By pikmike, history, 5 months 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 Ne0n25 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

 
 
 
 
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5 months ago, # |
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5 months ago, # |
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oh! Vovuh back!!

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5 months ago, # |
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yeahhh!! vovuh....big fan...so much excited!!

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5 months ago, # |
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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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    5 months ago, # ^ |
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    Yes. How many AC, then how many minutes (penalty).

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      5 months ago, # ^ |
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      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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        5 months ago, # ^ |
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        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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    5 months ago, # ^ |
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    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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      5 months ago, # ^ |
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      Yes.I thought that too

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      5 months ago, # ^ |
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      Can you please tell what's the difference between educational and div-2 round? I'm new here.

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        5 months ago, # ^ |
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        **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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codeforces is the best

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5 months ago, # |
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CF4

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5 months ago, # |
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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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    5 months ago, # ^ |
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    I think entire blog is almost copy pasted, maybe so

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:D

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Pretty excited for this contest !!

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

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cf5

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hope to get a decent rank this time

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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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When I find People Cheating in Rated Rounds

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    5 months ago, # ^ |
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    But after seeing that there code are same even comments are same my heart like "OOOH LALA";)

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Excited for vovuh div2 round after a long time.

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Good luck everyone

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IMG-20200625-165915

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

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

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    5 months ago, # ^ |
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    Spend some time on cf giving contests regularly and you will be able to answer this yourself.

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    5 months ago, # ^ |
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    He is the chosen one who sends all undeserving people like me from specialist to expert.

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

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Nobody:

Problem A in every Educational Round ever:

  • 1 < t < 1000

  • 1 < a, b, c, x, y, z < 10⁹

Seriously?

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Again B is easy than A -_-

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    5 months ago, # ^ |
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    Actually, in Educational round it doesn't matter at all)

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      5 months ago, # ^ |
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      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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        5 months ago, # ^ |
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        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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        5 months ago, # ^ |
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        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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    5 months ago, # ^ |
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    idea was simple but problem statement could have been worded better

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

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    5 months ago, # ^ |
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    I would suggest don't give up and keep trying :)

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      5 months ago, # ^ |
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      and what are you doing ?

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        5 months ago, # ^ |
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        Trying to motivate others. That's a good thing to do I guess. What are you doing?

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          5 months ago, # ^ |
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          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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    5 months ago, # ^ |
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    Hey xiaoshizhan

    It's just a smart brute force approach Check out detailed explanation: Link

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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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    5 months ago, # ^ |
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    i feel the same way, contestants are doing much better recently

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    5 months ago, # ^ |
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    todays div2D<<usual div2D

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

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    5 months ago, # ^ |
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    Off-topic:suddenly notice , after a long time you change your profile picture..

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      5 months ago, # ^ |
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      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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        5 months 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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    5 months ago, # ^ |
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    I'd be much obliged if you could share the solution logic of C here.

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    5 months ago, # ^ |
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    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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      5 months 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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    5 months ago, # ^ |
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    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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How to even think about the problem E.

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    5 months ago, # ^ |
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    Yeah man, the difference between D and E was huge today.

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      5 months ago, # ^ |
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      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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    5 months ago, # ^ |
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    My solution for problem E: https://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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    5 months ago, # ^ |
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    You can check out how to come up with the solution here :D

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

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    5 months ago, # ^ |
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    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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I had a hard time trying to crack C, can anyone please explain me their approach?

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    5 months ago, # ^ |
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    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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    5 months ago, # ^ |
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    Lets take an example

    --++----+--

    Assume after each '+' sign good sequence start (good sequence = +++--- or +- or ++-- i.e. +++....---...)

    Break it in good and bad sequence(continuous '-' without + before)

    For above example you can break it like [-- (++--) -- (+-) -] (where inside () its good sequence.)

    Suppose good sequence A = (++--) {- sign after that is 3 excluding in good sequence}. good sequence B = (+-) negative sign after B i.e 1

    After analysing it you will see we are incrementing bad sequence 1 by 1 and iterating good sequence in one go.

    so ans due to bad sequence = (n (n + 1) / 2) {i.e 1 + 2 + 3...} = 15 (as n = 5)

    ans due to good sequence = good sequence length * no of '-' after that i.e. ans due to A good subsequence = (4(length of A) * 3(no of '-')) = 12

    ans due to B good subsequence =. (2 * 1) = 2

    Overall ans = (15 + 12 + 2) + (length of string) {because we will iterate our string fully once just before breaking out of infinite while loop}

    = 15 + 12 + 2 + 11 = 40

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How to solve E?

Thanks in advance

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Test case 3 for E?

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    5 months ago, # ^ |
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    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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How to solve problem D?

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

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    5 months ago, # ^ |
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    How did you generate those values?

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      5 months ago, # ^ |
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      For $$$k = 0$$$ greedy algorithm. For other $$$k$$$ just check all $$$x$$$ from $$$0$$$ to $$$10^{9}$$$.

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        5 months ago, # ^ |
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        That means you checked for all values in your laptop, really clever idea

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          5 months ago, # ^ |
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          Yea. Precalc took ~$$$1$$$ minute.

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            5 months ago, # ^ |
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            Wow, I couldn't find $$$ans(150, 1)$$$ for twice as long XD

            1e9 operations per second or what?

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              5 months ago, # ^ |
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              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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                5 months ago, # ^ |
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                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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      5 months ago, # ^ |
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      For those following the thread

      Check out the detailed explanation of how to construct that number: Link

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    5 months ago, # ^ |
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    How did you map all the values, did you get any online tool to calculate that ?

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    5 months ago, # ^ |
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    How did you generate values?

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    5 months ago, # ^ |
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    Is that allowed? Or would it be considered cheating

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    Meanwhile setters: Am I a joke to you.

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    5 months ago, # ^ |
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    FBI wants to know your location

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    5 months ago, # ^ |
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    Codeforces be like :

    371.jpg

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    5 months ago, # ^ |
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    Can you prove it?

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    5 months ago, # ^ |
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    Those who are wondering about how to construct the numbers?

    Check out: Link

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How to solve E?

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5 months 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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    5 months ago, # ^ |
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    Or prefix sums while maintaining min

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      5 months ago, # ^ |
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      Isn't that exactly what Kadane's is?

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        5 months ago, # ^ |
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        Yes same thing but i used different implementation as Kadane's requires max and current sum

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    5 months ago, # ^ |
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    Can you help with some intuition on, how to solve F with kadane's ? Thanks

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      5 months ago, # ^ |
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      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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        5 months ago, # ^ |
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        I will definitely try to use the hints and to solve this problem. Thanks

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        5 months ago, # ^ |
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        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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          5 months ago, # ^ |
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          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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            5 months ago, # ^ |
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            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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        5 months ago, # ^ |
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        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://codeforces.com/contest/1373/submission/85147824

        Thanks in advance for the help.

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          5 months ago, # ^ |
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          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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            5 months 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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    5 months ago, # ^ |
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    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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What was test case 4 in problem D? ;-;

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I almost took 90 min to solve A question. Soo confusing.

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My brain: 30 minutes on A and 20 min on B+C -_-

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did anybody get a flow solution to pass for F?

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

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      5 months ago, # ^ |
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      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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        5 months 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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          5 months 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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    5 months ago, # ^ |
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    Its memory limit gets exceeded anyways with Push Relabel (with gap heurestics). https://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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Is think explation in C and D can be useful in figuring Problems more easily.

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How to solve D ? give some hints

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For me today B < A and D < C.

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    5 months ago, # ^ |
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    Really D < C?

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      5 months ago, # ^ |
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      D was easier to think than C. I wasted all of my time on C.

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

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

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    5 months ago, # ^ |
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    According to me b < c < a < d

    problem A just ruined my contest ;(

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      5 months ago, # ^ |
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      How was it so hard? I was tring a*x< ceil(b/x) + c, which is not easy to solve!!

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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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5 months ago, # |
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.

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5 months ago, # |
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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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    5 months ago, # ^ |
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    #define int long long
    
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      5 months ago, # ^ |
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      You better think before writte int or long long.

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        5 months ago, # ^ |
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        Yes sometimes long long can lead a just passing soln to TLE

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    5 months ago, # ^ |
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    But isn't 'overflow' in your name?

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    5 months 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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    5 months ago, # ^ |
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    I have same feeling with you

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

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Was not able to solve C..any help?

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5 months 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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    5 months ago, # ^ |
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    The problem can easily be reduced to maximum subarray problem

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      5 months ago, # ^ |
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      Crap, yes.... I feel really stupid now. I solved this in squared time and didn't recognise the similarity smh... Thanks!

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    5 months 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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    5 months 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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Problem D was really cute.

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    5 months ago, # ^ |
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    Can you share how you solved it?

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      5 months 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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        5 months 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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          5 months 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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          5 months 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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      5 months ago, # ^ |
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      you can find a detailed explanation here in the video

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How to solve F?

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I was not knowing that in Educational round, there is no point system. So I skipped A and solved B, C, D. I got a rank below 6500.:(

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Toughest Most confusing A ever.

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5 months ago, # |
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Why NlogN TLE'd on D?

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

Submission

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

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    5 months ago, # ^ |
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    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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      5 months 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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        5 months ago, # ^ |
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        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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How to solve D??

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

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    5 months ago, # ^ |
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    85011757

    Spoiler

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    5 months 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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    5 months 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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    5 months ago, # ^ |
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    You can check out my video editorial here

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I stuck on c for one and a half hour because of forgetting using long long...

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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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    5 months ago, # ^ |
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    lol..

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    5 months ago, # ^ |
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    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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    5 months ago, # ^ |
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    Guess it's the same person with two accounts

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    5 months ago, # ^ |
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    Here's another one

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Fuck it , I need a urgent editorial. Hell yaaa..

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

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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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    5 months ago, # ^ |
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    Is it because you are casting ad to int32_t at the end?

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      5 months 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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        5 months 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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when will the official analysis ?

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5 months 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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    5 months ago, # ^ |
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    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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      5 months ago, # ^ |
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      Wow, that's totally elegant, thanks a lot!

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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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        5 months 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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      5 months 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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If someone want a poorly coded and inefficient DP solution for E : 85057205

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Fucking D . Any solution? Waiting for help.

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    5 months 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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      5 months ago, # ^ |
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      thk you

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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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        5 months 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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    5 months ago, # ^ |
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    Some Hints:

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

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

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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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    5 months 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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What was the complexity of intended solution of F?

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It's really annoy :(

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Spoiler
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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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Did a semi brute-force solution for E and it worked. https://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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    5 months ago, # ^ |
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    At the first glance I read the last line as "Then do hookups".

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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://codeforces.com/contest/1373/submission/85035196

PS: They added the greedy tag.

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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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