$ as minimum beautiful song, then you must pick up to $$$k-1$$$ songs from $$$[x_{i+1}, x_{i+2}, ..., x_{n}]$$$. You can use std::multiset or something to pick $$$k-1$$$ most longest songs.

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

    When selecting a song, the total length and minimum beauty matters. So let's sort the songs by decreasing beauty, and process them like that. To make sure we have maximum length, we keep a multiset of the k-1 longest songs processed so far. Let's process song i. We do ans = max(ans, max_total_length*beauty[i]. If the multiset has size less than k-1, add this song's length to it and increase max_total_length. Otherwise, if this song's length is greater than the shortest length in the multiset, remove that length, subtract it from max_total_length, and add the current like previously.

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6 years ago, # |
  Vote: I like it +26 Vote: I do not like it

Please update the rating.

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

Please update the rating.

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

Have refeshed my page hundreds of times.

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6 years ago, # |
← Rev. 2   Vote: I like it +88 Vote: I do not like it

It is really disappointing that some 2100+ rated players make fake accounts before the contest in order to participate officially. For example take a look at this account: bigbigbigbigcat111 This account was made just before the contest and ranked 30. I think this account was made by bigbigbigcat111 who is unable to participate officially because his rating is above 2100. Making fake accounts and stealing rating from other participants is unacceptable. Please remove him from the list of official participants. Upd: I noticed that he has account: bigbigcat111 and bigcat111 as well. All accounts are from China and they have similar rating.

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

    Please, someone should steal bigbigbigbigbigbigbigbigbigcat111 to stop this unethical behavior!

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

Please do not refresh the ratings... I'm sad enough...

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6 years ago, # |
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...Why hasn't it been rated yet?I've been waiting for a long time.Just be a little faster,OK?Thanks!

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    6 years ago, # ^ |
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    Probably there's alot of cheaters in this round that's why the system hasn't started testing.

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

When will the editorial come out?

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    6 years ago, # ^ |
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    Probably after the System Testing gets completed I guess

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6 years ago, # |
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What about the System Testing, when it's going to start?

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6 years ago, # |
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Why is system testing is so late?

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6 years ago, # |
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System Testing ended.

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6 years ago, # |
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problem c was very nice,thanks for the problem author!!!

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6 years ago, # |
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When the editorial will be published?

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6 years ago, # |
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Hello!Can anybody expalin why my sol for problem C fails?Seems like it must work.Thanks! https://codeforces.me/contest/1140/submission/51763978

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

    • If you use tmp.erase(x), it will remove all the elements with value "x" from the multiset.

    • You have to remove this "if", because of the following case:

    			if (ans[ans.size() - 1] < ans[ans.size() - 2]) {
					tmp.erase(l[i].second);
					tmp.insert(a);
					mus += a;
					mus -= l[i].second;
				}

    Case:

    4 2
7000 5
1 5
8000 2
70000 1

    https://codeforces.me/contest/1140/submission/51766475

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

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

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

How can we solve D? I think I should solve prefix sum... T^T I will do not go to sleep when I solve D before!!

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