By awoo, history, 5 years ago, translation, In English

Hello Codeforces!

On Jun/30/2019 17:45 (Moscow time) Educational Codeforces Round 67 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

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

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

Good luck to all participants!

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 Radewoosh 7 220
2 244mhq 7 346
3 jqdai0815 6 120
4 square1001 6 136
5 HIR180 6 153

Congratulations to the best hackers:

Rank Competitor Hack Count
1 2014CAIS01 76:-1
2 Prateek 50:-1
3 takumi152 32:-2
4 Holland_Pig 28
5 sys. 37:-24
1268 successful hacks and 1683 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 arknave 0:02
C HIR180 0:06
D RUSH_D_CAT 0:22
E Noam527 0:06
F zhoutb 0:28
G Luma 0:46

UPD: Editorial is out

  • Vote: I like it
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| Write comment?
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5 years ago, # |
Rev. 3   Vote: I like it +16 Vote: I do not like it

The problems gonna be high-level because it's @PikeMike

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5 years ago, # |
  Vote: I like it +36 Vote: I do not like it

This group of people have held many educational rounds.Thanks a lot!

And I hope that the problems will be as nice as the past educational rounds'.

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5 years ago, # |
Rev. 3   Vote: I like it +82 Vote: I do not like it

[DELETED]

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

I'm looking forward to this round, cause I hope to turn blue:) Good luck to you all!

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

Just please prove your solutions :))))

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

I'm hoping to learn something new again in this contest. Because of the last educational round, I learnt the sparse table and binary lifting. Thank you, problem setters for nice educational rounds.

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

Is there any proof why my contribution is negative ? (without your upvote / downvote to this comment)

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

    If a comment has rating between 0 and -5, it's shown that its rating is 0. Your previous comments had rating 0, but maybe they were like -1 or -2 so, after some time, your contribution became -1. I guess that's the reason.

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

Do successful hacks add upto your total score?

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

    No,in educational rounds,you can only hack after the round ends for 12 hours.This will not effect your score.

    But if you hack others' solution successfully,and their ranks are higher than yours,your rank may become higher.

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

      But the possibility of that is quite low .. the hack must include a case the testers didnt notice ..
      I think it's for educational purpose only

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5 years ago, # |
Rev. 2   Vote: I like it -34 Vote: I do not like it

12hrs System test will be like

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

hope to have a contest !!

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

    hope to have a rated contest.

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      5 years ago, # ^ |
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      I hope, too. Moreover, it is important that this contest become rated even to prevent the contest from being unrated twice in a row.

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

All the misery was necessary

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5 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Educational Rounds and awoo.

Still a better love story than Twilight.
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5 years ago, # |
Rev. 3   Vote: I like it +16 Vote: I do not like it

I hope everyone gets what he deserves.

Good luck everyone!

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

When it clashes with India's batting.

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

if( Educational Round == awoo || Div 3 == vovuh ){ print("Contest gonna be lit."); }

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5 years ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

Sure about timings ? :p

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

Hope this one's rated :)

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

The contest start has been delayed for 10 minutes. I think these days so many contests are delayed.

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

Why did the time suddenly extend?

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

    It seems like new tradition of CF. I am observing it since last 4 contests.

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    5 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    traditions bro. Traditions.

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5 years ago, # |
  Vote: I like it +71 Vote: I do not like it

What's with all the delays nowadays?

Not to say that I don't appreciate this site, but why not just schedule it 10 minutes later beforehand so we can work on other stuff? Is something unexpected coming up?

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    5 years ago, # ^ |
      Vote: I like it +36 Vote: I do not like it

    Exactly I am yet to take my dinner today. And I delayed it for two hours for contest. If i knew it was 10 mins later then I could have managed taking my dinner as well before contest :(.

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

    They are waiting for 10k registrations.

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    5 years ago, # ^ |
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    This 10 minutes is the most boring time of day :3

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

again a delay ? ! great !

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

Finally codeforces scheduled delay takes its place

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

expecting nice problems!!

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

Every contest delayed for 10-15 minute . So , why contest time fixed 10-15 minute earlier rather than starting time ?

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

Delay by 10 minutes and CF rounds! better love story than Twilight :P

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

I wanted to say thanks for not having a delay, after contest :\

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

Delayforces

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

I hope this round is perfect.:)

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5 years ago, # |
  Vote: I like it +71 Vote: I do not like it

arisjo hacked pikmike's account and made this contest

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

The contest "codeforced" again.

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

Test case 3

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

If anyone needs help in E question.You can refer to F. Tree with Maximum Cost question. They both are quite similar :)

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

In my opinion, C harder than D and E.

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    5 years ago, # ^ |
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    D was definitely harder, it was hacked left and right.

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    5 years ago, # ^ |
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    I agree. In my opinion hard ad-hoc problems are always harder than problems that need a general algorithm or paradigm to be solved.

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

What is the 6th testcase for problem C?

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

A and B super easy. C, D, E, F need some "trick". I do not see the educational point.

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

    C, D and E needed some trick? I didn't use a single data structure, well known algorithm or "trick" in them and in fact I think that C and D were quite interesting.

    Edit: fuck my life.

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

      I got an AC with an horrible brute force, I expect this to be hacked. Worked about two hours on it and I have no real idea of how the solution should look like. Thats what I call "you need a trick".

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

    E did not need a "trick" in my opinion, it was an interesting DP on Trees problem.

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

      Maybe E was to hard for me. But, it is obvious that one need to find the "correct" first node. If you see how to find it, you found the trick.

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        5 years ago, # ^ |
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        no you can find the answer for all vertices on o(n)

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5 years ago, # |
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What is the approach of Problem E?

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5 years ago, # |
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E is very similar to 1092F - Tree with Maximum Cost. (I know being original is not a priority, I'm just saying in case it benefits some people).

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

    I solved that F problem without any help,but couldn't solve E now...

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    5 years ago, # ^ |
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    I did E using sliding through nodes only(Sliding Window??)

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      5 years ago, # ^ |
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      I read your solution could you explain in bit detail please.How are you managing window on tree, heard it first time,seems interesting!!

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

        Not actually windowing but if I find answer for any node say, 1 then for its children it will be just addition of a quantity a-b

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

          So you are using the concept of re-rooting basically and finding the effect of traversing one node below the given node.

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    5 years ago, # ^ |
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    Oh,I solved it,They are REALLY similar problem... interesting,of course...

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

What is "Unexpected verdict" for hacks?

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

How to solve D?

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

    int n; cin >> n;

    int a[n];
        int b[n];
    
        vsi dp;
        si tmp;
        dp.assign(n, tmp);
    
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            a[i]--;
            dp[a[i]].insert(i);
        }
        for (int i = 0; i < n; i++) {
            cin >> b[i];
            b[i]--;
        }
    
        SegTree st(a);
        bool fl = true;
        for (int i = 0; i < n && fl; i++) {
            if (dp[b[i]].size() != 0) {
                int ind = *(dp[b[i]].begin());
                dp[b[i]].erase(dp[b[i]].begin());
                if(ind == 0){
                    a[ind]=(1<<30);// is not necessary
                    st.update(ind , (1<<30));
                    continue;
                }
                int min = st.rmq(0 , ind-1);
                if(a[min] < a[ind])
                    fl = false;
                else{
                    //update segment tree and with INF value instead of delete
                    a[ind]=(1<<30); // is not necessary
                    st.update(ind , (1<<30));
                }
            } else fl = false;
    
        }
    
        if(fl) cout << "YES\n";
        else cout << "NO\n";
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5 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Hacks for C and D?

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

Testcase #6 for C. My approach : all segment of type 1 is increasing order. initial value = 1e7.(value greater than 1000) fill rest in decreasing order staring from 1000. Check if it violates type 0 segment rule. what is the problem?

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5 years ago, # |
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idk how but this code gives me correct output on terminal for test case 1, but I get wrong answer on test case 1 on codeforces, https://codeforces.com/contest/1187/submission/56341353. Can someone please tell me why this is happening, it happened to me so many times before.

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

    You forgot to return true in your chk() function.

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

Hackforces?

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

Problem E, in first test sapmle, why answer is 36? I think i even didn't understand the problem(

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

C -> whats wrong with my approach ? My approach : sort all pairs of type 1; take a initial value greater than 1000. for first segment of type 1 fill the segment with increasing value starting from initial value. for each segment except first. if starting point<=ending point of previous segment then do not reset previous value. continue increasing from there.

else if starting point>previous ending point reset to initial value.

fill rest in decreasing order staring from 1000. Check if it violates type 0 segment rule. what is the problem?

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

    Overlapping ranges... I think you need to check every single index, if index and index+1 must be sorted, or must be unsorted. Then you can build the array easyly.

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

      that is great thinking but I thought I handled overlapping segments. For type 1 segments I merged the segments . and for type 0 I checked is they are ok after building the array: 56343410 Can You give me a small hack..please?

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

        You might have ranges like

        1 1 10
        1 2 4
        1 6 12
        

        If you sort for 1, 2, 6, you will loose the 10 while checking if 6 is gt 4. So you decrease the counter on segment (6, 12), which violates rule for segment (1, 10).

        ... But I am not really sure about that... just a guess ;)

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

Today Problem A was a night mare to me . Got AC on 6th attempt.

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

There was DP on tree in E, isn't it?

P.S. Great contest!!!! Wish every contest gonna be like that.

UPD: C dropped

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5 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Lol what are all the hacks in D?

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5 years ago, # |
Rev. 2   Vote: I like it +34 Vote: I do not like it

Hack case for D:
1
5
5 4 1 2 3
1 2 3 5 4

answer must be YES

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

HACKFORCES at its best.

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

C. Maybe someone could help me? I don't understand, why my solution didn't work. What criterium to say "NO"? (In my sol I sort all conditions on left bound, connect them(conditions {1 2 5}, {1 3 6} will be {1 2 6}) and build for O(n) answer). Solution.

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

    After you build the array using the "1" facts, loop through all of the "0" facts to see if any of them are sorted.

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5 years ago, # |
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Can anyone tell how to solve B? Tried for like whole contest but Everytime time limit exceeded on test case 4.

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

    Build a pref matrix: matrix[letters][index], then use it to find the least indice that for each letter in the name, matrix[letter][index] >= frequency[letter] in the given string.

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

    First pre-compute the given string by taking a vector of vector of size 26 and insert there indexes . Then for each query make a frequency array for all 26 letters and iterate frequency array and if frequency is not 0 check for the minimum index needed for given frequency in pre-computed array . Here is my solution

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

    Your solution is probably $$$O(N\sum{t_i})$$$. You need to pre-process the string in order to quickly answer the farthest position needed for each name. You can do this by storing the $$$k$$$th occurrence of each letter. Then you just have to use this lookup table to find the answer for each name. Then the runtime is $$$O(N+\sum{t_i})$$$ (plus a constant factor for the size of the alphabet).

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

I am generating a large Test case for hacking for Problem D. I am generating Test cases in which n is of the order of $$${300000}$$$ and $$${t = 1}$$$. But when I try to hack someone it says Test Cases cannot be longer than 256KB. Why it is happening, In question it is mentioned that $$${n}$$$ can go upto $$${300000}$$$. Can anyone Tell me the reason.

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

    You should generate big tests with "generated input", in case you were just copying your test cases. I may be wrong

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    5 years ago, # ^ |
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    For testcases larger than 256KB you need to upload a generator i.e. a program which prints that testcase.

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    5 years ago, # ^ |
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    use generator for large test case

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

How to solve D? Some people suggested the answer was to count inversions of each individual number and check those, but then they got hacked.

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

    The key idea is the following: you need to sort subsegments of length 2, i.e. swap pair of consecutive elements if the first one is larger.

    Let's suggest that the element $$$b[1]$$$ is in the position $$$a[pos]$$$. If $$$min(a[1..pos]) < a[pos]$$$ then there is no way to move $$$a[pos]$$$ to $$$a[1]$$$. Otherwise, we can move it and, moreover, relative order of $$$a[1..(pos - 1)]$$$ doesn't change. So we "delete" $$$a[pos]$$$ and $$$b[1]$$$ and solve recursively.

    Of course, you need to write it in an efficient way.

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      5 years ago, # ^ |
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      but how can i get index "pos"?? will u pls share yur solution !!

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        5 years ago, # ^ |
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        Considering that solution as correct, you can make an array of vector of size 3*(1e5)+1 and store the indices of the input elements inside the vector. That way you can always obtain pos in O(1) time

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

      Consider this : 1 5 3 1 2 4 1 1 2 1 3 4 According to your solution, first element 3 can move to a maximum of 3rd index (consider 1-based indexing). But it can go till 4th index if we swap indexes 4 and 5. Also in that case relative order changes. Answer to this test shall be "YES". If I didn't get your code correctly please elaborate

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

        You should move not $$$a[1] = 3$$$ to the right, but $$$a[2] = b[1] = 1$$$ to the left. Then you get array $$$a = [1, 3, 2, 4, 1]$$$ and $$$b = [1, 2, 1, 3, 4]$$$. Next step is to move $$$a[3] = b[2] = 2$$$ to the position $$$2$$$, $$$a[5] = b[3] = 1$$$ to the position $$$3$$$ and that's all.

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

    Match indices of a to corresponding indices of b (this matching is unique if equal elements aren't reordered). Create an array pos where pos[i] = j means that a[i] is matched to b[j]. Then, if there is some pair (i, j) such that i < j, a[i] < a[j] and pos[i] > pos[j], then the answer is NO. So this gives a necessary condition that no such pairs exist. I didn't prove it's sufficient though :(

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

    Keffa2's solution this might help...

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

    First check that arrays are permutations, and get the map fuction $$$p$$$ such that if the $$$k$$$-th occurrence of value $$$v$$$ in $$$A$$$ is in position $$$i$$$, and the $$$k$$$-th occurrence of value $$$v$$$ in $$$B$$$ is in position $$$j$$$, then $$$p(i) = j$$$.

    Then, instead of "sorting a range", think of your operation as "swapping consecutive elements $$$A_i$$$ and $$$A_{i+1}$$$ if $$$A_i > A_{i+1}$$$", if you can get to an array by sorting ranges, you can also get to it by swapping consecutive elements, like that.

    You can notice that your operation allows you to "remove" already existing inversions, but you definitely can't create new ones. Then, finding a single "new" inversion present in $$$B$$$ is enough to answer NO, and, I haven't proved this yet, but I think that if you don't have any "new" inversion, then you can safely answer YES.

    You can check that you are not creating new inversions by going through the array $$$A$$$, and maintaining a segment tree that answers the maximum in a range.

    The idea is that $$$ST[v]$$$ is the maximum index (in $$$B$$$) in which an already seen value of $$$v$$$ (in $$$A$$$) ends up. When you are at position $$$i$$$, check if there was a previous smaller value $$$v$$$ ($$$v \leq A_i$$$) that ends up after $$$A_i$$$, if this happens, you can already say NO. Otherwise, just update the $$$ST$$$ accordingly.

    The code snippet might be easier to understand:

    for(int i = 0; i < n; ++i){
      int mx = ST.query(0, A[i]);
      if(mx > p[i]){
        ok = false;
        break;
      }
      ST.upd(A[i], p[i]);
    }
    

    You can also check my code: 56332559.

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

      Your claim about inversions can be proven by induction on $$$n$$$, the length of the given arrays. In the step you just have to notice that first $$$n-1$$$ values of the first array, on which we swap adjacent elements, can be sorted in the same way they are sorted in the second array, if we neglect $$$a_n$$$ (the arrays are indexed from 1). That is true by the hypothesis. And now it just remains to somehow deal with the $$$a_n$$$. But it is not hard to prove that it can go to its appropriate position. All you need to do is to notice that if there was an inversion that consists of $$$a_n$$$ in the first array before the change of the order of its first $$$n-1$$$ elements, then the same inversion exists after that change and vice versa.

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

    int n; cin >> n;

    int a[n];
        int b[n];
    
        vsi dp;
        si tmp;
        dp.assign(n, tmp);
    
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            a[i]--;
            dp[a[i]].insert(i);
        }
        for (int i = 0; i < n; i++) {
            cin >> b[i];
            b[i]--;
        }
    
        SegTree st(a);
        bool fl = true;
        for (int i = 0; i < n && fl; i++) {
            if (dp[b[i]].size() != 0) {
                int ind = *(dp[b[i]].begin());
                dp[b[i]].erase(dp[b[i]].begin());
                if(ind == 0){
                    a[ind]=(1<<30);// is not necessary
                    st.update(ind , (1<<30));
                    continue;
                }
                int min = st.rmq(0 , ind-1);
                if(a[min] < a[ind])
                    fl = false;
                else{
                    //update segment tree and with INF value instead of delete
                    a[ind]=(1<<30); // is not necessary
                    st.update(ind , (1<<30));
                }
            } else fl = false;
    
        }
    
        if(fl) cout << "YES\n";
        else cout << "NO\n";
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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

56341660 56338689

These submissions are very similar

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

    That's what happen when someone shares his solution with one friend, and don't share it with the other lol

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

please anyone explain B ?

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

    just store Kth occurance of every character like count[ch][k]=i (character ch appeared k times at index i)
    then count the occurance of every character from query string. for example in query string 'a' appeared 2 times 'b' appeared 3 times. so the answer will be maximum of count['a'][2] and count['b'][3]

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5 years ago, # |
  Vote: I like it +126 Vote: I do not like it

So this is what happened today:

  • I solved A, B, C, D, E and was finally about to be master *_*
  • Plot twist 1: 5 mins after the contest, my D got hacked. No master for me :(
  • Plot twist 2: A lot of people had their D hacked. I was going to be a master after all *_*
  • Plot twist 3: My C got hacked. Oops, not going to be a master :(

Requesting all hackers to add another plot twist here (By hacking other people's solutions, not mine :P)

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    5 years ago, # ^ |
    Rev. 2   Vote: I like it +26 Vote: I do not like it

    at the end, the editorial get hacked and the contest becomes unrated...

    happy ending for every hacked person :-)

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

The people who end hacking D starts hacking C. XD

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    5 years ago, # ^ |
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    What should be the condition for printing NO in problem C?

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      5 years ago, # ^ |
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      An unsorted fact is fully contained in a range that has to be sorted which is the union of overlapping sorted facts. You make the all the sorted ranges were non-decreasing and everything in between decreasing and check if it satisfies the unsorted facts.

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

        Thanks , earlier I was considering that even a partial overlapping will lead to answering NO, Also I have another question. The Problem statement mentions that some range will be sorted and some won't, but if the ranges does not cover the entire array , what can be said about the range which is not covered in the queries. Should that be sorted or unsorted. eg:- if array size is 6 and the ranges are (1-indexed) 3-4 4-6 what about 1-2 ?

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          5 years ago, # ^ |
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          Make them anything you want, it doesn't matter. You could just make anything outside of a sorted range unsorted so that you satisfy as many unsorted facts as possible, and then check if all the unsatisfied facts are true. If one of the facts aren't true, there's no solution since you're already making as much of the array unsorted as possible.

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

simple test hack for D -_-:

1 3 3 2 1 1 3 2

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

Before start hacking, the number of a test case of Problem D was 18, and there are 74 now. :O

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

    Pretty much means some solutions might be filtered out during the system test so don't be too happy if your solution hasn't been hacked (notes to myself... ahem).

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

    Lol, there are 152 test cases right now :)

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      5 years ago, # ^ |
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      How do you check that?

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        5 years ago, # ^ |
          Vote: I like it +11 Vote: I do not like it

        You can resubmit your solution and it will be tested on the full set of test cases

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

Many people have been hacked on D, including me.. We should use data structure to solve.

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

Any hint for F?

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    5 years ago, # ^ |
    Rev. 2   Vote: I like it +26 Vote: I do not like it

    Answer is $$$E[(1+\sum_{i = 1}^{i = n-1}(x_{i} != x_{i+1}))^2]$$$ = $$$E[(1 + \sum_{i=1}^{i=n-1}(x_{i} != x_{i+1})^{2} + 2\times \sum_{i=1}^{i=n-1}(x_{i} != x_{i+1}) + 2 \times \sum_{i!=j}(x_{i} != x_{i+1})(x_{j} != x_{j+1})]$$$.
    Now use linearity of expectations to calculate it.

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5 years ago, # |
  Vote: I like it +124 Vote: I do not like it

shrpy

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

Can anyone explain the idea of Problem F ?
Thanks in advance.

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

    Count expected number of group pairs.

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

C and D — R.I.P. It's gonna be very nice system testing))0)

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5 years ago, # |
  Vote: I like it +78 Vote: I do not like it

Do upvotes increase contribution points? If yes, can I get some upvotes just to get some contribution points? I currently have none :(

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

what is the hack in C? so many purples have been hacked

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

Wow, more than 600 solutions on problem D passed on pretests and now it's only slightly more than a 100. What's so wrong with this problem, did everyone try to do something greedy? My solution is stupid, what could possibly go wrong.

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

    even C is turning out like that

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

      C is a constructive task, so no doubts with a small amount of tests you can't really filter the small mistakes. But in D they even had multitests! It blows my mind.

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

    there was not strong tests for D , unlike E.

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5 years ago, # |
  Vote: I like it +19 Vote: I do not like it

I was rank bout 200 at the end of contest.But now,I am rank74. That's hackforces.

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

    Happiness comes to you quickly like a storm LOL :D

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

This round's D will like this problem,which has 588 tests

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

7 7 2 3 6 5 4 1 1 2 3 6 4 5 7

problem D, i excute some ac codes, result is YES, and unsuccessful hack

anyone can help explain? i think it is NO.

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

    Apply the operation to [5, 6] to swap them, [1, 2], [2, 3], [3, 4]... to move 7 to the end, and [5, 6], [4, 5], [3, 4]... to move 1 to the beginning. (numbers in brackets are indices, 1 indexed)

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

    7 7 2 3 6 5 4 1 1 2 3 6 4 5 7

    first sort [5,4,1] then it'll be 7 2 3 6 1 4 5 then [7,2,3,6,1] it'll be 1 2 3 6 7 4 5 then [7,4,5] it'll be 1 2 3 6 4 5 7 which is equal to B

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

Weak test cases for C and D

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

How to solve B? I think I have overcomplicated using the precomputation and binary search.

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

Okay, so I think you gave such weak tests on purpose.

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

Another 2 solve contest lol im gonnna drink some beer

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

When you go from -65 to -13

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

What is the idea behind C?

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

    Greedy. Lets make unsorted all the subarrays that don't required to be sorted. Note that if we have to intervals $$$[i_1,i_2]$$$ and $$$[j_1,j_2]$$$ that needs to be sorted, and if $$$i_2 >= j_1$$$ then the interval $$$[i_1,j_2]$$$ needs to be sorted as well. So lets find all actual intervals that needs to be sorted using the rule above. Suppose that those intervals are $$$[l_1, r_1],[l_2, r_2],...,[l_k, r_k]$$$. We can fill in them starting from the first one by assigning the $$$n$$$ to all the positions within the $$$[l_1, r_1]$$$, $$$n-1$$$ within the $$$[l_2, r_2]$$$ and so on. Finally check wherther all condition with $$$t==0$$$ holds. If not, there is no answer, otherwise you have already found one.
    It may seem that we need to fill in the segmends between sorted intervals somehow but actually they can be handled as a sorted intervals with length 1.

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      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it
      • Thank You But How we fill other that sorted intervals?
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        5 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Imagine that we need to produce an unsorted array of length 4. Its clear that intervals $$$[1;1],[2;2],[3;3],[4;4]$$$ needs to be sorted (yep, array of length 1 is allways sorted, but the goal is to see how the algorithm above will handle them). If we run an algo on this intervals we will get the following array: $$$[4,3,2,1]$$$. As you can see, all the subarrays of this array with $$$length > 1$$$ are unsorted.

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

          Hi Can you provide Me the code?

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

            56403000
            Sorry for the delay in my response, I was at work.

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

              Thank you bro. But i think you code is very complex. I saw some of the red codes and they were just like 20-30 Lines But i couldnt get them :(

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

    I use the Disjoint set to make the interval of sorted, then i test the unsorted case in it.

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

problems are good,but pretests are too weak

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

Hey awoo, Editorials please !!

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

Is there any simple solution for D which does not use segment tree?

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

Could someone check out my submission and tell me what do the diagnostics mean. I am not able to interpret it. Submission no — 56381811 Thanks in advance.

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

    The actual error is:

    Error: attempt to subscript container with out-of-bounds index 161, but container only holds 161 elements

    This is saying that you are trying to access element number 161 in a container (a vector, I think) that contains only 161 elements. This is an error because, in such a vector, the elements are numbered from 0 to 160.

    Looking at your code this might be in merge(), where you index with j+1:

                 for(;j<one.size();j++){
                      if(one[j+1].first <= one[j].second)
    
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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

for(int i=0;i<n;i++) { cout << cur << " \n"[i==n-1]; cur+=arr[i]; } can someone explain [i==n-1]; in above and for (i = 0; i < n; i++) cc[i] = (i % 2 == 0) == (s[i] == '(') ? '0' : '1' what does that == after (i%2==0) does?

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    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    • " \n"[i==n-1] is a way to print a space when i<n-1 and a newline for i==n-1, because " \n" is a string with two characters : " \n"[0] is ' ' and " \n"[1] is '\n'
    • the "==" tests if both conditions have the same result, so (i % 2 == 0) == (s[i] == '(') is equivalent to ((i % 2 == 0) && (s[i] == '(')) || ((i % 2 != 0) && (s[i] != '('))
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5 years ago, # |
  Vote: I like it +14 Vote: I do not like it

So where is the Editorials?

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

Was not able to solve any question but will try next time Wish me luck

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5 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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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In problem C, for sorted subarrays, most of the solutions are doing +1 for L[i] to R[i] — 1. Why not L[i] to R[i]? Why is R[i] not being taken into consideration?

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5 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).