By awoo, history, 2 years ago, translation, In English

Hello Codeforces!

On Nov/30/2020 17:35 (Moscow time) Educational Codeforces Round 99 (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 neal 7 223
2 jiangly 7 224
3 tute7627 6 190
4 mystery 6 195
5 noimi 6 213

Congratulations to the best hackers:

Rank Competitor Hack Count
1 MarcosK 41:-6
2 dapingguo8 34:-2
3 jerdno 12:-4
4 ARTpositive 10:-1
5 halyavin 9:-1
357 successful hacks and 769 unsuccessful hacks were made in total!

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

Problem Competitor Penalty
A edickLPM 0:00
B SSRS_ 0:03
C corol 0:03
D pajenegod 0:08
E neal 0:29
F jiangly 0:55
G jiangly 1:17

UPD: Editorial is out

 
 
 
 
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2 years ago, # |
  Vote: I like it -31 Vote: I do not like it

Hope this round will be good

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2 years ago, # |
  Vote: I like it -23 Vote: I do not like it

Another Educational Round. Hopefully I will do better like previous Educational Round. I think, The problem will be more interesting. Ops! I'm waiting for that.):

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

    I don't know why people giving me down vote! I just expressed my feelings! I really shocked about people's attitude. But I have done well today that is my satisfaction.

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      2 years ago, # ^ |
      Rev. 2   Vote: I like it +22 Vote: I do not like it

      People downvote your comment probably because they think this types of comments just fill the comment section and then one need more time to find useful content. No one wants to read 100 comment of people expressing their feelings!

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

        Thank you for knowing me. I didn't guess it before. Next time i wll keep it in my mind.

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

Super Excited !!

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

This is my chance to go up to master!!!

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

People reading this comment, hope you have a good contest. All the best!!

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

May i become specialist this time !!

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

This may be my last contest before my college final exam. After that I will open my brand new advanced mathematics book and my linear algebra book. Oh my god, brand new(╥﹏╥). I hope i can get good results all of them.

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

Maybe the last round before senior high school...
It's time to say goodbye to you now guys. Hope we'll all do our best in this round. :)

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

    Don't go :(

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

      I don't want to do that either but I have to :(
      After all, high school entrance examination is always more inportant than OI :(

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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    However, I come back now... (Eason_AC is my another account)
    That means I can enjoy the CF rounds again guys :)

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2 years ago, # |
  Vote: I like it +116 Vote: I do not like it

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

the comment section is shit

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2 years ago, # |
  Vote: I like it -31 Vote: I do not like it

Score distribution awoo ?

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    2 years ago, # ^ |
      Vote: I like it +70 Vote: I do not like it

    There is no score in Educational Codeforces Round

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

      Why? Also what is differnece between educational div. 2 and other div. 2.

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        2 years ago, # ^ |
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        In edu each problem scores 1 point, other div2 have different points per problem.

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

In the hope that I don't mess this round by silly and not required lengthy implementations.

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2 years ago, # |
  Vote: I like it -16 Vote: I do not like it

I have end semester exam tomorrow. Should I compete this round?

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

    What? So early in India? In China usually in January or February...
    Anyway, you'd better take a good rest as the end semester exam is more important than a CF round :)

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

I just pray and hope that it wouldn't be another Mathforces round.

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

When will be shown the difficulties of tasks from Codeforces Round #685 ? It's been more than a week ago!

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

I'm gonna participate so more chance to gain lost rating on sunday

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

Congratulations to the 100000000th submission!

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

Most uninteresting problem set ever

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    2 years ago, # ^ |
    Rev. 2   Vote: I like it -13 Vote: I do not like it

    nvm

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

      Also, example- Education round 99 problem C

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

        Yes, i will also agree that 'C' is uninteresting, but this guy is shitposting before solving them which is not fair. The only problem i found interesting was 'B'.

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

          can you share your solution please

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

            keep moving greedily and once you cross you destination you will come to know what step you should not have taken greedily that will be K=(finalDest-x)th step.

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

              can you explain the case for 7

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

                Keep jumping greedily till you reach a point >= x. If your current_pos-x is 0, then obviously answer is number of jumps. Now if its not, then there are two cases

                1. current_pos-x == 1
                In this case, you always need to jump back a step. So, answer is jumps + 1.
                

                The second case current_pos-x > 1 Since it is greater than 1, there is always a jump in range [1, jump], that you can decide to not perform and jump back. Example

                x = 7
                1 3 6 10
                lets stop jumping since 10>=7
                

                cur_pos = 10, jumps = 4 here, difference is 3 (10-7), so there is always a previous step,in which you decide to not jump forward and jump back. In this case, if you dont do the second jump (+2) and jump back (-1), you will end up at 7 in 4 jumps.

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

                  How would you prove the last section of your comment? Why in case we overjumped our x by 1, then we make a jump back, but when we overjumped by n greater than 1, there is ALWAYS a jump in range [1, jump] that i wanna not perform and jump back? Why if I overjumped by n greater than 1, i can't just jump back n times?

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                  2 years ago, # ^ |
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                  This is because the kth jump goes over k positions, and n-x is garanteed to be smaller than the last jump.

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

                  Ok, I understand that the distance of overjump is less than kth jump, but i don't understand why i shall do a jump back(and what's the count of them) somewhere in my jump path, instead of inserting my backjumps in the end of jump sequance?

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

                  See it that way: If you do a step back in ith step instead of jumping, then the contribution of this step back is i+1 positions to the left. (Because you did not jumped i and steped one back).

                  So if you are in the end y position to much on the right, then you kind of revert one jump. And because it should cause y positions, you choose the y-1 -th one. This works as long as y>1. If y==1 then there is no jump you could revert.

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

                  Because you want to minimize the number of steps. Changing one of the already taken jumps to -1 will cost nothing. But jumping back in the end will cost 1 move for jump.

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                  2 years ago, # ^ |
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                  Wow, great! As soon as I read your comment i firstly started to think that i misunderstood the task, then i read again the task, returned to this post and started to write that you're wrong, and in this moment i finally understood why it works. Thanx!

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

                  Consider 4 jumps [1, 2, 3, 4]. Here, the sum is 10. Now try replacing any of these by -1 and check what the sum will be. You will get all sums except 9. So you always a need an extra jump if difference is 1.

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

      Kid looks like its your first time solving D. Doesn't mean its an interesting problemset (atleast for me). Besides I left the contest long ago, and I dont judge someone's opinion based on their rating or whether they have solved B or not in a contest I like problems which are ds oriented not based on Math

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

        Looks like you got a burn?

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

          Yes burnt just like your rating graph

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

            You just literally said you dont judge anyone based on their rating and then went on to judge him on his rating graph. Hypocrite much

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

              lol what I said I don't judge someone's opinion/likeness towards a problemset on basis of his rating or whether he solved B or not lol

              He was mocking me for my comments I mocked him for his graph.

              These are unrelated. You need some humour man.

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2 years ago, # |
  Vote: I like it +100 Vote: I do not like it

Congratulations to ub33 for the 1e8-th submission!

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2 years ago, # |
  Vote: I like it -21 Vote: I do not like it

Is C's statement an english test? I can't understand it.

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

    C's statement is definitely hard to understand at first, need to google "ping-pong" after reading the statement

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      2 years ago, # ^ |
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      hm. later i solved it by guessing output from samples.

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2 years ago, # |
  Vote: I like it +40 Vote: I do not like it

Either I will live or geometry in this world !!

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

How to solve E ???

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

ez game pls dont hack me pitch

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

Use me as dislike button if you also didn't like this contest.

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

Solved C in 1 minute after the announcement ;)

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

It's the second time in a row that in a Educational Round problem C is easier than problem B, what is this :(.

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    2 years ago, # ^ |
      Vote: I like it +52 Vote: I do not like it

    ObservationForces

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

    I think C easier then A. Because during contest I solved A help with pretests but C was really apparent but for understand what it says on C i need to read multiple times maybe this becomes C harder than A.

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

    So it's a good reason to start solving C firstly the next time. I think cfpredictor is a suitable work for you :)

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

How to solve E?!

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

Sounds stupid.. But someone help me with B

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

    After you draw the steps (don't come back initially) you will realize that you can come back by not making some pth move in front, but by just making it backwards. suppose you are at point y such that y>x, you need to check if there exists some p, such that (p+1) = (y-x). this accounts for not making the pth move in front, i dont have proof, but since p can take all natural numbers, it's always possible.
    and yeas if(y-x)=1 then add one.

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

Solved Div.2 D for the first time.

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

How to solve c (please tell along with proof) ?

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    2 years ago, # ^ |
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    Ans is always (x-1,y), since the 2nd player wants to maximize he always gets y points by loosing the first x-1 games and now player 1 has only 1 stamina and uses it to serve and player 2 reflects it to get his first point and wins the rest y-1 points after that.

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

      how you can say that ? he also has to minimizes the apponent chance of winning can u please tell the proof why u r doing this little more ??

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        2 years ago, # ^ |
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        but first priority is to maximize his own points

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        2 years ago, # ^ |
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        The order is that in first place they want to win as much as possible. Only if they cannot win any more they will care for the other ones points.

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

          How the above strategy ensures that bob cannot play in such a way that Alice wins less than x-1 times and he wins y times ?

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            2 years ago, # ^ |
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            If you are not serving, its always optimal to lose the play and not lose stamina. This way you make the opponent lose stamina on the next try, since they compulsarily have to serve.

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

          what if alice doesn't return in the 1st move?

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

wasted 35 min on b then did c in 1 min then went to b for another 35 min to actually solve it. C must have been in place of A. C<=A<B<D

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

    C easier than A. Seriously?

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

      ok ,say C=A in difficulty,because, Ans is always (x-1,y), since the 2nd player wants to maximize he always gets y points by loosing the first x-1 games and now player 1 has only 1 stamina and uses it to serve and player 2 reflects it to get his first point and wins the rest y-1 points after that.

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

        I get C is easy but what did you find tough in A that you placed it after C? In A you don't even have to think anything.

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        2 years ago, # ^ |
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        in A ans is always n.size(), where n is the string in input. So A easier than C.

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

      Yes.

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2 years ago, # |
  Vote: I like it +80 Vote: I do not like it

I miss those times when Educational used to be great!

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2 years ago, # |
  Vote: I like it +18 Vote: I do not like it

Thank you for the new mathforces round !

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

Is it just me or the game thoery problems are a bit tough to understand?

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2 years ago, # |
Rev. 2   Vote: I like it +57 Vote: I do not like it

Strong dislike for C, it is completly based on observation and hence has no educational worth at all. That might be a nice problem, but simply misplaced in educational contest.

It is basically no programming problem.

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    2 years ago, # ^ |
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    Exactly. A,B,C were more about observation than implementation! -_-

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    2 years ago, # ^ |
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    The problem wasn't even interesting. Just trying to come up with possible ways of playing and hoping one of them is optimal.

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    2 years ago, # ^ |
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    Today's A,B,C were more like Atcoder's ABC contest A,B,C

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    2 years ago, # ^ |
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    Yes, C was very ambiguous for me

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    2 years ago, # ^ |
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    I won't say C was bad even though i was not able to solve it during contest . I didn't expected that observation will be so simple and was always thinking in complicated way .

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

The best boring ping pong in C.

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2 years ago, # |
  Vote: I like it +69 Vote: I do not like it

I don't like this round. Learned nothing from ABC.

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

How to solve B ?

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    2 years ago, # ^ |
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    It was observation that you calculate $$$s = 1+2+3\, ...\, + k\ge n$$$. Now see that if $$$s = n+1$$$ then answer is $$$k+1$$$, otherwise it is $$$k$$$.

    This is because if you overstep by some amount $$$o > 1$$$ and also we have $$$o < k$$$ so we can do the $$$-1$$$ step on $$$(o-1)^{th}$$$ move and $$$+j$$$ step on other $$$j^{th}$$$ moves and so effectively we have reduced the sum to $$$n$$$, so answer is $$$k$$$.

    But if $$$o$$$ is $$$1$$$ then you must step back on $$$(k+1)^{th}$$$ move, so giving total $$$k+1$$$.

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

      Can you explain me that how I reach to 7 in problem B

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        2 years ago, # ^ |
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        You can follow the sequence $$$0, 1, 0, 3, 7$$$ (step back on the second step).

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

          Oh!I understand this.Thank you so much.

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2 years ago, # |
  Vote: I like it +57 Vote: I do not like it

I might get a lot of downvotes here, but I felt that $$$C$$$ was a pathetic problem, with no learning at all.

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

What was the intended solution for problem D?? Was it DP or greedy??

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    2 years ago, # ^ |
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    greedy

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      2 years ago, # ^ |
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      I did dp, I guess there are more solutions.

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        2 years ago, # ^ |
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        can you explain the dp solution plzzz

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

          dp[index][valueofx][0/1] So you can see that you have n+1 elements in total, let's rename them 1 -> k (where k is the number of distinct elements) The solution will have n elements from your n + 1 elements. For example, if your normalized values are 1, 2, 3, 4, 4 and x = 5 (in total 6 elements), you can see that the solutions can be something like 1, 2, 3, 4, 4 or 1, 2, 4, 4, 5 etc. (you pick 5 elements). index — means the element you are at valueofx — the current value of x after you do the operations on the first "index" elements 0 / 1 — you haven't skipped / have skipped an element (which means if you are still taking the smaller n elements, or have already skipped 1, so your solution will end with the n+1th) dp holds the answer obv. Try solving yourself from here on, since it is not hard.

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          2 years ago, # ^ |
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          The states are (current position i, Max element from [0,i), current x).

          Transitions are simple, you either make the current element x or you don't. My submission

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            2 years ago, # ^ |
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            This is a way easier solution, but performs worse. But for this problem, n^3 is fine as well

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              2 years ago, # ^ |
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              Yup, I noticed the memory limit was 512 MB which was unusual from a general 256MB. So I decided to go with this since its pretty easy to implement

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                2 years ago, # ^ |
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                If it would have been 256MB, you could have only kept the last 2 lines of dp, because you only need them. And had like dp[2][500][500]

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                  2 years ago, # ^ |
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                  that would work in iterative solution right? because in recursive dp, we can't remove the state $$$pos$$$ as far as i can think.

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

                  Isn't the current state unique for (i,x)? Submission

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

    I used a greedy approach: First check whether the original array is already sorted or not. If it is sorted, print 0. Otherwise traverse the original array. Swap the value at current index with 'x' if it's greater than x and increment your answer by 1. At every step of the traversal check whether the Array has become sorted or not. The low constraints allows this. If it has become sorted print your answer. At the end of the traversal if the array is not sorted print -1.

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

    It was a greedy solution. I solved it this way:

    • Notice that $$$x$$$ is monotonically strictly increasing.
    • If at any point, for any $$$i$$$ starting from $$$1$$$ to $$$n$$$, $$$a_{i \dots n}$$$ is sorted, break out.
    • If for any $$$i$$$, $$$a_i > x$$$, do the operation. As an operation can only lower down the value of an element, so it is optimal to lower down the values when possible starting from the left.
    • Finally, at the end check if $$$a$$$ is sorted, if it is, then print the moves you took, else print $$$-1$$$.

    Update: Cleaner and faster code.

    C++ Code: 100058194

    Time Complexity: $$$O(n)$$$
    Auxiliary Space: $$$O(1)$$$

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2 years ago, # |
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Problem C was really ambiguous. Strongly dislike it.What was the point of it?

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2 years ago, # |
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Maybe my complexity analysis was wrong. But I think I probably submitted an O(n^3) solution to the problem D where the sum of n over all test cases was <=500. Am I missing something in my complexity analysis? I would be glad if someone helped out.

My submission: 100048016

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2 years ago, # |
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Got VERY stuck on B ;(.

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Interesting problem С, liked it. But, in my opinion, problem B is more difficult than problem C :(

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neal's stream is running now, feel free to join :)

https://codeforces.com/stream/71

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2 years ago, # |
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I can't believe D is just mostly implementing what they given, I overthought that so hard D:

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Someone, please help me with the logic of problem D. Thank You in advance!!

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    2 years ago, # ^ |
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    Hey, I iterated on the input array and every time check whether input array is equal to sorted input array if that's the case print answer and return else if the current element is greater than X then swap it. Here is the implementation of the same.

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2 years ago, # |
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how to solve b

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

    Find the cnt where cnt*(cnt+1)/2 is just greater than x Then find by how much it is greater than x,let it be y=(cnt*(cnt+1)/2)-x; Now just we have to remove the step y which can be done without any extra step. if y>1 so ans = cnt and if y==1 ans=cnt+1

    Let's say x=7.So cnt=4; And the guy moved 1+2+3+4 steps. Then his final position is at step 10; y=(1+2+3+4)-7=3 Only if he could move 3 steps behind then the problem will be sorted. So we just need to remove the step size 3. So lets do Step 1: +1 position =1 Step 2: -1 position =0 Step 3: +3 position =3 step 4: +4 position =7

    So we get the minimum number of steps as cnt=4

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

    For B, let's use only +i as long as sum < X.

    +1  +2  +3  +4  ...  + K  =  S  (>=X)
    
    If we switch a "+i" to "-1", then S updates to S-i-1
    That is great, because for i = 1, 2, 3.. it is -2, -3, -4 ... -K on S with no more jumps (result = K), and we could reach any S, S-2, S-3, ... except S-1, in which case we need an additional "-1" jump (result = K + 1).
    
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2 years ago, # |
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How the answer of B for n=12 is 5 ?

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How to solve B?It took me an hour and a half to solve problem B,but I did not succeed in solving。

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    2 years ago, # ^ |
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    Find the cnt where cnt*(cnt+1)/2 is just greater than x Then find by how much it is greater than x,let it be y=(cnt*(cnt+1)/2)-x; Now just we have to remove the step y which can be done without any extra step. if y>1 so ans = cnt and if y==1 ans=cnt+1

    Let's say x=7.So cnt=4; And the guy moved 1+2+3+4 steps. Then his final position is at step 10; y=(1+2+3+4)-7=3 Only if he could move 3 steps behind then the problem will be sorted. So we just need to remove the step size 3. So lets do Step 1: +1 position =1 Step 2: -1 position =0 Step 3: +3 position =3 step 4: +4 position =7

    So we get the minimum number of steps as cnt=4

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2 years ago, # |
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MarcosK Make a nice hacks for E

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    2 years ago, # ^ |
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    Doesn't it mean that someone did bad tests?

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      2 years ago, # ^ |
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      They don't need to have perfect test btw you can resubmit if you think your solution bad while accepted this means no one says pretest are perfect

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        2 years ago, # ^ |
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        What? What is your point? Do you wanna say that it's normal that in all tests $$$t \leqslant 10^3$$$ while $$$t \leqslant 10^4$$$ is in statement?

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This was my first contest ....can someone please send me solution for strange functions A question ? Pls it will be of great help ...thanks

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    2 years ago, # ^ |
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    You simply have to print the length of the input for this question

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2 years ago, # |
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The constraint of $$$E$$$ is $$$1\le t \le 10^4$$$,but the maximal $$$t$$$ for the tests are $$$1000$$$ only...

That's why there can be a lot of hacks in $$$E$$$...

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

    Out of curiosity, was the limit for $$$t$$$ originally $$$10^3$$$ and not $$$10^4$$$ awoo? I had a solution without ternary/binary search in mind, but I saw my solution passed in only 483 ms, so I assumed that it most likely wouldn't be hacked :P

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2 years ago, # |
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Is it codeforces or ObservationForces?

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2 years ago, # |
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This was a really nice round, but I have to complain about problem E. I hacked more than 40 submissions (including mine), just because the complexity is $$$O(log(10^9) * 4!)$$$, and all of them exceeded time limit because of having 10^4 subtests. I don't get the decision of the setter about it.

Besides that, thank you for the round, i loved the problems!

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    2 years ago, # ^ |
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    Theoretical computer scientist: That is O(1) XD

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    2 years ago, # ^ |
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    My solution is $$$O(4!)$$$ if we allow such notation. Initially, I didn't intend to make limits so tight and wanted to allow sub-optimal solutions but, I'm sorry, I messed up with tests.

    P.S.: it's a little strange that something that looks like $$$t \cdot \log^2$$$ doesn't pass with $$$t \le 10^4$$$.

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    2 years ago, # ^ |
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    Wait, that sounds weird. 2 sec TL, log(10^9) is ~30, 4! is 24, then for all tests we have 7200000 (7M) total. How that could be TL?

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      2 years ago, # ^ |
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      In most of the codes, the constant factor is not small. For example, some codes perform more than one binary search. My hacking method was:

      Open submission. Look for a binary/ternary search. If there is one, send a hacking attempt with $$$t=10^4$$$. Successful hacking attempt.

      Of course it didn't work in 100% of the submissions, but I did more than 40 hacks in that way :P

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        2 years ago, # ^ |
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        That’s all pretty cool, but math is math: constant factor should be very big to jump from 7M to TL. Hack is hack, good for you, just very weird.

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2 years ago, # |
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problem C was much much easier than b.

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2 years ago, # |
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any tutorial for b please

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A~D: No programming, just observation and math...

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    2 years ago, # ^ |
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    well , I solved D with DP , is there another solution?

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      2 years ago, # ^ |
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      We will iterate over the given array a from beginning. If a[i] <= x, we can't change the value. If a[i] > x and a[i] should be changed to sort the whole array, we should swap a[i] and x — if we don't, it cannot be corrected later. After each swap, I checked the array is sorted. If then I escaped the loop and printed # of swaps.

      If the array is not sorted after all swap, print -1. We can solve the problem by this greedy approach.

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        2 years ago, # ^ |
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        That's problem B, not problem D.

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          2 years ago, # ^ |
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          Oh, it's my mistake. I will edit the comment.

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2 years ago, # |
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No offense to the writers but I don't really think this contest is educational.

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Adhoc-Forces crossed 100 million+ submissions. wow!

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How to solve Problem E?

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    2 years ago, # ^ |
    Rev. 3   Vote: I like it +25 Vote: I do not like it

    Essentially you are compressing the x and y co-ordinates each to two distinct values, and the problem has many similarities to making all elements of an array equal. In the array problem, you compress all elements to be equal to the middle value, or in the case of an even-length array, one of the middle values.

    In this problem, we want to compress the co-ordinates such that x and y deltas are minimal. If we sort our starting X co-ordinates (X1,X2,X3,X4) and starting Y co-ordinates (Y1,Y2,Y3,Y4), ideally we will compress them such that all x co-ordinates are between X2 and X3 inclusive, and all Y co-ordinates will be between Y2 and Y3 inclusive. The optimal length of the square, L, will be the max(Y3-Y2,X3-X2), since we don't want to compress the square more than necessary.

    Suppose we cannot fit the whole square inside the bounding box X2 to X3 and Y2 to Y3. Then we want as much of the square as possible to be inside that bounding box, and as much of the rest of the square to be inside the bigger bounding box X1 to X4, Y1 to Y4. It can be seen quite easily by drawing an example that any shift outside of these bounding boxes is sub-optimal.

    So we can say the bottom left corner of our square is at (min(X2,X4-L),min(Y2,Y4-L)). [Note the we equivalently could have positioned the top right corner at (max(X3,X1+L),max(Y3,Y1+L))].

    Therefore we have the four co-ordinates of the optimal square. From there we can try all 24 permutations of starting points to corners, and find the optimal one. There may be a neater way to do this final step, but I brute forced it as 24 is not many.

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      2 years ago, # ^ |
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      I dont get it how here the connection between x and y is done. Since it is a square x and y size of resulting rect must be same. But the logic separates the x from the y coordinates :/

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        2 years ago, # ^ |
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        The logic combines the x and y results, though.

        • Length (both horizontally and vertically) = max(Y3-Y2,X3-X2), so the length is determined by the x and y co-ordinates

        • Bottom left corner of square is at (min(X2,X4-L),min(Y2,Y4-L)), so the position of the square is also determined by the x and y co-ordinates

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

      This is my approach for E after taking some ideas from jimm89 :

      int x[4], y[4], old_x[4], old_y[4];
          for(int i = 0; i<4; i++){
              cin>>x[i]>>y[i];
              old_x[i] = x[i];
              old_y[i] = y[i];
          }
          sort(x, x + 4);
          sort(y, y + 4);
          int b_a_u = x[3] - x[0]; // The upper limit for difference of x2 and x1 where x1 and x2 are defined below
          int b_a_l = x[2] - x[1];
          int d_c_u = y[3] - y[0];
          int d_c_l = y[2] - y[1];
          int side = min(b_a_u, d_c_u); 
          int x1 = max(x[0], x[2] - side);
          int x2 = x1 + side;
          int y1 = max(y[0], y[2] - side);
          int y2 = y1 + side;
          cout<<findmin(old_x, old_y, x1, x2, y1, y2)<<"\n";

      Here, x1, x2, y1, y2 are choices for the lines that define the square (4 lines x = x1, x= x2, y = y1, y = y2) and findmin() checks the minimal distance to corners in all permutations. https://codeforces.com/contest/1455/submission/100091962

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        2 years ago, # ^ |
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        Looks like a nice variant on the same ideas. Nicely done.

        My in-contest submission was a little messy as I was very short on time, and a little panicked. Afterwards I tidied it up.

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

    Here is a case-work based solution (no binary search) to problem E: https://codeforces.com/contest/1455/submission/100073456.

    Edit: I'm not sure if this solution is always correct.

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Guys look at the topic(contest) rating, going to zero XD

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How to solve E?

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2 years ago, # |
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Me after reading the solution of problem C

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Coded an n^3 solution for D using DP, is it hackable? 100060462

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    2 years ago, # ^ |
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    Problem statement guarantees:

    The sum of values of n over all test cases in the input does not exceed 500.

    So (sum of n) * 500^2 operations should be fine.

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    2 years ago, # ^ |
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    no problem with the given constraints n^3 solution passes

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Is the round difficulty below average ?

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G was nice, thanks authors! Sadly, I needed 20 more minutes to solve it :)

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Solving A-D feels like guess some conclusion and just try it...Even don't need to think seriously. Am I too impulsive?

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    2 years ago, # ^ |
      Vote: I like it +63 Vote: I do not like it

    I own none of these problems' ideas but I personally find the observation problems the hardest of them all. They tend to make me think a lot more than the standard-ish ones and usually way more than it takes people of my rating :( The fact that you don't need to think seriously just shows that you have better intuition I guess. So you can enjoy your "free" rating and the guys like me can enjoy practicing more in this kind of problems :P

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      2 years ago, # ^ |
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      I also got immediate ideas for A, C, and D but didn't believe them in the first place so spent a lot of time proving them :(

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I can't understand why so many people downvote a contest. Just for the problems? ok but also the writers and the testers make some effort to gives us these contests. And about the problem : Easy A like always , a easy math problem for B , C an observation one and D greedy (I Didn't understand that N <= 500 but that is not an issue). If A-D were easy why just 300 make at least 5 problems? You can say , was a huge gap between D and E and I agree , but that s not a reason to downvote a contest. I enjoy every contest (and I would like to see more DP and graph problems in div2 and Educational contests)

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    2 years ago, # ^ |
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    problem D I solved it DP lol. I didn't even think. I just wrote the code and AC that is why $$$N<=500$$$ so dumb people like me could write a standard DP solution :/

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      2 years ago, # ^ |
      Rev. 2   Vote: I like it +9 Vote: I do not like it

      N=500 you can get accepted with O(N^3) if I am not mistaken. I think N=5000 was better but that's the most irrelevant thing in this contest

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        2 years ago, # ^ |
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        i guess it was intended that authors wanted N^3 dp solution to pass, that's why n=500, just doubling N would have cut dp solution.

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          2 years ago, # ^ |
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            2 years ago, # ^ |
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            i am wondering that why it doesn't depend on $$$prev$$$ state in dp. can you please elaborate on this?

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              2 years ago, # ^ |
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              I'm very bad at proof and explanation but here's a try.

              I thought of it this way, dp(i,x) is the min number of steps needed to sort [0,i]. Obviously, the configuration we are considering might be invalid so i return inf in that case.

              I did not consider prev, because I'm swapping x and a[i] in every move. so if x=a[y] for any y at idx i, I can be sure that I've sorted from [0,y] and infact used moves from [0,y]. From [y+1,i] I have not used the moves. This is enough to uniquely define the state of the array.

              I might be wrong, if so please correct me.

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      2 years ago, # ^ |
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      Wish I was dumb like you!

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    2 years ago, # ^ |
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    People didn't like problems so they downvote, what's wrong with that? — Errichto
    source

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Good contest. Finally I can be expert

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hi all, in problem B,if we consider only the first operation jumping to y+k, then we can do the following jumps,

on 1st jump 0 to 1

on 2nd jump 1 to 3

on 3rd jump 3 to 6

on 4th jump 6 to 10

so we get the following series of 1 3 6 10 15....

now what i did is stored this series upto the given constraint of x which is 10^6 in a vector a. Then if x is just greater than or equal to any a[i] I calculate ans=i+1+a[i]-x where i+1 is the jumps to reach a[i] and a[i]-x is to move back to x by using the second operation y-1;

now after this I noticed this that for x=4 following is the optimal operation

1st jump 0 to -1

2nd jump -1 to 1

3rd jump 1 to 4

then i noticed if we do this kind of operation for any x(except 1 2 3) the total number of jumps will be x-1;

so at last i took min(ans,n-1).

is my approach wrong?pls help here is my code

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

    You seem to have made a few false assumptions, just trying to get your formula to match the example solution.

    The main problems are that the min with n - 1 doesn't really play a role for larger n, as the answer starts to increase much slower than $$$n$$$. The other one is that the answer will be the same for multiple consecutive $$$n$$$, where a[i] stays the same, but your formula does not account for that.

    Thus, as the second test shows, for example with input $$$7$$$, a[i] is $$$10$$$, i is $$$3$$$, and it is possible to reach $$$7$$$ with the sequence $$$0, 1, 0, 3, 7$$$ ($$$4$$$ jumps), but your formula gives $$$3 + 1 + 10 - 7 = 7$$$, and with the minimum $$$min(7 - 1, 7) = 6$$$.

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Problem G is really nice.Thanks!

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    2 years ago, # ^ |
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    What should be the problem rating for E,F,G in the contest according to you?

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      2 years ago, # ^ |
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      Sorry ,I haven't read E&F.For me G is about 2600.

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I can't understand why people are writing wrong code and then hacking theirselves like this guy — U_Atoev

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    2 years ago, # ^ |
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    They are most likely coping someone else's code and testing their hacks on the dummy account.

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I personally believe that if problem C was given as Problem A I would have come up with a solution much faster, I just thought since it was a C problem, it can't be so simple!

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Can we apply BFS in B problem??

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Solved D with DP and used only two states — [index][x]. I thought that x and prev are inter-related. I also tried [index][prev] but it was wrong. Is it wrong and just a coincidence with test cases? https://codeforces.com/contest/1455/submission/100080134

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    2 years ago, # ^ |
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    (i,x,prev)
    (i,x) They are inter-related according to me. How is it wrong?

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      2 years ago, # ^ |
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      Can you please explain why both these states are inter related? What redundant information are we providing in 3D DP?

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        2 years ago, # ^ |
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        Does this help?

        In other words, if prev!=v[i-1] then it's obvious that it has been swapped till i-1. you will see that the prev does not matter when you have fixed x because it's v[i-1] if not swapped, and previous x if swapped.

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          2 years ago, # ^ |
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          Thanks for your explanation. I think I got the gist.

          For any state to be uniquely defined. You need 3 information here: - current index - x on the current index - Maximum value till now on [0, curr_index-1]

          Since you only proceed further when you are convinced that [0,curr_index-1] is sorted, you can derive the 3rd information from the first 2.

          That is, it will always be arr[curr_index-1] (considering you have made swaps in the array wherever needed).

          Thus the 3rd condition is redundant and can be removed only because of the virtue of having a sorted array.

          I hope I got it right?

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            2 years ago, # ^ |
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            Yeah that's pretty much what I meant.

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    2 years ago, # ^ |
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    Oh, that's a nice idea to remove the index from the state as it is not providing us any useful information, just telling on which step we are. So, it makes sense to just focus just on values at step and step — 1

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anything on when ratings will be displayed?

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

I'm curious why Educational Round doesn't system test instantly after finishing the hacking phase.

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

Next time add/change notes or info in the main problem page too. In problem C, changes didn't show up after refreshing the page.

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

    There was no change in problem statement. Announcement for the explanation and that was writing on problem statement.

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

where is the editorial?

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

Where are the tutorials?

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2 years ago, # |
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Can we solve 2nd using dp?

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2 years ago, # |
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Can anyone tell me why am i wrong ? :'< https://codeforces.com/contest/1455/submission/100029658

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2 years ago, # |
  Vote: I like it -50 Vote: I do not like it

Below average problem set.

Cf standard going downward day by day

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    2 years ago, # ^ |
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    Why, then, could not solve problem E, since the standard is easy?

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

      The answer is in your comment XD The guy who cannot solve E,solved ABCD within 50 minutes! and no ds recquired till D! Neither a hard implementation problem nor any requirement of hard observation or ad-hoc logic till D -_- so you want to treat problem D as B in previous difficulty standard, Nice. Clap clap

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

when will the rating changes will be published ???

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

what about ratings update?! hacking phase is annoying.

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    2 years ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    Yeah Even I thinks why there is 12-hour hacking phase.

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

      probably timezones, so people don't have to stay awake at 3AM to make hacks.

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

It's annoying to check and find that the rating is not changed yet. update: Oh finally! the rating changed.

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

By when shall we expect ratings to change?(Don't know why I'm getting so excited for having a -30 :( )

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    2 years ago, # ^ |
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    My case is same as yours :( and I am also excited

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

Why the rating hasn't been updated yet?

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2 years ago, # |
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Has it ever occurred that after a round has been conducted the ratings are not updated and as a result the whole round gets canceled? Just Curious.

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

    As long as nothing serious happened during the contest,(no wrong checkers, no wrong problemsetter's solution, no wrong/very misleading statement) then there is no reason to make the round unrated.

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

I don't know why does it take so much time to give us the rates !

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

    eliminates those who copy code

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

Why codeforces takes long time for the rating?what's going on.

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2 years ago, # |
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Will the editorials be published for this contest?

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2 years ago, # |
Rev. 5   Vote: I like it -8 Vote: I do not like it

bad luck :(

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

    I bet one cannot write something more stupid than this XD XD

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

    How is it possible that his solutions were skipped and not yours? Weird! I mean ideally submissions of both the contestants should be skipped. MikeMirzayanov

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

      No. Not both. First submission shouldn't be skipped(at least in normal Div 2's). I will leave it as a practice for the reader why first submission should be accepted.

      Hint: Think about hacking within contest rooms.

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    2 years ago, # ^ |
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    No use claiming such thing here. They are just too irresponsible to check whether or not did you really copied the solution or not. Once the MOSS has flagged you then even if you are trully innocent, they will just don't give a shit.

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

Ratings updated

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

After 6 months finally expert, excited

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

Have you noticed codeforces just crossed 100M submissions during this(Edu 99) contest?

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

Thanks for the contest for +191 and pupil :)

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2 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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2 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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2 years ago, # |
  Vote: I like it -33 Vote: I do not like it

worst contest ever :/

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

Girl say:Can you open this jar for me? Boy say:Download Java