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By awoo, 3 years ago, translation, In English

Hello Codeforces!

On Oct/27/2020 17:35 (Moscow time) Educational Codeforces Round 97 (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 6 or 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 the participants!

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 Um_nik 7 149
2 jiangly 7 193
3 hank55663 7 214
4 neal 7 231
5 uwi 7 250

205 successful hacks and 712 unsuccessful hacks were made in total!

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

Problem Competitor Penalty
A Drew_is_me 0:01
B Drew_is_me 0:03
C IgorI 0:03
D xb0nS 0:09
E vinfat 0:17
F SunshinePie 0:28
G tfg 0:13

UPD: Editorial is out

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| Write comment?
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3 years ago, # |
  Vote: I like it +76 Vote: I do not like it

Is it only me or other people also feel that educational rounds are harder than div2 rounds(considering they have more number of questions)?

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    3 years ago, # ^ |
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    +1

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      3 years ago, # ^ |
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      +1

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        3 years ago, # ^ |
        Rev. 3   Vote: I like it -69 Vote: I do not like it

        why are you guys getting downvotes

        EDIT: why am I getting downvotes i just wanted to know -_-

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          3 years ago, # ^ |
            Vote: I like it +115 Vote: I do not like it

          They could have simply upvoted the first post instead of spamming by adding that reply +1.

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          3 years ago, # ^ |
            Vote: I like it -51 Vote: I do not like it

          cause they could use ++ instead of +1

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    3 years ago, # ^ |
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    I know it would be controversial, coming from a newbie, however adding a point system in Educational rounds similar to Div2 rounds, but in a compressed manner, would have fared better. Experts would have focused on harder problems upto their level, while the newbie/pupils start from the easier ones and gradually peaking up...

    The current system follows the number of solved problems and time penalty. While it's not bad, but why not something similar to Div2 round point system?

    Apology in advance, if I am missing some obvious reasons.

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

      I just see this as a different kind of Div2 event.

      There are Div2-only rounds (not Div1+Div2), which follow the said scoring system. This is just another way of scoring. I don't see why Div2 can't have both formats.

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3 years ago, # |
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[cf comment sections these days: no offense ](cf)

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    3 years ago, # ^ |
    Rev. 2   Vote: I like it +27 Vote: I do not like it

    I don't know why but it turns out that the people are desperate to make my situation as the meme guy lol!

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3 years ago, # |
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One more Educ from awoo!

Hope, that Educ will be as good as last one from you)

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    3 years ago, # ^ |
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    The comment is deleleted)))

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      3 years ago, # ^ |
        Vote: I like it -26 Vote: I do not like it

      Ikr, and this actually defeats the purpose of 'comment section', it should be used to express your opinion or thoughts but those comments are often downvoted

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        3 years ago, # ^ |
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        I wrote my first comment not to be upvote. It's just my feelings

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3 years ago, # |
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can i go to 2000 rating QAQ

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3 years ago, # |
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why does people love spamming CF announcements with "is it rated?" it's so annoying & frustrating !

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3 years ago, # |
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comment deleted, figured it out

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3 years ago, # |
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Upvote if you lose some rating because of this round :(

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3 years ago, # |
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In one of the examples, the range of cans that the customer could buy was between 3 and 4. How could we sell a pack with 5 cans to him (according to explanations?

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    3 years ago, # ^ |
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    If the customer wants 3 cans 5 // 3 is 0 3 % 5 is 3 so the customer will buy 3 cans one by one which is smaller than 5/2 = 2.5

    (same thing if the customer wants 4 cans)

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3 years ago, # |
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Wtf is pretest 2 in C

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    3 years ago, # ^ |
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    If you did greedy, it wont work.

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      3 years ago, # ^ |
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      Its DP. Difficult but very nice kind of DP!

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        3 years ago, # ^ |
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        DP wasn't difficult. The difficult part was to prove that DP works.

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          3 years ago, # ^ |
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          Isnt the very notion of dp would work in every problem even if it greedy (ignoring the part that dp solution would very slow relatively)

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            3 years ago, # ^ |
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            How do you know that you can only assign it increasingly? Why isn't there a case that a number with ti=4 is placed at 2 and then one with ti=3 and then one with ti =4 snd then more values.

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              3 years ago, # ^ |
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              I agree but I think the more hard part was to come up that its dp not greedy, which I did not during contest

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              3 years ago, # ^ |
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              The proof was the constructive part. To me, constructive logic comes more easily than the DP part. The DP wasn't difficult, yes, but it took a lot of time for me to realize that it can be done by DP :(

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3 years ago, # |
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How to solve C ? My greedy solution didn't work :( .

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    3 years ago, # ^ |
    Rev. 3   Vote: I like it +20 Vote: I do not like it
    Spoiler
    Brief Explanation

    Submission

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      3 years ago, # ^ |
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      I struggled with C, but the moment I read knapsack I instantly had the solution.

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      3 years ago, # ^ |
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      can you explain the base case? why you check time <= 2*n + 10

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        3 years ago, # ^ |
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        since the cooking time lies between 1 to n (inclusive); so for every T > 2*n you will have difference of at least 'n' for each dish . which will make the answer worse .. so there is no point in going ahead of 2*n .

        PS:- he has kept additional 10 minutes for his convenience .

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

    it's dp. look at constraints, they gives hint in itself.

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    3 years ago, # ^ |
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    You can solve using Dynamic Programming . Sort the given array . The corresponding array of times will be also sorted and will be of size n .

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    3 years ago, # ^ |
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    $$$dp[i][j]$$$ — min answer having placed the first $$$i$$$ dishes in the first $$$j$$$ minutes.

    First lets sort the dishes in non-decreasing order of $$$t_{i}$$$ so we can iterate on time in ascending order.

    $$$dp[0][0]$$$ is clearly $$$0$$$.

    Clearly at any state (where $$$i \lt n$$$) we can place the dish now, in which case $$$dp[i + 1][j + 1]$$$ = $$$min$$$($$$dp[i + 1][j + 1]$$$, $$$dp[i][j]$$$ + unpleasantness), where unpleasantness is of removing the $$$i$$$-th dish at the instant j or $$$abs(j - t_{i})$$$.

    Or we can just not place it, in which case, $$$dp[i][j + 1]$$$ = min($$$dp[i][j + 1]$$$, $$$dp[i][j]$$$)

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

      Apologies in advance if my comment seems stupid. I used a similar logic as yours. However, my dp[i][j] represented the optimal cost if time taken is i and the no. of objects picked are j, i represented the time and j represented the number of elements picked. so my dp transitions were dp(i,j)=min(dp[i-1][j],dp[i-1][j-1]+mininum of abs(A[k]-i) provided k is permitted. I have stored permitted ks in a vector of unordered_maps. However, I am getting WA. Can you please explain where my approach goes wrong. Thanx in advance.

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      3 years ago, # ^ |
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      I believe the 2nd equation should be dp[i+1][j+1] = min(dp[i+1][j+1], dp[i+1][j])

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        3 years ago, # ^ |
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        Oops, yeah it should have been $$$dp[i][j + 1]$$$ = min($$$dp[i][j + 1]$$$, $$$dp[i][j]$$$), since we don't place a dish in this step. Thanks for pointing that out.

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

      How do you get this good of a intuition with DP? It seems so obvious when someone tells use this DP and during the contest I cannot think of any transitions or states.

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        3 years ago, # ^ |
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        once you master Recursion.

        then just by adding few lines of code you can code TopDown Dynamic programming(recursion + storing answers at every recursion call).

        and if you starts with base case and use recursion logic, like how the solution builds up in call stack, you will be able to figure out BottomUp Dynamic programming.(base case + iteration)

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          3 years ago, # ^ |
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          How are you making sure that no time is used twice?

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      3 years ago, # ^ |
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      can we solve it using dp but without sorting the dishes?? i think this question would become like max biparate matching with one end as time and other as dishes.

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3 years ago, # |
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.

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    3 years ago, # ^ |
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    Dynamic Programming!

    Sort the intitial oven times

    Try giving time t to current index i note their absolute difference move forward for the remaining suffix and find the minimum sum of absolute difference of timings it can be done using bottom up dp easily.

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    3 years ago, # ^ |
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3 years ago, # |
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How to solve B ?

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    3 years ago, # ^ |
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    .

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      3 years ago, # ^ |
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      But why does it work?

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        3 years ago, # ^ |
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        .

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          3 years ago, # ^ |
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          fxcking greens with no reasons but able to solve B miraculusly fxck

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            3 years ago, # ^ |
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            .

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            3 years ago, # ^ |
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            This was not "miracle". I solved the same way but the intuition is tough and even tougher to put it in words. I am not even interested to try explain that and that does not imply somebody solved it with magic.

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            3 years ago, # ^ |
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            .

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      3 years ago, # ^ |
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      What is "group of contiguous unordered elements" here?

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        3 years ago, # ^ |
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        .

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          3 years ago, # ^ |
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          The code is clear, but I still got no clue why this gives answer.

          In this 96877997 there is an explanation based on pairs "00" and "11", and the number of such pairs. Somehow that relates to your explanation, does it?

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            3 years ago, # ^ |
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            .

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              3 years ago, # ^ |
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              Ok, thanks!

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                3 years ago, # ^ |
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                I solved it in a similar manner, but ig the approach was kinda more logical..

                We can assuredly say, that any set of consecutive characters cannot be allowed.. Also, we can argue that k 1s are incorrectly placed only when k 0s are incorrectly placed as well.

                So, we just have to count the set of incorrect occurences. And it would require half steps to correct them.

                One of the ways of doinng so can be to count the consecutive set of characters+(bool)(s[0]==s[n-1]).(Since in an even lengthed sequence, they cannot be same..) You can check my submission here for reference-96899666

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

            Let me try. Let's divide the string into : S = "A" + "B" + "C".
            A = alternating
            B = not alternating
            C = rest of the string
            Here B can either be a series of 1s(111..) or 0s(000...). We are going to pick a substring D = "B" + "some part of C" and reverse that. B = 111 then D should end at some position i such that s[i] = 0. you keep the first 1 of B, so "111" becomes "1 + 010... + 11". And we are going to do it again and again. So if you analyze at the end we will have
            S = "X" + "Y" + "Z" where,
            X = alternating
            Y = series of 1s/0s
            Z = series of 0s/1s (opposite of Y)
            And at the end when you are only left with Y and Z, it will take 1 operation for each 0s/1s.

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

          @Uvaish_Zafri

          Amazing Brother, positive and helpful people like you are very rare in comments, thanks for the explanation <3

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    3 years ago, # ^ |
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    Another way is to count number of consecutive ones. Let this be K. Therefore, add K — 1 to the answer. Do same for zeroes. Answer is max of answer for one and answer of zero. 96876793

    This is mainly because when we solve for 111000, we see that we actually need 2 moves.

    First move => 100110.
    Second move => 101010.

    Try for higher strings like 11110000, you will notice the answer is always K — 1 for K consecutive ones.

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3 years ago, # |
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Am I the only one who misread problem D and tried to solve thinking it was DFS for most of the contest T_T

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

    I found D way easier than B (maybe because I'm more familiar with graph theory) Basically just give the nodes greedly and maintain the height (because if you try to store the tree you'll mle)

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      3 years ago, # ^ |
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      I think C was easier than B and D was tougher than C.

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        3 years ago, # ^ |
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        I feel complete opposite. B seemed pretty trivial to me. D was standard. I still can not understand C even after reading the comments. Maybe C was easy for those who are decent in dp.

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          3 years ago, # ^ |
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          yeah C is trivial dp. if you are familiar with dp then it's just read and implement but i read it very late when unable to solve D, then tried B, and finally read C.

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      3 years ago, # ^ |
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      To clarify, I didn't think the solution was DFS, I read BFS in the statement then proceed to solve it if the process was DFS

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      3 years ago, # ^ |
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      I second this

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3 years ago, # |
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how to do B ??

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    3 years ago, # ^ |
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    just count consecutive zeroes and ones and ans is max of both ,I learned the hard way :(

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3 years ago, # |
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Isnt B much harder than D or im missing something obvious?

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3 years ago, # |
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RIP my ratings

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3 years ago, # |
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How to solve B?

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    3 years ago, # ^ |
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    Here is an intuitive solution that worked for me.

    For each consecutive pair of 1s, u have to insert a 0 between them and u will always have a "smart way" to do that in a single move(considering the fact that we have equal zeroes and ones). The same holds for pair of zeroes as well.

    Doing it for a single pair of 1s, also do the job for a single pair of 0s. So count the consecutive 0s(00)-> cnt0 and 1s(11) -> cnt1, then max(cnt0, cnt1) will be the answer.

    Hope it helps.

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    3 years ago, # ^ |
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    At this kinds of problems, you usually want to search for alternative statements that are more helpful. We can see that we could reframe it as: given a string s consisting of n/2 0's and n/2 1's, lets define f(s) = the number of i for which s[i] = s[i + 1], 1 <= i < n.

    To make s alternating, we want to make f(s)=0.

    A reverse operation changes f(s) with at best 2. So f(s) can be decreased with at most 2 in one operation. A lower bound for the answer to this problem is ceil(f(s)/2).

    Now we only have to prove that you can always decrease f(s) by 2, excepting the case where f(s) = 1.

    if f(s) > 1, then we will always have at least a pair of consecutive 0's and at least a pair of consecutive 1's. So we can just reverse the substring cotaining one of this 1's and one of this 0's and decrease f(s) by 2.

    For example s = 01001011,f(s) = 2. we have the 0 pair (3,4) and the 1 pair (7,8). So if we reverse (4,7) we get 01010101.

    If f(s) = 1, than the string will have to be either 0101..0110101...1010 or 1010....1001...0101. So you can reverse a prefix/suffix and decrease f(s) by 1, arriving at f(s)=0.

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      3 years ago, # ^ |
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      Thank you so much for such a detailed explanation!

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3 years ago, # |
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Guys, where can I hack? I don't see any locks..

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    3 years ago, # ^ |
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    Educational rounds do not have the same hacking feature as div2 or div1. You can hack after the contest, but you'll not get any points for that.

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      3 years ago, # ^ |
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      So it was incredibly stupid to hack myself lmao.

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3 years ago, # |
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how to solve G? is nsqrtnlogn supposed to pass? (I had tle on test 17)

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    3 years ago, # ^ |
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    nlogn is intended, nlog^2 or nsqrt can pass if written carefully enough but I highly doubt sqrtlog can pass.

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    3 years ago, # ^ |
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    Just make Aho-Corasic, and calculate link2(v) = link(v) but only with ends of strings. Now u just go left-to-right in some string of query. After adding some char, you have node V, u must check every suffix, that's end of some string, of this node. U can do it with link2(v). It works O(q * sqrt(5*10^5)), bcs u have only sqrt(5*10^5) different lengths of strings;

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      3 years ago, # ^ |
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      How the O(sqrt(5e5)) comes?Can you explain further. My solution with HLD is O(log^2 n).

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        3 years ago, # ^ |
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        you have maximum fail depth when calculating query value only if size of n strings must be strictly increasing.

        For example : a, aa, aaa, aaaa, aaaaa ....

        so n is sqrt(5e5) in worst testcase

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          3 years ago, # ^ |
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          Oh, I didn't notice this during the contest. So I built the FAIL pointer into a tree and then solved it by HLD+segment tree.

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3 years ago, # |
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problem 2 ~~~~~ string solve(){ // CALM DOWN : — )

ll n;
cin>>n;

string s;
cin>>s;

ll l=0;
ll r=n-1;
vector<char> ans1(n);
bool t=0;

for(int i=0;i<n;i++){
    if(t){
       ans1[i]=('0');
       t=!t;
    }else {
       ans1[i]=('1');
       t=!t;
    }
}

ll tl=l;
ll tr=r;
ll last=0;
bool rev=0;

while(l<r){

    while(l<r and s[l]==ans1[tl]){
       l++;
       if(rev==0){
         tl++;
       }else tl--;
    }

    while(l<r and s[r]==ans1[tr]){
       r--;
       if(rev==0)
         tr--;
       else tr++;
    }



    if(l>=r) break;

    last++;
    swap(tl,tr);
    rev=!rev;
}


l=0;
r=n-1;
tl=l;
tr=r;
ll last1=0;
rev=0;

while(l<r){
    while(l<r and s[l]!=ans1[tl]){
       l++;
       if(rev==0){
         tl++;
       }else tl--;
    }
    while(l<r and s[r]!=ans1[tr]){
       r--;
       if(rev==0)
         tr--;
       else tr++;
    }

    if(l>=r or l>=n or r<0) break;

    last1++;
    swap(tl,tr);
    rev=!rev;
}


ret(min(last,last1));

ret(""); } ~~~~~ why this code is giving TLE .... i have created both the ans array and then used 2 pointer one pointer at left and another at right and i swap the pointers please if any one could help

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    3 years ago, # ^ |
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    Please put code in spoiler tags

    like this
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    3 years ago, # ^ |
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    code looks very dirty... plz polish up

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      3 years ago, # ^ |
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      can you explain how to do this question with two pointer??? thanks for help

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        3 years ago, # ^ |
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        i can't understand what your code exactly works..

        i don't solve this problem using not two pointer but greedy.

        so i recommend to read the tutorial of this problem.. main idea of that is gorgeous in low rating problem

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3 years ago, # |
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WTF is test 11 in E?

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3 years ago, # |
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Very nice problemset.

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3 years ago, # |
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Even D was much easier than B.. what the hell is answer to B ?

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    3 years ago, # ^ |
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    whats the solution of D???

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      3 years ago, # ^ |
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      D can be solved with this observation.

      Spoiler
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      3 years ago, # ^ |
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      Use queue and push the levels into the queue. When you find a smaller number, pop and push the popped level + 1 in the queue. Keep track of max level pushed. Print the max level pushed.

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    3 years ago, # ^ |
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    • We may assume the array is circular.
    • An alternating array has no monochromatic adjacent pair (like 00 or 11).
    • If the array is not alternating, there are always two adjacent pairs 00 and 11. We can remove both of them in one operation.
    • Every operation removes at most two adjacent monochromatic pairs.
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      3 years ago, # ^ |
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      Man I went for this exact approach other the fact I still dont get why we assume it to be a circular array ? Please help on this one ...

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        3 years ago, # ^ |
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        Just for convenience. Otherwise the third statement is false.

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3 years ago, # |
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What is the expected time complexity for E?

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    3 years ago, # ^ |
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    O(NlogN)

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    3 years ago, # ^ |
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    I solved it in $$$O(n log(n))$$$, if LIS is required to solve it I don't think it can be any better.

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      3 years ago, # ^ |
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      Can you explain your approach?

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        3 years ago, # ^ |
        Rev. 7   Vote: I like it +105 Vote: I do not like it

        First of all, if $$$a_{b_{i + 1}} - a_{b_{i}} \ge b_{i + 1} - b_{i}$$$ isn't true for all $$$i \lt n - 1$$$ there is no way to place numbers between them, so the answer can't exist.

        Now lets mark the closest left and right $$$b_{i}$$$ for each index. (for any position b/w $$$b_{i}$$$ and $$$b_{i + 1}$$$, the closest left one is $$$b_{i}$$$ and the closest right one is $$$b_{i + 1}$$$, so we can mark it in $$$O(n)$$$). Lets call these $$$left_{i}$$$ and $$$right_{i}$$$.

        Now clearly for the increasing condition, for any $$$i$$$, $$$a_{i} \ge a_{left_{i}} + (i - left_{i})$$$ so we can actually place some valid numbers in this range (if not we can't even place $$$a_{left_{i}} + 1, a_{left_{i}} + 2, \ldots, a_{i} - 1$$$ between $$$left_{i}$$$ and $$$a_{i}$$$ while keeping it increasing). Similarly we must also have $$$a_{i} \le a_{right_{i}} + (right_{i} - i)$$$. If either of these don't hold, we will have to update $$$a_{i}$$$. So lets set $$$marked_{i} = 1$$$.

        Now in each range $$$(b_{i}, b_{i + 1})$$$, all values with $$$marked_{i} = 0$$$ are individually valid. We want to take a maximum subset of these such that they are increasing and values can be placed between them. Now for increasing the first thing that comes to mind is longest increasing sequence, but if we just do LIS we can end up with issues in cases like this (which Sample 4 thankfully contains):

        3 0
        2 1 3
        

        Here the longest increasing sequence may give us $$$(2, 3)$$$ but this isn't actually valid, we can't place any digit between them such that it forms an increasing sequence. So we also need that for any $$$i, j, (i \lt j)$$$ we pick, $$$a_{j} - a_{i} \ge j - i$$$ must hold. This can be simplified to $$$a_{j} - j \ge a_{i} - i$$$. Or in other words, we just want to find the longest non-decreasing sequence of $$$a_{i} - i$$$ in each segment.

        Adding the sizes of all these sets together will give us the total number of values which don't need to be changed, so the answer will just be $$$n$$$ minus this total size.

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          3 years ago, # ^ |
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          Perfect!! Thanks ExplodingFreeze.

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          3 years ago, # ^ |
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          i was almost there, unfortunately i didn't knew nlogn approach for LIS :(((.

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          3 years ago, # ^ |
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          shouldn't it be longest increasing subsequence of ai-i instead of longest non-decreasing??

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            3 years ago, # ^ |
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            nope

            1 2 3 4 5 becomes -> (1-1) (2-2) (3-3) (4-4) (5-5)

            0 0 0 0 0

            if 1 2 3 4 5 is good then we should count longest non-decreasing of ai — i

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              3 years ago, # ^ |
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              But why do we consider a[i]-=i for eg in one of the editorials it is written [2,5,3,1] we can' take 2,3 as LIS but after doing above operation it solves the problem. I am not getting the exact problem and solution by this.

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          3 years ago, # ^ |
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          Thanks! It really helped!

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          3 years ago, # ^ |
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          Thanks a lot for this explanation. During virtual I got stuck on the point where :
          while finding LIS , we must keep in mind that a[j] >= a[i] + (j-i)
          It didn't strike me that we can move that j over to the other side of the equation and started thinking it might require something like convex hull trick to work. -_-

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In problem D , for first vertex 1 i considered largest increasing sub-sequence starting from i=2 as it's children . For example if input was 1,2,3,5,6,4 . I considered 2,3 as child of 1 and 5,6 as child of 2 and 4 as child of 3 . What is wrong in this approach or in my solution

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    3 years ago, # ^ |
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    Your approach is correct Maybe you glitched your code or failed maintaining the height ?

    Here is my code which AC https://codeforces.com/contest/1437/submission/96920232

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

    For this test case-

    8 1 2 3 8 7 6 5 4

    Answer should be 3 your code is giving 1 . I might be wrong

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      3 years ago, # ^ |
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      you are right.Thanks a lot for your help . There was very silly coding mistake .

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    3 years ago, # ^ |
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    1 as root

    2 3 5 6 can all be childrens of root

    4 can be a child of 2

    so overall the height is $$$2$$$

    The easy way to describe the answer is to calculate the number of increasing sequences, and in each sequence count the number of elements in it.

    In each node, you can put an entire increasing sequence as the children of it. and now the number of nodes for the next level increases by the number of elements in the increasing sequence.

    for example this: $$$1, 7, 8, 5, 6, 2, 3, 4$$$

    $$$[7, 8]$$$ $$$[5, 6]$$$ $$$[2, 3, 4]$$$

    so $$$[5, 6]$$$ will be a child of $$$7$$$ , and $$$[2, 3, 4]$$$ will be a child of $$$8$$$. and now for the next level you have $$$5$$$ nodes namely $$$[5, 6, 2, 3, 4]$$$.

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      3 years ago, # ^ |
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      i did same can but it is giving wrong on 273rd input in test case 2. Can you check.96912116

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        3 years ago, # ^ |
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        Your code break on this such case:

        $$$n = 10$$$

        $$$1, 2, 3, 6, 5, 4, 10, 9, 8, 7$$$

        Your code output $$$4$$$, while the correct answer is $$$3$$$

        How the answer is $$$3$$$?

        well at depth $$$0$$$ you have $$$1$$$

        and at depth $$$1$$$ you have $$$2, 3, 6$$$

        and at depth $$$2$$$ you have $$$[5]$$$ under $$$2$$$, and $$$[4, 10]$$$ under $$$3$$$, and $$$[9]$$$ under $$$6$$$.

        and at depth $$$3$$$ you have $$$[8]$$$ under $$$5$$$, and $$$[7]$$$ under $$$4$$$, and you still can put many more..

        so the answer is $$$3$$$

        Hope that helps !

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3 years ago, # |
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I try to hack C but it returns this: FAIL Expected EOLN (test case 1, stdin, line 3, why?

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    3 years ago, # ^ |
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    I hacked my solution, ez.

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      3 years ago, # ^ |
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      Sorry Graphter lol.

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        3 years ago, # ^ |
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        Why are you sorry ?

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          3 years ago, # ^ |
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          I hacked cuz I thought a lot of people would have had it wrong, while it was just the 2 of us and it was thus stupid, since it probably wouldn't have been hacked lol.

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            3 years ago, # ^ |
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            Although my solution was hacked, it wasn't hacked by you. Am I missing something ?

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              3 years ago, # ^ |
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              Oh really it wasn't? I thought it was cuz we were the only ones hacked. Oops I'm seeing wrong lol. That's even worse if I only hacked myself lmao.

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            3 years ago, # ^ |
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            you haven't hacked him.

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Can someone explain how to prove the answer to B is just half the number of positions where $$$s_{i} = s_{i + 1}$$$ cyclically? It intuitively feels correct as we can fix at max 2 such positions in each operation but I have no clue how to prove that we always can.

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    3 years ago, # ^ |
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    If there are at least $$$2$$$ positions to fix, there are always indices $$$x, y$$$ such that $$$s_x = s_{x+1} = 0$$$ and $$$s_y = s_{y+1} = 1$$$.

    Proof by contradiction: suppose wlog that $$$s_i = s_{i+1} = 0$$$ is false for each $$$i$$$.
    Let $$$z_i$$$ be the position of the $$$i$$$-th $$$0$$$. Then, the inequality $$$2(j - i) \leq z_j - z_i \leq 2(j - i) + 1$$$ holds.

    But there are two positions to fix, so $$$z_{i+1} - z_i = 3$$$ should hold for two indices $$$i = a, b$$$, $$$a < b$$$.

    Then $$$z_{b+1} - z_a = (z_{a+1} - z_a) + (z_b - z_{a+1}) + (z_{b+1} - z_b) \geq 3 + 2(a + b - 1) + 3 = 2a + 2b + 4$$$.

    But $$$z_{b+1} - z_a \leq 2(a + b + 1) + 1 = 2a + 2b + 3$$$ (contradiction).

    So $$$x, y$$$ exist, and you can fix $$$2$$$ positions by reversing either $$$[x + 1, y]$$$ or $$$[y + 1, x]$$$.

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    3 years ago, # ^ |
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    My solution was also intuitive , although I didn't consider it as a cycle. I counted positions where $$$s_{i}=s_{i+1}$$$ and $$$ans = (cnt+1)/2$$$. No idea why this works either but it just seemed right and it passed. Was able to solve B in just 4 mins. Felt really good after failing terribly in the Techno Cup :)

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    3 years ago, # ^ |
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    Hey I didn't participate in the round, so don't know if this solution is correct. But maybe can help with intuition.

    Keep a count of number of chunks of 1s(c1) and number of chunks of 0s(c0). When we reach n/2 chunks each of 0s and 1s, we are done. In one operation if you reverse a sub-array starting and ending in different bits you can increase both c0 and c1 by 1 potentially (potentially coz that won't happen on the ends). It can be proved that while there are more breakable chunks of 0s and 1s there are sub-arrays ending in opposite bits that break these chunks, incrementing both c0 and c1 by 1. You may need one extra operation if you have an almost alternating sequence but with just c1 = n/2 — 1 and c0 = n/2 or vice versa. With this you can see that you will need max(n/2 — c0, n/2 — c1) reversal operations.

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3 years ago, # |
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What is the intended solution of B and how to prove it? I... I just believe...

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    3 years ago, # ^ |
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    consider the case 111011100000

    1**1101110**0000 : flip the selected part, it becomes : 1**0111011**0000

    101**110110**000 -> 101**011011**000

    10101**10110**00 -> 10101**01101**00

    1010101**1010**0 -> 1010101**0101**0

    the selected part in each step is from the first occurrence of 11 to first zero after all ones, which by flipping reduces consecutive 1s by 1 and eventually become zero. So, the answer would be no of 1s for which (a[i]==a[i-1] && a[i]==1).

    Same goes for 0 as well and minimum of those two will be the answer.

    code
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If I submit two solutions for the same problem, first one can be hacked while the second one can't be. Then if after the 12-hour hacking phase, the first one gets hacked but second one is still is accepted, will my second submission be considered for total score or not ?

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    3 years ago, # ^ |
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    Most likely yes: the first will incur time penalty, and the second, along with its time, will be counted towards the score. Generally, the submissions would be (re)judged as if all the hack cases were there from the start.

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3 years ago, # |
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Yay solved A, beware CF for when i become red

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3 years ago, # |
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nice testcases...

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3 years ago, # |
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I wish I could ask "how to solve G?" instead of "how to solve A?" :) anyway, How to sole A?

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    3 years ago, # ^ |
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    if((2*l)>r)-YES else-NO

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      3 years ago, # ^ |
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      but why?

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        3 years ago, # ^ |
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        We are basically trying to exploit the fact that for all m>n, n % m =n. Let us say l=14 and r=26

        We have to find a number x such that it gives a no larger than or equal to x/2 for all numbers from l to r, modulo x. Like 14 % x => x/2 .. 15 % x >= x/2 , up to 26 % x >= x/2 . It is clear that the number should not exceed 2*l , as on taking modulus, any number x larger than l would return l itself. Hence after x=2*l, l itself will give a remainder smaller than x/2.

        If by any means, the number r lies beyond 2*l ; we are certain there exists no solution x , as l % x < x/2 i.e. l < x/2 .

        P.S. In the above example, 28 would be an optimal answer

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    3 years ago, # ^ |
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    Lets assume that $$$l = \frac{a}{2}$$$, now $$$a=2l$$$.

    If $$$(a>r)$$$, we know that all elements in $$$(l,r)$$$ are greater than or $$$\frac{a}{2}$$$ as $$$l = \frac{a}{2}$$$, therefore whatever the customer buys, he will have to buy $$$a$$$ cans which is what we wanted so answer is YES. If $$$(a<=r)$$$, He can just buy $$$a$$$ cans and he doesn't need extra cans as $$$a$$$ lies in $$$(l,r)$$$.

    Why assume $$$l = \frac{a}{2}$$$? Because it is optimal and it gives us the biggest window.

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3 years ago, # |
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Did anyone solve problem C, greedily ?

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    3 years ago, # ^ |
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    No. Now I know trying to solve a dp problem greedily is dumb.

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The key education that I seem to get from these Educational Rounds is how to get my ass kicked. How did people go about spotting the pattern for B — tried so many approaches but none of them worked.

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    3 years ago, # ^ |
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    It did using two pointers. I'm not good in explaining or proving things. It was mainly intuition. You can see my solution . Edit: Why downvotes?

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    3 years ago, # ^ |
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    I just counted the number of consecutive 0's, and that is the answer. Because, the number of consucutive 1's/0's are present is == number of minimum swap required.

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3 years ago, # |
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Hello for question D I used a greedy approach which I have seen others use. But my code WA and I have no idea why. Can someone please look: https://codeforces.com/contest/1437/submission/96922858 ? Thank you!

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    3 years ago, # ^ |
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    For god's sake it's not a question, it's either 'problem' or 'task'

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      3 years ago, # ^ |
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      Depends on which English do you speak. In Indian English, anything that requires your mind to solve is tagged a "question" . I only came to know that "question" and "problem" have different meanings after coming to cf.

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    3 years ago, # ^ |
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    1
    17
    1 17 2 16 3 15 4 14 5 13 6 12 7 11 8 10 9
    

    Maybe check your last for-loop?

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3 years ago, # |
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Can anyone please explain me why greedy will fail on problem C.

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    3 years ago, # ^ |
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    Can you explain why greedy will work in problem C!

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In problem e : "In one operation, you may choose two integers i and x (1≤i≤n, x can be any integer) and assign ai:=x. This operation can be done only if i does not belong to the set b." in the first example :

7 2

1 2 1 1 3 5 1

3 5

So I understand that I can never apply such operation on the elements with indices 3 and 5. So a[3] = 1 for always. Why the output is not -1?

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    3 years ago, # ^ |
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    x can be negative

    4 possible moves are : (1, -1), (2, 0), (4, 2), (7, 6)

    Array after these moves : -1, 0, 1, 2, 3, 5, 6

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    3 years ago, # ^ |
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    Because you can change value to negative.

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Did someone else solve C in O(N*logN) with the solution for 713C in this blog? Thanks zscoder for such a useful trick.

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in problem A possible a always be r + 1 ? why..

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    3 years ago, # ^ |
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    If possible, we want to fit the whole range [L, R] within [A/2, A-1]

    We can include R for any A > R, but to also include L we have to extend it as left as we can, means we need smallest A > R

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Here's a more formal explanation of solution to problem B. Let's say you need to make the input string S equal to T 101010.... The procedure for converting a string equal to the other target string i.e. 010101... is quite similar. Now, let's xor our input string S with the target string T. Say this string is R. It's easy to analyse that R shall have even number of ones. Now, reversing a substring in S is equivalent to reversing the same substring in R. Also, on reversing a substring of even size, with all ones, changes all the ones to zeros. So, basically, the best strategy for us is to bring all the ones closer to each other and then use one operation to change all ones to zeros. Thus, the ans is number of blocks of consecutive ones in R

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C has a greedy tag too. How to solve it greedily?

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    3 years ago, # ^ |
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    Sorting the array before doing the dp is a greedy observation.

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      3 years ago, # ^ |
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      Oh. I see. I didn't notice. Thanks for pointing out.

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    3 years ago, # ^ |
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    Uhh I think B is veryyyyyyyyyyyyy different the one you posted

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      3 years ago, # ^ |
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      The key idea is the same in both problems.

      we will compare the actual string with 010101.. and 101010.. and check which will give the minimum result. The only difference in today's B was we can select contiguous subsequence of any length. while in the problem similar to B, we can select only one element.

      submission for today's B 96936502

      submission for similar to B https://atcoder.jp/contests/abc124/submissions/14613891

      you can check yourself.

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    3 years ago, # ^ |
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    How can you forget C. It's just knapsack.

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How to approach C with DP?

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    3 years ago, # ^ |
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    First sort all the dishes in increasing order, then define dp state as dp[i][t] = the minimum unpleasantness possible in taking out the first i dishes in that order within time t.

    There are two possibilities here:

    1. You can take out dish i at time t, in that case dp[i][t] = dp[i-1][t-1] + abs(t[i]-t).

    2. You don't do anything at time t, in that case dp[i][t] = dp[i][t-1].

    Pick the minimum of these two.

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3 years ago, # |
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https://codeforces.com/contest/1437/submission/96883973 https://codeforces.com/contest/1437/submission/96876758 please check ... those submissions are doubtful. i think, it's a Plagarism case.

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3 years ago, # |
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What is the expected time complexity for F? My solution works in O(NlogN) (only because of sorting) and I don't realy know how to solve this problem with different aproach, but N<=5000 and time limit is 4s.

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    3 years ago, # ^ |
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    That's really interesting, can you describe your solution? My solution (which is the model one for this problem) uses dynamic programming with $$$O(N^2)$$$ states.

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      3 years ago, # ^ |
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      Оно немного непонятным может быть, поэтому я лучше на русском)
      Я писал дп с N-1 состоянием, где dp[i] — количество способов расставить первых i рыбаков, если мы уже както *хорошо росставили остальных n-i рыбаков. Сразу стоит сказать что если a[n-1]*2>a[n], то ответ 0.

      Теперь, как мы пересчитываем dp[i]:
      Если мы хотим поставить a[i] на первую позицыю, то для начала надо росставить все числа от L[i]-того до (i-1)-го надо поставить на любые позицыи кроме первой(L[i] — такое минимальное j, что a[j]*2>a[i]). Количество способов ето зделать — (n-i)*(n-i+1)...*(n-L[i]-1). А теперь нужно еще росставить числа от 1 до L[i]-1, при том что мы уже *хорошо росставили тех кто от L[i] до R[i], а ето по опредилению — dp[L[i]-1].

      И еще мы можем не на самую первую позицыю ставить, но тогда количество способов ето зделать — (n-i)*dp[i-1]. Вот решение с етой идеей, но произведение я тут считал, так сказать, в тупую, так что там O(N^2) 96921154. А вот уже за линию — 96930477. Если непонятно(а скорее всего, учитывая что писал ето я — непонятно), я постараюсь ответить на вопросы.

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        3 years ago, # ^ |
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        If someone needs the translation, I can try to do that, but I don't but I don't guarantee)

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        3 years ago, # ^ |
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        Почему мы можем расставить всех от L[i] до (i-1)-го на ЛЮБЫЕ позиции? Почему это ничего не ломает в конструкции?

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          3 years ago, # ^ |
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          Когда мы считаем dp[i], мы считаем что суфикс из (n-i) рыбаков росставлен так, что бы самое левое число было хотя бы в 2 раза больше a[i].

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    3 years ago, # ^ |
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    Very cool. I haven't yet read your solution in detail, but I believe that the model solution that BledDest mentions with the $$$N^2$$$ states is this:

    Let's first sort the numbers. For every person $$$i$$$, we know that there is a certain prefix of people such that each element $$$j$$$ of the prefix has $$$2*a[j] <= a[i]$$$. These people can freely be placed once person $$$i$$$ is placed, since they are guaranteed to become sad due to $$$i$$$. Let' call this the sad prefix of $$$i$$$, and say it has length $$$pref[i]$$$.

    Let $$$dp[i][j]$$$ represent the number of (suffix of) emotional sequences which start with person $$$i$$$ and $$$j$$$ elements from the sad prefix of $$$i$$$ has already been used. That is, they must not appear in the generated sequence once again.

    The transitions can now be writted as: $$$dp[i][j] = (pref[i]-j) * dp[i][j+1] + sum[suf][j+1]$$$

    Here, $$$suf$$$ represents the least index of any element with value $$$>= 2*a[i]$$$. $$$sum$$$ maintains the suffix sum of the DP.

    This works because we can choose to select one of the available items in the sad prefix and then the remaining sequence generated will be $$$dp[i][j+1]$$$, or we can choose to continue to another number which will be happy.

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      3 years ago, # ^ |
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      Yes, the model solution does exactly the same.

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3 years ago, # |
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Is there a greedy solution for problem C? If no can you prove it?

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3 years ago, # |
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Editorial when?

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3 years ago, # |
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Who else is waiting for editorial ???

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3 years ago, # |
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Hi all I am unable to construct a case for which my code for problem D fails. Can anyone please help? Link to submission: https://codeforces.com/contest/1437/submission/96924627

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3 years ago, # |
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.

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3 years ago, # |
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B's solution was leaked on telegram . Cheaters pls go to codechef long challenge ;leave codeforces.

PS:I saw it after the contest.
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3 years ago, # |
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can someone help me in testcase 12 in problem E, I want a smaller case to see where the wrong is.

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3 years ago, # |
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Wow A and B, especially B were a lot trickier than i expected...

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3 years ago, # |
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Since nobody has said this yet — I have used Hungarian algorithm for C.

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3 years ago, # |
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Suppose we have alternate problem for A. That is, only customer who visited are L and R. then what will the solution?

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    3 years ago, # ^ |
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    I only come up with an $$$O(\sqrt n)$$$ solution.

    Note that $$$l\%a = l - \lfloor \frac l a \rfloor * a$$$, thus the constraints become $$$l \geq (\lfloor \frac l a \rfloor + \frac 1 2) a$$$

    Now we can see for those $$$a$$$ that $$$\lfloor \frac l a \rfloor$$$ are same, you only need to consider the smallest $$$a$$$.

    For a fixed $$$l$$$, $$$\lfloor \frac l a \rfloor$$$ only have $$$2\sqrt l$$$ different values.(You can find proof and the way to iterate here) So there are $$$2\sqrt l + 2\sqrt r$$$ possible $$$a$$$, just iterate through all of them and check.

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3 years ago, # |
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Drastic increase in competition on codeforces:- I just feel that in these days who have rating around 1700 would definitely be a candidate master around 8 months back, And now a days becoming a expert and retaining it becomes difficult.

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3 years ago, # |
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Could anyone tell why is this logic getting hacked for the 1st problem ? HERE

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    3 years ago, # ^ |
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    Use Double instead of Float. Some people used float and I hacked it. Tast case is

    1
    500000000 1000000000
    

    answer is NO but your solution result is YES

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3 years ago, # |
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Stupid hacking question: If I open a solution and do not do anything after looking at it, will I lose points?

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    3 years ago, # ^ |
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    No

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    3 years ago, # ^ |
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    Hacking has no effect on the standings of Educational round and Div3 rounds until and unless you the one being hacked. And you created this account 20 minutes ago just to ask this question? Man i feel sorry for you :(

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3 years ago, # |
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What makes you instantly feel that problem C is a dp problem? (Except the constraints).

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    3 years ago, # ^ |
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    Can you tell the time for a given dish without caring about other dishes? No, the time for a dish has to be adjusted keeping in mind the values of all other dishes as well. Thus, you need to use dp. In general, if you ever get confused, just check if the decision on one element is dependent upon other elements or not. If not, you can find the answer greedily, else use dp.

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    3 years ago, # ^ |
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    Because I learned 0/1 knapsack. This question is exactly similar to 0/1 knapsack.

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      3 years ago, # ^ |
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      Can you please elaborate on how did you reduce problem C to binary knapsack problem?

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        3 years ago, # ^ |
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        At first sort the array in non-decreasing order.
        Let, $$$dp[i][j]$$$ = the least possible unpleasant value till time $$$i$$$ considering 1st $$$j$$$ elements in the array.
        $$$dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1] + abs(t[j] - i))$$$
        We update dp like this in 0/1 knapsack.

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          3 years ago, # ^ |
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          Yeah, I solved this in the same way. But I was not able to relate this with the knapsack problem. Anyway, thank you!

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      3 years ago, # ^ |
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      Can i get the resource link from where u learned ?

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3 years ago, # |
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How will the graph for problem D look like for this test case :

Test Case
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3 years ago, # |
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https://codeforces.com/contest/1437/submission/96958667

can anyone suggest me a correction for my code. for A

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    3 years ago, # ^ |
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    Used more precise double and got AC

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    3 years ago, # ^ |
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    Its not necessarily to use float or double.

    Your code can look like this

    Code

    and you will get AC (96867627)

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3 years ago, # |
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96923678 1:57

96925007 1:59

I found 2 contestants with the same code

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3 years ago, # |
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When will the ratings be updated??

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3 years ago, # |
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Anyone explain how to solve E or explain the solve function in this submission 96887072?

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3 years ago, # |
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how to solve D? https://codeforces.com/contest/1437/submission/96974806 on what test case my solution fails?

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    3 years ago, # ^ |
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    You just need to make one small change. You should increase "clevel" with 1, if parent is non zero. This got accepted: 96976307

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3 years ago, # |
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Can anyone please explain the jiangly's solution for problem C 96873242? He used only one state ( dp[n] ).

Update: After the explanation from comments below, I think the idea is during the i'th itereation, dp[j] stores the minimum value to place j elements before time i. In next iteration, For updating dp[j] we only need value of dp[j-1] of (i-1)'th iteration. (And since we are going right to left dp[j-1] will contain the value corresponding to time (i-1)).

Therefore, Transitions:
dp[j] (value to put j elements till i'th time) = min (dp[j] (placed j elements during (i-1) time) , dp[j-1] + abs(t[j-1]- i))
cost of placing j'th element ( 0 based indexing in t) at i time.

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    3 years ago, # ^ |
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    He is traversing from right to left. So at any particular position, the left ones contain the dp values of time i-1 and all the others contain the answer for time i. Since we only need to consider smaller values hence it won't make any difference if the larger ones contains values for time i. If you traverse from left to right then it won't work cause your dp will have values for both i-1 and i.

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      3 years ago, # ^ |
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      I traversed from left to right with dp[n]: https://github.com/actium/cf/blob/master/1400/30/1437c.cpp

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        3 years ago, # ^ |
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        Yeah I guess in your case it will work because dpj in its ith iteration means what would be the minimum value if you place the ith element between time i to i + j. Its a bit different from that solution, there dpj in ith iteration means what would be the mimumum value if the jth element is placed at time i.

        As long as you don't mess your dp you can iterate anyway :P

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      3 years ago, # ^ |
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      Thank you kumaraditya1999.
      Actium Can you please explain your approach?

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        3 years ago, # ^ |
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           --1--2--3--4--5--6--7--> t
        2:   1  0  =  =  =  =  =
        2:   x  1  1  =  =  =  =
        4:   x  x  2  1  =  =  =
        4:   x  x  x  2  2  =  =
        4:   x  x  x  x  3  3  =
        5:   x  x  x  x  x  4  4
        

        x means time point is reserved for a previous dish, = means the unpleasant value is unchanged

        Look at how unpleasant values are being changed. Obviously you do not need to keep all rows in the dp table, and you can shift every next row 1 position left. Then you can keep only n values and maintain them.

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How to solve C using flows?

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    3 years ago, # ^ |
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    I think it can be done with something similar to "assignment problem". And it can be done with flows.

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3 years ago, # |
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I am not understanding which it is happening this or not that in first question my offline compiler"visual studio code" is giving differenet answer than that of codeforces compiler.

Can anybody help me out in this.

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3 years ago, # |
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Terrible contest

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3 years ago, # |
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Can someone explain me how to do C. Thanks:)

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deleted

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3 years ago, # |
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My unofficial editorial for this contest.

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Now i am doubting if this round is rated or not?

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can someone tell me why max time is 2*n In question C ?

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    3 years ago, # ^ |
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    Edit: I initially misread the query.

    Think of the following case.

    $$$t_1 = t_2 = ... = t_n = n$$$

    It is optimal to take out the last dish at $$$t = 2n-1$$$. If you go further, you are guaranteed to leave unnecessary empty time slots behind.

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    3 years ago, # ^ |
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    Consider the one dish with the smallest t[i] value. It is never optimal to take that one out of the oven after t[i]. Because if we would do, a better solution would be to take it out one minute earlier.

    So, consider the next smallest t[i_2]... again, it is never optimal to wait after t[i_2]. By induction follows that it is never optimal to take out the last one later than 2*n.

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      3 years ago, # ^ |
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      hmm gotcha ! btw do you suggest writing forward Dp(updating yet-to-come states on the go) or backward dp(updating previous states on the go) ?

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        3 years ago, # ^ |
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        I do whatever seems simpler to me. That depends on the problem statement.

        For complecated dp I find it most often simplest to have a function that gets all state as parameters and calls itself recursive. Then add memoization.

        Once that code is written it is often simple to write the imperative code as well.

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3 years ago, # |
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When will the tutorial be uploaded?

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is it rated??

If yes, when rating is going to updated??

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Tutorial and rating should have been updated by now.It is way beyond limit

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The problem D could be solved using Two pointers as well.In my below AC code , cnt0 is for cnt of the nodes of height=cureentheight-1 which has no child. And Cnt1 is for cnt of the nodes of height=currentheight which has no child. Here is my solution https://codeforces.com/contest/1437/submission/96985177

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where is the editorial....its too late

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This is the first time I turned blue, a memorable game! Hope everyone can have good luck and high rating!

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3 years ago, # |
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For Problem C, If there is a single dish and its cooking time is say 50, shouldn't the answer be 0 as we can just wait and take it out at 50th minute? Why is it not the answer?

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    3 years ago, # ^ |
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    how do u know its not the answer ??

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    3 years ago, # ^ |
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    The time for each dish ranges from 1 to n. Thus, if there is only one dish, the cooking time for that dish must be 1 and the case you stated is not a valid input for the problem.