Educational Codeforces Round 90 [Rated for Div. 2]
By awoo, history, 4 years ago, translation, In English

Hello Codeforces!

On Jun/25/2020 17:35 (Moscow time) Educational Codeforces Round 90 (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 Geothermal 7 130
2 ksun48 7 143
3 300iq 7 147
4 vepifanov 7 168
5 Radewoosh 7 175

Congratulations to the best hackers:

Rank Competitor Hack Count
1 EduPeres 40
2 Grey_Matter 39:-3
3 lx430621 26:-1
4 killa_vanilla 25:-5
5 checkingagain 17:-1
351 successful hacks and 375 unsuccessful hacks were made in total!

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

Problem Competitor Penalty
A Geothermal 0:01
B ksun48 0:01
C ksun48 0:03
D Noureldin 0:09
E Geothermal 0:22
F ElOrdyh 0:15
G dario2994 0:29

UPD: Editorial is out

  • Vote: I like it
  • +550
  • Vote: I do not like it

Comments Comments (388)
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4 years ago, # |
  Vote: I like it +335 Vote: I do not like it

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4 years ago, # |
  Vote: I like it -60 Vote: I do not like it

yeahhh!! vovuh....big fan...so much excited!!

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

can somebody explain me the way of scoring in educational rounds? I think you will only get scores due to the number of solved problems regardless of their difficulty. am I right?

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

    Yes. How many AC, then how many minutes (penalty).

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

      And bye how many minutes do you mean the total time spent to achieve the score or in some other way?

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

        If you solved any problem in 40 minutes.Then your penalty+=40.and every wrong answer penalty+=10 Generally..If both solved same no of problem then standing will be sorted with penalty.

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

    Good to know, I have participated in a few educationals but did not know this one. :O What I noticed about them is that they tend to be harder than normal Div2 rounds.

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

      Can you please tell what's the difference between educational and div-2 round? I'm new here.

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        4 years ago, # ^ |
          Vote: I like it +10 Vote: I do not like it 
        **Educational round
1.Here only have penalties.here a wrong submission give you +10 penalty.you solved a ques in 20 minutes with 1 wrong submission .then penalty is (20(in 20 minutes you solved that)+10(for 1 wrong submission))=30.
2.Standing firstly sorted by no of problem solved..if that same then penalty(whose penalty less he/she go up than others) .

**Div 2
1.Every problem have a score and you can gain the score by solved that.Time going and every problem score slightly(2/4/6 per minutes) decrease..and a wrong submission decrease of that problem score by 50.
2.Standing sorted by your gained score.
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4 years ago, # |
  Vote: I like it +385 Vote: I do not like it

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4 years ago, # |
← Rev. 2   Vote: I like it -131 Vote: I do not like it

CF4

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

Just curious, in educational rounds announcement, why its mostly written like "6 or 7 problems". Is the number of problems and problems itself not decided till few hours before the contest? Or is there something, which I am unaware of?

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

    I think entire blog is almost copy pasted, maybe so

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4 years ago, # |
  Vote: I like it +604 Vote: I do not like it

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

:D

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

Pretty excited for this contest !!

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4 years ago, # |
  Vote: I like it +176 Vote: I do not like it

![ ](WhatsApp-Image-2020-06-19-at-6.41.44-AM.jpg)

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4 years ago, # |
  Vote: I like it +121 Vote: I do not like it

cf5

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4 years ago, # |
  Vote: I like it -18 Vote: I do not like it

hope to get a decent rank this time

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

I have one question!, Know the Educational rounds are more than the div1 and most of the Educational rounds have 6 or 7 problems It is a good idea to add one hard problems and make educational to div1 ? :)

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4 years ago, # |
← Rev. 3   Vote: I like it +57 Vote: I do not like it

When I find People Cheating in Rated Rounds

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

    But after seeing that there code are same even comments are same my heart like "OOOH LALA";)

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4 years ago, # |
← Rev. 2   Vote: I like it -26 Vote: I do not like it

Excited for vovuh div2 round after a long time.

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

Good luck everyone

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4 years ago, # |
  Vote: I like it -17 Vote: I do not like it

IMG-20200625-165915

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

More than 20k people registered for this contest! Looks like it will be a fun contest.

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

Who is this vovuh and why are people so excited about him?

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

    Spend some time on cf giving contests regularly and you will be able to answer this yourself.

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

    He is the chosen one who sends all undeserving people like me from specialist to expert.

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

If I get WA1 on A and I can't proceed with the contest, will my rating drop?

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

Nobody:

Problem A in every Educational Round ever:

  • 1 < t < 1000

  • 1 < a, b, c, x, y, z < 10⁹

Seriously?

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4 years ago, # |
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Again B is easy than A -_-

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

    Actually, in Educational round it doesn't matter at all)

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

      matter brother..If a,b swap then my penalty only for A B will be (6+17)=23..But now it was (11+17)=28.

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

        If everyone finds A difficult than B, than it is actually difficult, and if everyone else finds it difficult, then all would take time to do that question and eventually your rank would be good even though you had more penalty

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

        And currently, mine is $$$(6 + 7)$$$ = $$$13$$$, but if they would have swapped $$$A$$$ and $$$B$$$, it would have been $$$(1 + 7)$$$ = $$$8$$$. But that doesn't matter much, rating would have been affected by $$$1$$$. Focus on improving skills brother!

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

    idea was simple but problem statement could have been worded better

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

OK, E is above my level. Go to bed instead.

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

    I would suggest don't give up and keep trying :)

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

      and what are you doing ?

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

        Trying to motivate others. That's a good thing to do I guess. What are you doing?

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

          I am also trying to motivate others.

          If you read my comment properly, I was trying to embrass you so that you go and focus on your work.XD

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

I am feeling that nowadays competiton is increased so much, not able to secure rank under 2k from past 3-4 contest.Contestant now even solving problem D like B, please tell me what do you think.

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

    i feel the same way, contestants are doing much better recently

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

    todays div2D<<usual div2D

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

I wonder how people solved C that easily :/ it was really hard for me

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

    Off-topic:suddenly notice , after a long time you change your profile picture..

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

      I wanted to put it after I reach expert but I guess that will take a long time .. I'm so disappointed about my performance

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        4 years ago, # ^ |
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        don't worry brother..be continue your practice your desired day come quickly.Wished you Good luck <3

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it 
    set answer, low , n to 0

iterate the string
if '+' n++, else n--
if n < low, answer += string.position, low = n

print answer + string.length

    i got WA and spent so many times to realize ans can't be store in int

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      4 years ago, # ^ |
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      Thats were im wrong too but i changed the int to long long in the last 8 minutes of the contest

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

    I think with enough practice on this types of problems, it would become intuition. I used to find these types of problems really hard, and after solving and understanding them they become easy to solve.

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

How to even think about the problem E.

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

    Yeah man, the difference between D and E was huge today.

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

      Yeah true I was trying to apply digit dp for 1 hour in the contest .XD and than after the contest i saw that people did it with brute force ...facepalm.And so it became typeforces for most experts.

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

    My solution for problem E: https://mirror.codeforces.com/contest/1373/submission/85070127

    The key insight is that K<=9 so you can only overflow at most once. For example if you pick the last digit to be x%10=7 with K=5, then it must look like this:

    (front    ) 7
(front    ) 8
(front    ) 9
(front + 1) 0
(front + 1) 1
(front + 1) 2

    So for every fixed starting digit and K, the sum of all digits is:

    sumOfOnesPlace + numNoOverflow * sumOfDigits(front) + numOverflow * sumOfDigits(front + 1) = N

    Where front is the variable we want to solve for and the rest are constant.

    If numOverflow is 0, you can solve for:

    sumOfDigits(front) = (N - sumOfOnesPlace) / numNoOverflow

    If front doesn't end in a 9, you also know that sumOfDigits(front) + 1 == sumOfDigits(front + 1), which again lets you solve for

    sumOfDigits(front) = (N - sumOfOnesPlace - numOverflow) / (numOverflow + numNoOverflow)

    Once you know a target for the sum of digits of front, you can greedy it.

    This isn't complete because I didn't cover all the cases, but I am guessing other cases are impossible via proof by AC. :)

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

    You can check out how to come up with the solution here :D

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

Did 0 based indexing ruined time in anyone's D?

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

    Yeah bro I also initially wrote code for 1 based indexing and after checking for test case 1 I felt something is wrong & and read the question again and realised that it is 0 based indexing

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

I had a hard time trying to crack C, can anyone please explain me their approach?

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

    void solve() { int n,i,j=-1; ll ans=0; string s; cin>>s;

    n=s.size();

vector<int> pre(n,0);
pre[0]=s[0]=='+'?1:-1;

for(i=1;i<n;i++)
{
    pre[i]=pre[i-1]+(s[i]=='+'?1:-1);
}


for(i=0;i<n;i++)
{
    if(pre[i]==j)
    {
       ans+=i+1;
       j--;
    }
}

ans+=n;

cout<<ans<<endl;

    }

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

How to solve E?

Thanks in advance

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

Test case 3 for E?

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

    Nevermind, figured it out:

    Sometimes it is optimal to put an 8 as the first digit after the 9s and 1s digit, rather than the remainder of the excess sum divided by 9.

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

How to solve problem D?

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4 years ago, # |
  Vote: I like it +190 Vote: I do not like it

$$$O(1)$$$ solution for E.

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

    How did you generate those values?

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

      For $$$k = 0$$$ greedy algorithm. For other $$$k$$$ just check all $$$x$$$ from $$$0$$$ to $$$10^{9}$$$.

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

        That means you checked for all values in your laptop, really clever idea

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

          Yea. Precalc took ~$$$1$$$ minute.

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

            Wow, I couldn't find $$$ans(150, 1)$$$ for twice as long XD

            1e9 operations per second or what?

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

              You can iterate by $$$x$$$ from $$$0$$$ to $$$10^{9}$$$ with fixed $$$k$$$ and store the sum of digits of numbers. If this sum appears first time, you found the answer for this sum and our $$$k$$$.

              This huge brute forse is needed only for $$$k=1$$$, for other $$$k>1$$$ it is enough $$$10^{6}$$$ candidates.

              To speed up the whole brute force, you can do transition from $$$x$$$ to $$$x+1$$$ by $$$O(1)$$$ instead stupid $$$O(log(x))$$$.

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

                So, it's only 1 loop from $$$0$$$ to $$$10^9$$$, I get it. At first I thought you did that for all $$$k$$$, hence the question..

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

    How did you map all the values, did you get any online tool to calculate that ?

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

    How did you generate values?

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    4 years ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

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

    Is that allowed? Or would it be considered cheating

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it 
      'Allowed'
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      4 years ago, # ^ |
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      I mean it isn't technically cheating, more of a hacky solution

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

        It is not cheating at all.

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

          Why do i hear your comment in petyr baelish voice in my mind XDXD

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

    Meanwhile setters: Am I a joke to you.

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

    FBI wants to know your location

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    4 years ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

    Codeforces be like :

    371.jpg

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

    Can you prove it?

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

How to solve E?

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4 years ago, # |
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I ended up using Kadane's for both D and F; I feel like it made sense for D, but was there an alternate solution to F that I missed?

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

    Or prefix sums while maintaining min

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      4 years ago, # ^ |
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      Isn't that exactly what Kadane's is?

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

        Yes same thing but i used different implementation as Kadane's requires max and current sum

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

    Can you help with some intuition on, how to solve F with kadane's ? Thanks

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      4 years ago, # ^ |
      ← Rev. 3   Vote: I like it +21 Vote: I do not like it

      So, not sure if it is correct, but this was the idea I had: if you consider the bipartite graph of households and connections (NOT cities and stations, i.e. the first sample has 9 households and 9 connections), and have an edge between a household and a connection if their city and station are adjacent, then the problem becomes finding whether or not this graph has a maximum matching equal to number of households.

      The problem is that this graph is way too big to construct (it can have 2e15 nodes), so you need to use some general arguments. First observation is that if there is any subset $$$S$$$ of households such that its neighbor set is smaller than $$$|S|$$$, then it is not possible. If all subsets $$$S$$$ of households have neighbor sets that are greater than or equal to $$$|S|$$$, then we claim that it is possible. I didn't prove this, but it sort of feels like the proof will be similar to Hall's Marriage Theorem if it is true. Someone can correct me if I am wrong about this.

      Now, we obviously cannot check all subsets. However, we notice that if we are trying to find a subset $$$S$$$ such that the size of the neighbor set is smaller than $$$|S|$$$, we will always be able to use all the households from some contiguous set of cities (why? go through the proof, it's a good exercise).

      So, we are trying to find some contiguous group of cities such that the sum of their households is less than the sum of corresponding network connections that cover those houses. I'll leave this as another exercise to make the connection to Kadane's. You have to do quite a few modifications to the algorithm.

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

        I will definitely try to use the hints and to solve this problem. Thanks

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

        How did you know it is guaranteed that the sum over all connections is equal to the sum over all households? I mean, the samples do match the argument but how were you so sure so as to proceed?

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

          Not sure I totally understand your question. The sum of all connections isn't necessarily equal to the sum of all the households; in test 3 on the sample case the sum of connections is greater than the sum of all households, and it is still not possible.

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

            Oh, I misread your comment earlier. I thought it was perfect matching you were talking about. Now it makes sense, thanks.

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

        Using you hint i tried to solve this with kadane's kind of approach, But its getting TLE on test — 9. Can you please help me figure out why is this getting TLE ? https://mirror.codeforces.com/contest/1373/submission/85147824

        Thanks in advance for the help.

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

          I fixed your solution by adding one nifty line at the beginning of main:

          85152881

          The idea is that cin is actually quite slow at reading in input, and this only really matters when you are reading in something on the order of ~1e6 elements. It's generally accepted practice to include this line at the beginning of main, or otherwise use scanf for all your reads. If you look at the top 5 people in the standings you'll see they did this.

          If you add this line, though, you should NOT use scanf/printf while also using cin/cout. Choose one and stick to it, because the addition of this line essentially allows these operations to occur asynchronously and it will totally mess up your program in certain instances.

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            4 years ago, # ^ |
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            Thanks a ton for the help. I will remember this from now own.

            Feels so good to to see it getting accepted.

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

    my solution: i simply assigned all of the capacity of the i-th edge to the i-th node and greedily give the next node the cumulative excess capacity while paying attention to the capacity of the edge. i traversed through the graph twice since it is a cycle and checked for validity on the third pass. 2 passes might be enough but i was being safe and did not want to waste time on checking for correctness. in all honesty i am surprised this simple solution worked.

    edit: fst :(

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4 years ago, # |
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What was test case 4 in problem D? ;-;

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4 years ago, # |
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I almost took 90 min to solve A question. Soo confusing.

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4 years ago, # |
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My brain: 30 minutes on A and 20 min on B+C -_-

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

did anybody get a flow solution to pass for F?

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    4 years ago, # ^ |
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    I completely ignored flow because the limit was so large. Is it possible to pass?

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

      worst case runtime of Dinic is very bad, but it's very often misleading, since it can run wayyyy faster on certain types of graphs. On bipartite graphs, Dinic runs in O(sqrt(|V|)|E|) time in the worst case (edit: jk this is wrong, it still runs quick though), so I thought it was worth a shot. I think you can derive a solution by analyzing the augmenting paths of the graph.

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        4 years ago, # ^ |
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        I know that when Dinic's algorithm is applied to bipartite matching, the time complexity reduces. Its called Hopcroft-Karp algorithm. However, I have no idea when it comes to just finding a maximum flow on bipartite graph, not matching. AFAIK, the major reason why such bound holds is because the capacities are unit, which is not the case for this problem.

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          4 years ago, # ^ |
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          https://en.wikipedia.org/wiki/Dinic%27s_algorithm

          in bipartite matching, you have both unit capacity edges and a bipartite graph. each of those restrictions individually can speed up Dinic, since each allows you to find blocking flows very fast, which reduces the number of iterations you have to do on the graph.

          edit: oh i see what you mean, the runtime i posted above doesn't always apply. AFAIK it still is true that it runs fast on bipartite graphs though.

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    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it +15 Vote: I do not like it

    Its memory limit gets exceeded anyways with Push Relabel (with gap heurestics). https://mirror.codeforces.com/contest/1373/submission/85062675 (Total 2 * n + 2 nodes are needed to model the problem into flow, with capacity taken in long long).

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4 years ago, # |
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Is think explation in C and D can be useful in figuring Problems more easily.

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4 years ago, # |
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How to solve D ? give some hints

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4 years ago, # |
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For me today B < A and D < C.

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    4 years ago, # ^ |
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    No way C is more difficult than D, unless you coded some weird solution.

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    4 years ago, # ^ |
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    It took me more time to figure out correct solution for C than D.

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    4 years ago, # ^ |
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    According to me b < c < a < d

    problem A just ruined my contest ;(

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4 years ago, # |
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Always 'A' makes me so Panick...

Was staring at problems for 1hr 12 minutes. Zero solved! And then solved D->C->B->A. Honestly This was the difficulty for me. As soon as I saw D I got the solution.

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

F*uck int, F*uck integer overflow, int data type should be removed from C++, got AC on D just after the contest

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    4 years ago, # ^ |
      Vote: I like it +26 Vote: I do not like it 
    #define int long long
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      4 years ago, # ^ |
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      You better think before writte int or long long.

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    4 years ago, # ^ |
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    But isn't 'overflow' in your name?

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    4 years ago, # ^ |
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    I always use long long no matter whether it should be.

    Though long long is slower than int and some kinds of problems will return TLE when you use long long instead of int, but the timelimit on CF is far more loose, so I just use long long every time to avoid the integer overflow.

    BTW, I'm very sympathetic to you, I can understand your feeling because I had many times when some stupid mistakes blew my whole contest up just for dropping 50+ ratings.

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    4 years ago, # ^ |
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    I have same feeling with you

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    4 years ago, # ^ |
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    Also, I got WA in D because of int. Yes, data types can cause havoc.....

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4 years ago, # |
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Was not able to solve C..any help?

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4 years ago, # |
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How do you solve D? I came up with O(N*N) DP but it will TLE and I have no idea how to optimise. One observation I have is that there is no point in reversing a subarray of odd length as it'll yield the same sum. My DP solution was to go through all possible lengths of even subarrays and find the maximum value that can be obtained as sum at even positions upon reversing that subarray (which I do in squared time). Also, E seemed very interesting but apart from a few basic observations I found nothing. So, any suggestions on E are welcome too!

I found the problems very interesting, thanks for the round!

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    4 years ago, # ^ |
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    The problem can easily be reduced to maximum subarray problem

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    4 years ago, # ^ |
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    We can solve D simply by building (sum of odd indices — sum of even indices) for every prefix, and some simple calculations.

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    4 years ago, # ^ |
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    Here is my dp solution to problem D which is very different from Kaden's algorithm.

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4 years ago, # |
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Problem D was really cute.

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    4 years ago, # ^ |
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    Can you share how you solved it?

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      4 years ago, # ^ |
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      This is the link to my solution... First observation is that the array that you want to reverse has to have even length in order to get the elements swapped. And you can start considering that the initial array is the best..and then you can use kadane's algo for finding the best subarray to reverse. Reversing an even length array means in fact subtracting from the result the elements that were previously in the sum and adding the others.

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        4 years ago, # ^ |
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        Can it be solved using two pointers ?? considering the largest even subarray such that the sum of elements at odd position is greater than sum of elements at even position and adding the remaining even sums at both ends ??

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          4 years ago, # ^ |
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          I don't really know. Maybe it is possible. That was the only idea i had..and thankfully it worked

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          4 years ago, # ^ |
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          I tried to implement the same idea during the contest but failed. If you find someone's code using this method, let me know.

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      4 years ago, # ^ |
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      you can find a detailed explanation here in the video

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4 years ago, # |
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How to solve F?

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4 years ago, # |
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Toughest Most confusing A ever.

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4 years ago, # |
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4 years ago, # |
← Rev. 2   Vote: I like it -35 Vote: I do not like it

Why NlogN TLE'd on D?

After seeing the constraints, I assumed it should have passed.

Submission

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    4 years ago, # ^ |
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    You mean why $$$O(n^2)$$$ TLE while $$$O(n)$$$ passes?

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    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it -10 Vote: I do not like it

    Your solution isn't NlogN it is O(N^2) . Since number of even length subarrays is (N * (N + 1))/4 -> O(N^2)

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      4 years ago, # ^ |
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      Can you explain why? I was thinking it is NlogN, so wasn't optimizing during contest.

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

        you have commented in your code that " so we can check every len of 2, 4, ... , n using tc = Nlog2N."

        it would have been Nlog2N if it was 2,4,8,16...N but here it is 2,4,6,8,10....N

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4 years ago, # |
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How to solve D??

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    4 years ago, # ^ |
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    check this. It's a link to my comment where i explained a little

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    4 years ago, # ^ |
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    85011757

    Spoiler

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    4 years ago, # ^ |
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    I solved D without kadane. My approach is that store prefix even -prefix odd in one vector and prefix odd indexed values -and prefix even indexed sum values in another vector. and then try reversing maximum difference of odd indexed -even indexed sum values sum even length subarray if the array ends at odd index and vice versa for subarrays ending at even index. Try thinking on building intuitions. My submission link is :- 85056177

    Hope I would be hacked!

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    4 years ago, # ^ |
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    Check out the detailed explanation for it here, you can read more about Kadane's algorithm if you have further doubts :)

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    4 years ago, # ^ |
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    You can check out my video editorial here

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4 years ago, # |
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I stuck on c for one and a half hour because of forgetting using long long...

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4 years ago, # |
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85052253 is ...weird. Who would if(a==165) that deliberately, if not for other accounts to hack?!

Also some of the other A problem Hacks too. Weird.

This one

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

    And for his first try on problem A he wrote (a==7) but it didn't work :D. I also accidentally reviewed these two solutions in hacks section and was confused :DD

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    4 years ago, # ^ |
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    Guess it's the same person with two accounts

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    4 years ago, # ^ |
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    Here's another one

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4 years ago, # |
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Fuck it , I need a urgent editorial. Hell yaaa..

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4 years ago, # |
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Could anyone please confirm, Shouldn't the verdict for this would have been Runtime Error signed integer overflow? Instead it gave Wrong Answer

https://mirror.codeforces.com/contest/1373/submission/85049895

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4 years ago, # |
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Can someone point out the mistake in my submission for D? 85049220 I used maximum subarray approach. Don't know where its going wrong. Test case is too big to understand.

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

    Is it because you are casting ad to int32_t at the end?

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      4 years ago, # ^ |
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      Mistakes like these are always remembered :( Just changed that part and Boom 85058120 Thanks a lot. Would have never figured that out. Also, a silly question, I use: define int long long int because of which I have to use int32_t and stuff. How do you use library functions then? Because when I use max with just ad, it says no matching function calls. So, how do I use (any)library functions on long long int variables?

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        4 years ago, # ^ |
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        Just ensure that you typecast the other value to long long int as well.

        For example
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4 years ago, # |
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when will the official analysis ?

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4 years ago, # |
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vovuh I'm curious, was the following solution intended for E:

  1. If $$$k = 0$$$, greedy.

  2. If $$$k = 1$$$, notice some patterns, come up with a rule.

  3. If $$$1 < k <= 9$$$, write a brute force up to something on the order of 1e6 (possibly, with optimizations).

I also noticed many users did precalc (calculated all the answers offline). However, the vast majority did something totally different (idk what), so I wonder, whether the straightforward solution was intended or not. Thanks.

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    4 years ago, # ^ |
      Vote: I like it +103 Vote: I do not like it
    1. Iterate over the last (rightmost) digit
    2. Iterate over the number of 9s before it.
    3. Calculate the needed sum of left digits, check it to be non-negative and integer.
    4. Construct left part greedy to fit that sum.

    And find the minimum value among them. No special cases.

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      4 years ago, # ^ |
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      Wow, that's totally elegant, thanks a lot!

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      4 years ago, # ^ |
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      In your code, how did you know to insert an 8 in the middle when lft is greater than 8?

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        4 years ago, # ^ |
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        I need to put as large digit as possible, but I can't put a 9 there since I fixed the number of 9s. So here comes an 8.

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      4 years ago, # ^ |
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      As far as I can tell, you don't even need to iterate over the number of 9's

      If the rightmost digit would wrap around from 9 to 0 in your sequence, then the number of 9s before the last digit is always 0

      else the number of 9s is as large as possible

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4 years ago, # |
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If someone want a poorly coded and inefficient DP solution for E : 85057205

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4 years ago, # |
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Fucking D . Any solution? Waiting for help.

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    4 years ago, # ^ |
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    Here's one solution — 1.Standard problem to be used. Maximum sum subarray of even size (you can google and first geeks link can help). 2. Observation : reversing an odd sized subarray is useless. Problem is reduced to find a subarray of even size, where sum of odd indices elements are more than even indexed elements . 3. Actual solution - sum all even indexed elements , call it sum_1 , multiply all even indexed value by -1. Find the maximum sum of even sized subarray in this modified array, call it sum_2 . Ans = sum_1 + sum_2

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      4 years ago, # ^ |
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      thk you

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      4 years ago, # ^ |
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      by the way thank you for aur sharing your approach . but can please explain why "multiply all even indexed value by -1" is done (how is it helpful??).

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        4 years ago, # ^ |
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        If after this modification, there exists a even sized subarray with positive sum, that means of you reverse this array, you will get most benefited.

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

    Some Hints:

    Spoiler
    My Attempt
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    4 years ago, # ^ |
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    I think after see the solution you understood all.Just Using prefix count

    solution
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4 years ago, # |
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Hello!
I am MrPupsik. Help me to solve problems pls.

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4 years ago, # |
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Can anyone help me understand where my attempt to problem C goes wrong?
I think that this might be due to some overflow error, but i am unable to see where it occurs, as almost everything is long long.

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    4 years ago, # ^ |
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    I guess that ans = ans+it-vc.begin()+1; causes pointer overflow which leads to undefined behavior? As changing the line to ans = it-vc.begin()+ans+1; (submission 85085422) or ans += it-vc.begin()+1; (submission 85085571) solve the issue.

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4 years ago, # |
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What was the complexity of intended solution of F?

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4 years ago, # |
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It's really annoy :(

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4 years ago, # |
← Rev. 4   Vote: I like it +37 Vote: I do not like it

My solution for the E:

If k = 0 we can build the number, we simply put the largest digit to the left whenever we can.

If K = 1 and n is an odd number, we simply put an 8 at the end of the number, so when putting it together it would be:

XXXXXX9

XXXXXX8

But if N <17 we can check with brute force.

If K = 1 and n is an even number, we simply put 89 at the end of the number, so when putting it together it would be:

XXXXX90

XXXXX89

But if N <26 we can check with brute force.

The last case is when we have k> = 2 in this case we can search for it with normal brute force, because at most it will have 6 digits

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4 years ago, # |
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Did a semi brute-force solution for E and it worked. https://mirror.codeforces.com/contest/1373/submission/85047398. Precalculate the answer for all numbers that are I9JK, I99JK, I999JK etc.. Where I,J,K can be any digit and there is a bunch of 9s in between (0 up to 20). Then do lookups,

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    4 years ago, # ^ |
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    At the first glance I read the last line as "Then do hookups".

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4 years ago, # |
← Rev. 3   Vote: I like it +17 Vote: I do not like it

I saw the tags for problem F, and I did not see the greedy tag. My solution is only based on a greedy approach. I do not know how to prove its correctness. If it is wrong, feel free to hack it.

The solution is here: https://mirror.codeforces.com/contest/1373/submission/85035196

PS: They added the greedy tag.

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4 years ago, # |
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What's the reason, so one has accepted while other not?[submission:85006751][submission:84997771] (Thanks in advance)

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    4 years ago, # ^ |
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    Solution is failing due to this line input=sys.stdin.readline At end of each testcase, readline will take newline character '\n' along with input, which gets accepted in else part of logic, increasing value of on. To correct this behaviour, strip the '\n' character at end
    st = st.rstrip('\n')

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What was the name of website which kept track of all codeforces rounds and show us which questions were solved by us and which are left? It also kept track of rounds by there format. Eg:- Different categories for Educational,Div1,Div2,Global rounds. It tells us for this round we have solved A,B,C and left with D,E,F,G.

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4 years ago, # |
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I really liked problem C as it tests your ability to read, understand and reason about a piece of pseudocode. Very useful practice and skill.

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    4 years ago, # ^ |
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    Yes true. If I encounter such piece of code in the industry, I would never think to optimize it. Tried it out here just because I knew there was a solution

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4 years ago, # |
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Can somebody explain me why n*log(n) F solution gets TLE? I think I'm missing something, as I would expect it to easily pass. Thanks

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    4 years ago, # ^ |
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    sorry, n*log(b0), but question stays the same

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    4 years ago, # ^ |
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    This is what you need: ios_base::sync_with_stdio(0);

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

      Indeed that was the case, now it passed. My main language is java, I only tried c++ because I was confused that I got TLE. That being said I'm not sure how to fix it in java (will try to google it, or if you know, please let me know).

      Explanation: city[n] can get connection only from station[n-1] and station[n] (don't forget about modulo — but that is clear). So if I provide X connections from station[0] to city[1] I need to provide remaining connections for city[1] (city[1] — X) from station[1]. If there isn't enough I need to increase starting X. If there is enough I can take remaining connection from station[1] (call it Y) and give it to city[2] and again get remaining connections for city[2] (city[2] — Y) from station[3] (if there isn't enough start with higher X).

      If I was able to do it for whole cycle I need to check if station[0] has enough free connections (station[0] — X) for remainder of city[n]. If yes it is solvable, if not I have to decrease starting X.

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        4 years ago, # ^ |
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        Yes, nice solution, I did something quite different, for each subarray of cities, I checked whether it can be satisfied by the stations which enclose it. Solved it using sparse table

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

    Could you explain your solution?

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4 years ago, # |
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Does there exist a testcase that (the ans)%100+k>99? I just iterate 0~99-k for the last two digits of ans

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

    I checked every possible n and k combination with my solution, and none of them have ans%100 + k > 99, so it doesn't look like it, but I don't have a reason as to why yet.

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

    sorry, i think its wrong

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      4 years ago, # ^ |
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      Me too. But I can't find the testcase whose ans%100+k>99 or prove that it doesn't exist

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        4 years ago, # ^ |
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        no, your solution is correct (but not sure for larger $$$n$$$), that comment is for my rev. 1

        i also use the same solution and just intuition, now i try to prove

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

i refreshed the page 3 times in problem C .. i thought it was a bug showing the editorial solution xD

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4 years ago, # |
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When you click contests, you can't access it. Only an active contest is visible.

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

https://mirror.codeforces.com/contest/1373/submission/85001977 Could someone pls tell me why my code submission 85001977 for C can work perfectly for n=500(testcase4) but for testcase5 n=197, it shows TLE??? How can this happen. I implemented the same code they provided in the question, just a small change I brought was to make an array where I could store previous val of curr and the index where curr ended. And in then next iteration started at that index only, so effectively my code had complexity of O(n). Tell me if I'm wrong also how could that thing happen which I stated above.

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    4 years ago, # ^ |
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    that would result in tle brother as the iteration for the string would repeat and would pass the time limit.

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      4 years ago, # ^ |
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      No it wouldn't repeat. Only one iteration takes place. From 0 to no. of '-'s. I kept the record of last iteration and carried on from there for the next iteration. Had it been that case, why did my code didn't show TLE for n=500 but shows for n=197

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    4 years ago, # ^ |
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    500 and 197 is number of testcases not the string.length

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      4 years ago, # ^ |
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      yeah i got that later but I've tested.... and in fact sum of |s| wont go more than 10^6. My algo iterates over this only once. Why should it show TLE??

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

Enjoyed solving the problems :)

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4 years ago, # |
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How the answer for the third test case ( problem E) is 4 . Why not 3?because starting from 3 till 9(3+4+5+....+9) gives me 42 .

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4 years ago, # |
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Wrong ans in C for 2h because 'int' expect 'long long'=))))

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4 years ago, # |
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What's the binary search solution for F? I think the possible amounts of connections that can be given to the first city by $$$b_1$$$ lie in a contiguous range, but I can't think of how to apply binary search to find any number in this range.

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    4 years ago, # ^ |
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    Yes. It's possible to do so. You can binary search on how much of the connections from $$$b_1$$$ you assign to $$$a_2$$$. Basically you assign a potential number, and then perform greedy and see if it works.

    The greedy strategy is like this: If you assign $$$x$$$ connections to $$$a_2$$$, then you still need $$$y = \max(0, a_2 - x)$$$ connections from $$$b_2$$$ for $$$a_2$$$. That means you only have $$$y$$$ connections available from $$$b_2$$$ for $$$a_3$$$. Be greedy, so you assign all of those to $$$a_3$$$. Now you need additionally $$$a_3 - y$$$ connections from $$$b_3$$$, and so on...

    I think of it in a flow/pipes analogy. You just push as much water forward to the next one, which then can push more water forward, and so on. Or in the rope interpretation (see a few comments below) fixate on the of the rings and push the others as far as possible.

    Now let's think about the extremes. If you assign $$$0$$$ connections to $$$a_2$$$. Then you will probably be not enough, and at one point you will no be able to fill all required connections.

    The other extreme is exactly the opposite. If you assign all $$$b_1$$$ connections to $$$a_2$$$, then you will do a lot better connections. This is the best case for all the cities $$$a_2, a_3, \dots, a_n$$$ and you will satisfy their needs (unless of course it's not possible at all). The only problem that occurs is that you don't have enough connections remaining for $$$a_1$$$. You might end up with a demand for $$$z$$$ connections for $$$a_1$$$, and you don't have any connections left since you all assigned them all to $$$a_2$$$.

    Now if you assign $$$b_1 - 1$$$ connections from $$$b_1$$$ to $$$a_2$$$. What happens with the demand? It can get one higher, but it also can stay equal (if some of the pushed connections we assigned greedily are unused). Which means in that case we are better, since we still have one unused connection remaining. If we assign less and less connections to $$$a_2$$$, the difference between the demand and the unused connections will shrink, until it is zero (in that case we have found a solution), or until we gone to far and we can't fulfill and demands of the middle cities any more.

    To summarize: if remaining demand for $$$a_1$$$ can't be satisfied with the unassigned connections of $$$b_1$$$, we have assigned too many connections to $$$a_2$$$. Or if the assignment breaks somewhere and some of the other cities miss some connections, we have assigned (pushed) to few connections. Using these rules you can perform a binary search.

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      4 years ago, # ^ |
      ← Rev. 3   Vote: I like it +24 Vote: I do not like it

      Now that I think about it, there is no need for binary search at all.

      First assign all $$$b_1$$$ connections to $$$a_2$$$, and push as much as possible. This leaves the demand $$$z$$$ for $$$a_1$$$.

      Now assign $$$b_1 - z$$$ connections from $$$b_1$$$ to $$$a_2$$$, and push again greedily. This one will be a solution if it exists.

      Solution: 85077333

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      4 years ago, # ^ |
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      Thanks!

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4 years ago, # |
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What is the correct solution for A. Many people are getting A hacked.

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

.

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4 years ago, # |
  Vote: I like it +51 Vote: I do not like it

I have a physical proof of problem F at https://mirror.codeforces.com/topic/79862/

Let us consider a circle segmented to n arcs with n nodes N1, N2, ... Nn. the length between Ni ans Ni+1 is ai(the number of households in the i-th city.) and there is a small ring in arc_i, note as Ci. there is a rope with origin length bi between Ci and Ci+1. the rope can be stretched if neccesary.

Now if we can arrange all the rings so that no rope need to be stretched, there is a solution.

We just let all rings move freely. if no rope are stretched, it is ok. If some ropes are streched. there are 2 conditions:

a. all ropes are streched, it is condition 1; b. some continues ropes are streched, while the rope before and after them are not. In this case, the start and end ring of these ropes must be at the node. so it is condition 2.

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

Video Editorial For Problem — D . Hope you like it.

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4 years ago, # |
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In problem D, I multiplied the numbers at even positions by -1 and found the sub array of even length whose sum is largest.Since this will give me the difference between the original sum of elements at even positions and max sum at even positions after reversing,I'll just add the difference to the original sum. But somehow i got WA at tc 3. Can anyone please help me find why it didn't work? I'll really appreciate it. Here is the link to my solution : https://mirror.codeforces.com/contest/1373/submission/85045234

PS: It isn't kadane, it's just the name of the functions and was too lazy to change it again.

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    4 years ago, # ^ |
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    Maybe because in your function kadane your array dp is int and also the answer is int, but it can be long long (around $$$10^{14}$$$), and so it overflows?

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      4 years ago, # ^ |
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      Wow thanks brother,it worked. I don't know why i didn't think of that during the contest.

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

I would like to bring to notice these 2 users who wrote solutions so that their code can be hacked. Please ban these accounts; who create new ids just to disturb the competitive spirit in code forces rounds issafake and orkunisitmak

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    4 years ago, # ^ |
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    Too much yapping for a specialist,dont you think?

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      4 years ago, # ^ |
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      At least better than cheaters and newbie ig

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4 years ago, # |
← Rev. 3   Vote: I like it -8 Vote: I do not like it

Well I solved D in a different way, I have used dynamic programming ,first we know that only reversing the order of the l,rth border matters where l(mod 2)!=r(mod 2).

Next wherever this is the case we see that the odd positions interchange with the even position upon reversing. That means there is a net change of value of sum of elements in odd indexes — sum of elements of even indexes.

That means we need to find the maximum such value where all elements in range of l,r. To this extent let dp[i] denote the max such value where the right border is the ith index.Now transitions are as follows

if(i%2==0)

dp[i]=max(dp[i-2]+a[i-1]-a[i],0);

else

dp[i]=max(dp[i-2]+a[i]-a[i-1],0);

as a[i] is on odd index in this case, and we need to find the value of sum of elements in odd indexes — sum of elements of even indexes.

finally the answer is max of all dp[i]+ the sum of numbers previously on even indices from 1->n.

q.e.d

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    4 years ago, # ^ |
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    Fun fact- Kadane’s algorithm counts as dynamic programming as well!

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4 years ago, # |
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How long before the editorial drops?

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    4 years ago, # ^ |
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    It'll be available right after the end of the hacking phase

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4 years ago, # |
  Vote: I like it +24 Vote: I do not like it

Solutions to A-F are available at the end of my screencast of the round for people who are interested.

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4 years ago, # |
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Educational round without editorial, how ironic! We need editorial!!

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4 years ago, # |
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Is there a correct greedy solution for F? Without binsearch

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    4 years ago, # ^ |
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    See if it helps my submission.

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      4 years ago, # ^ |
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      My solution is the same:) During hacking time I was thinking that my solution is wrong, but lol it passed. For me it was unbelievable, because I've spent on solving about 5mins

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4 years ago, # |
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My solutions for problems A-E.

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4 years ago, # |
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I failed to finish E during the contest, but if anyone's interested, here is a non-bruteforce solution I was working on, the version I tried to make during the contest actually failed only on two test cases in the entire input space lol.

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4 years ago, # |
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Hey please can someone help me out by explaining how to solve E.

Also , are Educational Rounds harder than usual ones ?

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    4 years ago, # ^ |
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    I explain it at 2:11:35 here

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    4 years ago, # ^ |
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    https://mirror.codeforces.com/blog/entry/79277?#comment-650107 I found this easy to understand and easy to implement.

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    4 years ago, # ^ |
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    It is possible to precompute the solution for all possible n and k in a separate file and then paste the entire array into your solution and read the answers from it for each test case like this: https://mirror.codeforces.com/contest/1373/submission/85074779

    For k = 0, you can use greedy to precompute the answers.

    For fixed k >= 1, it suffices to check x from 0 to 1e9.

    The reason you only need to check x from 0 to at least 1e9 for k >= 1 is because when k = 1 you can use x = 999,999,998 to get a total digit sum of 80 + 81 = 161 > 150 = maxN, and when k > 1 you can achieve an even larger maximum sum. So by checking all x up to 1e9 you can achieve n up to 161, which is enough. I guess it can be proven that you also wont miss any n by not going above 1e9.

    The values of digitSum(x) from x = 0 to 1e9 can be stored in a vector and can be computed in linear time using a simple recurrence relation. To efficiently answer sum queries (f(x) + .. + f(x + k)) you can turn the vector into the vector of its prefix sums.

    In the end, the dp array can be calculated in O(9 * 1e9) time, which takes 10 seconds on my laptop.

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

    The non-bruteforce solution sketch is as follows.

    1. Let's express $$$f(x + 1)$$$ in terms of $$$f(x)$$$: $$$f(x + 1) = f(x) + 1 - 9g(x)$$$ where $$$g(x)$$$ is the number of tail nines in $$$x$$$.
    2. Because $$$k < 10$$$ the set $$$(g(x), g(x + 1), ..., g(x + k))$$$ will contain at most one non-zero item.
    3. Based on these two observations, we can express the sum $$$f(x) + f(x + 1) ... + f(x + k)$$$ as $$$f(x) \times (k + 1) + \sum_{i=0}^k i - 9p$$$ for some non-negative $$$p$$$, and that must be equal to $$$n$$$.
    4. We can rewrite that as $$$f(x) \times (k + 1) = n - \sum_{i=0}^k i + 9p$$$ and find values of $$$p$$$ for which $$$k + 1$$$ divides $$$n - \sum_{i = 0}^k + 9p$$$. From that, we get our candidate $$$f(x)$$$ values, too.
    5. Now $$$p$$$ defines which tail digits we can use. If $$$p = 0$$$, we can use any such digit up to $$$9 - (k - p)$$$. If $$$p \ne 0$$$, we can only use $$$9 - (k - p)$$$. However, this is further limited by $$$f(x)$$$, we can't take a digit that is already greater than $$$f(x)$$$.
    6. After that we just greedily add more digits to the number with one additional thing in mind: this construction requires that $$$g(x + (k - p)) = 1$$$, so we can't have a nine in the second position.
    7. We print $$$-1$$$ when there is no suitable $$$p < k$$$.

    The code is here.

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

Solution to F

For Each network station, at index i
    Fill its left city (city at index i) as much as possible
    If there is still capacity left, fill its right city ( city at index (i + 1) mod n) as much as possible
    
For Each House at index i: 
    fill the house to full using its left station (index (i - 1) mod n) 
        (if unable to make it full now, there is no solution)
    replace as much flow as possible from its right station (index i) with flow from its left station

There is a solution iff all the cities are full at this point

Submission link: 85057681

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

A well coded O(n*sqrt(n)*log(n)) solution could get AC on G?

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

Solution for G using max prefix sum segment tree:

First, note that for each pawn $$$i$$$ we can easily calculate the minimum row $$$r_i$$$ it has to get to. Then, let's pretend that we can place pawns on top of each other, and the question then becomes "how many rows must be added to the board so that all pawns are on the $$$k$$$'th column and no two pawns are on top of each other?" If a pawn can reach $$$(k, r)$$$, it can reach $$$(k, r')$$$ for all $$$r' > r$$$. Thus, it is only of interest to us what the earliest row each of our current pawns can get to is, and then we will simply propagate them all forward until no two are on top of each other.

I think it's easier to think of the new problem in terms of arrays: We can imagine the special column as a vector of length $$$n$$$ containing all $$$0$$$'s. Each time a pawn gets added, we add a $$$1$$$ to the index of the earliest row it can get to (where we add new indices if needed, never going over 2n indices). Now, given some array, how do we tell whether we need to add more rows? Playing around with it a little, if we assume our array is either of length $$$n$$$ or has no trailing 0's (i.e. the last entry is nonzero), then we need to add new rows to our array if there is some suffix sum that exceeds the number of elements contained in the suffix. For example, if $$$n=4$$$, the array $$$[0, 0, 2, 0]$$$ is fine but the array $$$[0, 2, 2, 0]$$$ is not since the suffix $$$[2, 2, 0]$$$ has sum 4, which is greater than the number of elements it has (3). The number of rows we need to add is actually exactly the number that the sum exceeds the number of elements by.

If we reverse the array and look at 1-indexed prefixes instead, then we are essentially looking for any $$$p_i > i$$$. This is identical to looking for $$$p_i-i > 0$$$, so we can actually initialize our array to be -1's rather than 0's and now our problem becomes looking for any $$$p_i > 0$$$! If we can find the maximum $$$p_i$$$ with our segment tree above, then this becomes an easy query.

Now, all we need to do is keep track of the maximum $$$r_i$$$ of all pawns currently on the board to keep track of where our prefixes should start from.

There is also apparently a solution that uses RMQ and lazy segtree, but not sure what the idea is. Anyone care to share?

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    4 years ago, # ^ |
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    My solution for G: (I could not AC within contest, ACd after that)

    Every point has a time range when it is active. We load this into a segment tree, and then traverse the segment tree. That way we can simulate adding points for ever query. We need another aditional DS, where we can do Arr[index]++ where we add a +1 to index. We will only add +1 to a previously 0 location, so we also need to have a binsrch thing to find the first 0 position(note that when we go return from a node, we need to do Arr[index]-- as well). Complexity is mlog^2n though.

    click

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    4 years ago, # ^ |
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    There is also apparently a solution that uses RMQ and lazy segtree

    I think this is what you are looking for https://youtu.be/Nuym8ejFH_w

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4 years ago, # |
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After 6 failed attempts, I am still unable to figure out, why is code for Problem C failing? Can anybody help me. Please!! Here is a link to my submission

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    4 years ago, # ^ |
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    I think there is some problem with your logic. I am explaining my logic, just crosscheck.
Basically the terminating condition for the outer loop is not getting negative current, so the number of times the outer loop will run =  (minimum prefix sum (taking '+' as +1 and '-' as -1) + 1).
The terminating condition for inner loop for each cur = i is prefix sum = (-i-1), because then the cur will go negative .
  So,
  let minimum prefix sum = s
  and index[i] represents the first index where prefix sum becomes -i;
  and ans = 0
  then we can compute the ans as:-
  for(i = 1 ; i <= (-1*s) ; i++) // number of times the outer loop will run
   {
   ans += (sum[index]+1) // number of times the inner loop will run
   }
also since the terminating condition for the outer loop is not getting negative value of cur, therefore it will run for one more time and this time the inner loop will run completely, therefore we will have to add n(length of string) to ans.
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4 years ago, # |
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common man , still no editorial , do they prepare it after the contest ?

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

Video solutions for A,B,C,D,E....Hope you will like it!!!

Problem A

Problem B

Problem C

Problem D

Problem E

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4 years ago, # |
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Hey, is there any official editorial for this contest?

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4 years ago, # |
← Rev. 3   Vote: I like it +19 Vote: I do not like it

If I uphack some solutions after the open hacking phase, but before systest, will the hacked solutions fail in systest?

upd: Yes they fails

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    4 years ago, # ^ |
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    They do fail, but not necessarily by your hack, at least in this Submission it failed systests on test 48 which is not your uphack. So I am not sure that uphacks after the open hacking phase are incorporated to the systests at the end of the contest.

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4 years ago, # |
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After wasting so much time on C, I found that I was using int instead of long long int. Now i solved C and D both. Hope I could solve it during the contest only.

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4 years ago, # |
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come on why I can't write more than 1e9 in first problem. :(

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4 years ago, # |
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Does anyone explain why it works? it's my code... but idk why it works.... problem F 85046506

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4 years ago, # |
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For the first time, I'm finally able to solve C problem in a contest XD

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4 years ago, # |
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I like to read cf comments more than FB memes.

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    4 years ago, # ^ |
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    Seems like Codeforces Watcher application can help you with this time spending.

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4 years ago, # |
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I can't find the editorial for this contest, Is it just me or everyone?

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4 years ago, # |
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I am a newbie. I solved 3 questions. Still, I got +82. Don't know how the rating works.

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    4 years ago, # ^ |
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    Rating changes depend on your rank,not how many questions solved.when the questions are tough,even solving even a few can drastically increase ratings.

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4 years ago, # |
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Anyone tried to solve D using dp approach? I need some idea about it.

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    4 years ago, # ^ |
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    I think Kadane's algorithm is dynamic programming only (just a memory efficient version of it) .

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4 years ago, # |
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Why they make A so lengthy?

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4 years ago, # |
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When will the editorial be out ?

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4 years ago, # |
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still no editorial...

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4 years ago, # |
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Yet Another question similar to D

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4 years ago, # |
  Vote: I like it +51 Vote: I do not like it

Why does editorials always take so much time in Educational Rounds?

I think it should be part of contest preparation so that the editorial is ready before the contest ends.

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4 years ago, # |
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I think they forgot to publish the editorial ..it's so much annoying..

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

Where is the editorial for this round

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

Hola community, ~~~~~

Your code here...

I was developing a software and ran into error, help is highly appreciated .. please reply fast vector<vector<pair<int,int>>> Cluster; for(auto cluster : Cluster){ vector<pair<int,int>> left_data_logins, right_data_logins; for(auto login : cluster){ if(value<mean_of_all_values) left_data_logins.push_back(login); }}

~~~~~ executing the above peice of code is giving SIGSEV error, i do not know if data_types are not compatible . please help!!

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4 years ago, # |
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Why did you change tests after the contest and I've got WA? That's unfair... Is not better to do strong tests?

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    4 years ago, # ^ |
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    This is called hacking. Tests didn't change but more tests were added before final testing phase.

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

    In contest time ,any solution judge by few test case to avoid long queue..And After contest always judge that solution by more extra test case. It's the rule.Not unfair at all.

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

Are the problem setters waiting for the corona vaccine to post the editorials?

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

I received a notification that "Your solution 84984892 for the problem 1373A significantly coincides with solutions izhang05/84984892, diegorestrepo68/84988994". I did not share my code with anyone and I did not use any online IDEs such as ideone.com. Given that this is a simple problem that only requires a few lines of code, it's not surprising that there may be multiple submissions that are similar. There's a limited number of possible solutions and implementations for this problem.

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

    I suggest you to use own template to avoid this type of coincides.

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

      I didn't use any templates. This was my code:

      include <bits/stdc++.h>

      using namespace std;

      int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { long long a, b, c; cin >> a >> b >> c; if (a < c) { cout << 1 << " "; } else { cout << -1 << " "; } if (c < b * a) { cout << b << "\n"; } else { cout << -1 << "\n"; } } return 0; }

      There are only so many ways to express "a < c" and "c < b * a".

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        4 years ago, # ^ |
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        i see both two code..i understood you are not cheater..but system are not human being, they just check code matched or not..so suggest you to use own template..Then so much change happens always from other code.

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

why is the editorial released so late in educational rounds?

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4 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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4 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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