dreamoon_love_AA's blog

Codeforces Round #631 (Div. 1 and Div. 2) — Thanks, Denis aramis Shitov!
By dreamoon_love_AA, history, 2 years ago, In English

Hello, everyone! Codeforces Round #631 will be held at Apr/03/2020 17:35 (Moscow time).

The problems are almost from me(dreamoon_love_AA), except one problem of which the idea is from drazil and developed by me. Also we want to thank 300iq for helping me prepare the round, isaf27, tmt514, rowdark, yp155136, wangyenjen, n_dao107 and zj713300 for testing this round, and MikeMirzayanov for Codeforces and Polygon.

This is my third time organizing a problemset for a Codeforces round (my previous rounds: Codeforces Round #292, Codeforces Round #320).

Good luck and have fun!

UPD: Below is a message for you from MikeMirzayanov:

About two weeks ago, we completed the crowdfunding campaign dedicated to the 10th anniversary of Codeforces. Community help inspires and provides resources for the development and operation of the platform. Thanks! With this round, we want to say thank you to Denis aramis Shitov, for his significant contribution and support. It is valuable, important and very nice when people from the community lend a helping hand and congratulate. Thank you, Denis!

UPD2: Also thanks to Shik for testing the round!

UPD3: Score distribution:

  • Div2: 500 1000 1500 1750 2500
  • Div1: 500 750 1500 2000 2500

UPD4: link to tutorial

UPD5:

I won't say sorry that the pretests are not strong. Because I don't think the pretests must be strong in every contest.

But I still want to give congratulations to the winners:

div1:

  1. Um_nik
  2. tourist
  3. Petr
  4. ecnerwala
  5. ffao

div2:

  1. 3u_my_light
  2. hydd
  3. fafafafafafa
  4. timmyfeng
  5. FrijolitoAzul

Among these winners, I want to especially congratulate Um_nik. I am quite excited when I see him solve all problems.

Tags 631
 
 
 
 
  • Vote: I like it
  • +656
  • Vote: I do not like it

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

It's going to be a great round. Really Excited :)

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +140 Vote: I do not like it

    But, be careful of test 1

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it +20 Vote: I do not like it

      This is not problem Generally, the test 1 is example in question and you will not get penalty even if you make a mistake on test 1.

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it +169 Vote: I do not like it

        Joke has gone above your head.

        → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
        ← Rev. 2   Vote: I like it +54 Vote: I do not like it

        Yes but please everyone check their code answers on samples otherwise it will make the queue too long and reduces the quality of contest.

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it +11 Vote: I do not like it

          Indeed you are quite right

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

    I see you know what meme is good

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    i wish god will save us from queue because there is lot of registration during quarantine.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    ohh really?

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it -63 Vote: I do not like it

    Bitch,dreamoon is just rubbish and his contest is as rubbish as him.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Don't judge the quality of contests if you haven't taken them.

      → Reply
»
2 years ago, # |
  Vote: I like it +15 Vote: I do not like it

How many problems?

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Usually, it is a 6 Problem and 2 Hour round.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +140 Vote: I do not like it

    I think there will be 5 problems for both division.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it +11 Vote: I do not like it

      Thus this round must be relatively hard...unsatisfactory results ending up usually

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it +42 Vote: I do not like it

        But when I still studied in university, 5 problems round is more often than 6 problems. So I determine use 5 problems.

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it -31 Vote: I do not like it

          Wow! It's actually very exciting to see you back! And the problems are definitely gonna be extremely high-quality and hard to solve as well. Don't think it's wise to take part in such a round, cause oftentimes I failed to get my rating higher(sad face Anyway, wish the round a complete success!

          → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
          ← Rev. 2   Vote: I like it -30 Vote: I do not like it

          U are really good manager[user:dreamoon_love_AA] ,Managing love and coding same time.

          → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
        ← Rev. 2   Vote: I like it -8 Vote: I do not like it

        .

        → Reply
    • »
      »
      »
      2 years ago, # ^ |
      ← Rev. 2   Vote: I like it -22 Vote: I do not like it

      You mean for each division?

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        What was wrong with the question — for both divisions there were 10 problems. :)

        → Reply
»
2 years ago, # |
  Vote: I like it +21 Vote: I do not like it

wow Setting Problem After such a long time

→ Reply
»
2 years ago, # |
  Vote: I like it -39 Vote: I do not like it

You have one of the most interesting rating graphs on codeforces!! dreamoon_love_AA

Please explain the trend!!

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +32 Vote: I do not like it

I hope problem statements will also be short and sweet just like the announcement :)

→ Reply
»
2 years ago, # |
  Vote: I like it +181 Vote: I do not like it

Didnt invited sorry_dreamoon?

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it -21 Vote: I do not like it

I wish the time could be earlier,I am not able to be there.o(╥﹏╥)o

→ Reply
»
2 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Hope, everybody is going to enjoy the contest. :)

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

It will be interesting to see whether tourist can regain his throne in this contest!!!!

→ Reply
»
2 years ago, # |
  Vote: I like it -67 Vote: I do not like it

it better be virus themed questions

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

I am excited for the contest , in this quarantine time it is just annoying and boring without cp contests and football matches .

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

can anyone suggest how to practice so that I am able to complete more than 2 problems in a contest?

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    there is no secret to it. just solve more tasks

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Solve problems which are a little harder than your comfort zone. Like you can easily solve 1000 rating problems, then try to solve 1100-1200 rating problems. And don't spend more than 30 minutes on a problem if you think you are stuck see the tutorial, understand, and solve it.And of course practice a lot.

    I am following this approach and it's helping me. happy coding.

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

      Actually you can spend more than 30 minutes on a problem. You can spend as much time as you wish until you have some progress on it. Trust my words ;)

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

Problem setting tab shows you have arranged 3 contests before. Round 272,292 and 320. Then it should be your 4th contest. Why are you not considering round 292?

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

    If you click the link of Round 272 provided in this blog, it opens up the page for Round 292.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    Actually, I don't consider round 272 as const organized by me. It's the problemset of round 272 is not chosen by me. I just only provided the hardest problem to my friend.

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

Wow! This announcement is great! I think this round is going to be a great round.

→ Reply
»
2 years ago, # |
  Vote: I like it +86 Vote: I do not like it

dreamoon_love_AA here I come!!!

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +114 Vote: I do not like it

to understand the meme for none one piece fans

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

Auto comment: topic has been updated by dreamoon_love_AA (previous revision, new revision, compare).

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

this is good ....

"Knowing Is Not Enough; We Must Apply. Wishing Is Not Enough; We Must Do.".......B)[user:hajer qandeel]

→ Reply
»
2 years ago, # |
  Vote: I like it +34 Vote: I do not like it

tourist vs MiracleFaFa. Who will win ?

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

Excited! Looking forward to your problems.

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it -39 Vote: I do not like it

.

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

Who is Denis Aramis Shitov????

→ Reply
»
2 years ago, # |
← Rev. 4   Vote: I like it -52 Vote: I do not like it

I am not begging for followers(friends) on codeforces.

Edit: I did not even edit my original comment uh.

→ Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    Why are you begging for followers(friends) on codeforces?

    Edit: Why would you even edit your original comment uh?

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

Feeling excited. Codeforces making this lockdown time easy to pass by holding the contests for programmers.

→ Reply
»
2 years ago, # |
  Vote: I like it +31 Vote: I do not like it

Let us see if this round can get 25000 registrations

→ Reply
»
2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

All the best to everyone who will participate in this round. I hope to become a specialist after this contest.

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +15 Vote: I do not like it

Thank You Denis Aramis Shitov

→ Reply
»
2 years ago, # |
  Vote: I like it +9 Vote: I do not like it

tourist vs. MiracleFaFa It will be exiting!

→ Reply
»
2 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Thank you mr. Denis Aramis Shitov

→ Reply
»
2 years ago, # |
  Vote: I like it -8 Vote: I do not like it

→ Reply
»
2 years ago, # |
  Vote: I like it +16 Vote: I do not like it

I hope I can do well in this round

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

is it rated?

→ Reply
»
2 years ago, # |
  Vote: I like it -6 Vote: I do not like it

Time is not good for me!

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

Thanks !!! We need competitions like this more and more .I wish everybody good luck.

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

Wow！It seems good！

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

Will it be a Mathforces Round?

→ Reply
»
2 years ago, # |
  Vote: I like it -11 Vote: I do not like it

I hope its going to be an interesting round

→ Reply
»
2 years ago, # |
  Vote: I like it +21 Vote: I do not like it

Looking forward to seeing concise and crap free problem statements.

→ Reply
»
2 years ago, # |
  Vote: I like it +29 Vote: I do not like it

Thank you Mr.Denis Aramis Shitov.

→ Reply
»
2 years ago, # |
  Vote: I like it -10 Vote: I do not like it

umm..what we know so far is that there will be at least 2 problems

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it -28 Vote: I do not like it

.

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

thank you Denis aramis Shitov and to all other contributors.

→ Reply
»
2 years ago, # |
  Vote: I like it +21 Vote: I do not like it

Thanks, Danis and other contributors. Because of you, we have problems to solve for free.

→ Reply
»
2 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Friendly reminder: watch out for FST.

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

What's the score for each problem?

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +33 Vote: I do not like it

.

→ Reply
»
2 years ago, # |
  Vote: I like it +15 Vote: I do not like it

May God bless us from queue.... All the best guys!!!!!!!!!

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

Hope I will turn blue after this contest.

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

Hope this contest does not have long queues like the last rated contest

→ Reply
»
2 years ago, # |
  Vote: I like it +19 Vote: I do not like it

When the problem scroring will be posted?

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

I think , this contest will be very intersting

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

good luck for every one)

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

Why hasn't the score distribution been published yet?

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

hoping to do well in it and get some +ve rating . let's see how it goes .

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

Please post the score distribution of the problems.

→ Reply
»
2 years ago, # |
  Vote: I like it -11 Vote: I do not like it

I think will be 500-1000-1250-1500-2000

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

Auto comment: topic has been updated by dreamoon_love_AA (previous revision, new revision, compare).

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

Is it rated ?

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

Enjoy the contest

→ Reply
»
2 years ago, # |
  Vote: I like it +16 Vote: I do not like it

site seems fast today

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

lets do well guys ..best of luck..

→ Reply
»
2 years ago, # |
  Vote: I like it +129 Vote: I do not like it

noExplanantionsForSamplesForces

→ Reply
»
2 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Did anyone else too missed the contest because they binge watched Money Heist ? :/ Guess I'm gonna give a virtual one now !

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +56 Vote: I do not like it

Is this the second april fools round?

→ Reply
»
2 years ago, # |
  Vote: I like it +86 Vote: I do not like it

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +93 Vote: I do not like it

    The authors likely spent some time writing the statement. Why do you think they can come up with a better explanation now in a shorter time? Besides that, it is much harder to understand your question than to understand the statement. $$$x$$$ is a number that is given to you in the input, what more do you want?

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it -87 Vote: I do not like it

      If they didn't say anything it would be better than this answer

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it +82 Vote: I do not like it

        No, I think it's an excellent answer.

        It lets you know the answer is written in the statement, the statement is clear (in the author's opinion) and you do not need any more information. As opposed to "oops, we left this part out" and "yes indeed, it's unclear, we will edit". Not answering creates a lot of uncertainity. "No comment" is also a reasonable answer, but I think CF wants to be friendlier than that in easy problems.

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it -19 Vote: I do not like it

          Not gonna lie you're right it's just a joke I don't why I can't understand it even though alot of people solved it

          → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +35 Vote: I do not like it

    Lolwut, do you really think this is a reasonable question? This is definitely the appropriate answer given that thousands of people didn't have a problem with it.

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

Is this round bit hard!!!

OR only i feel so?

→ Reply
»
2 years ago, # |
  Vote: I like it -24 Vote: I do not like it
The comment removed because of Codeforces rules violation
→ Reply
»
2 years ago, # |
  Vote: I like it -19 Vote: I do not like it
The comment removed because of Codeforces rules violation
→ Reply
»
2 years ago, # |
  Vote: I like it +126 Vote: I do not like it

Great contest, I hate it.

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it -6 Vote: I do not like it

    I hate this contest! one of my worst performances so far could not even solve B!

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

      Yeah, I also only solved 2. But the questions were really nice. I am waiting to see where I went stupid in my C and editorial for D.

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        how to find efficiently(better than O(k)) that all the number between 1 to k exist or not in a array of size k??

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          keep a count of the unique elements, as well as the maximum seen so far. Check if those are equal.

          → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          I used a set, if set.size() == max(set) then its a valid permutation

          and keep in mind to not add already added element

          → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
        ← Rev. 2   Vote: I like it -11 Vote: I do not like it

        C made my heart go up and down.

        → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I thought problem B was harder than problem C but in fact more people solved problem B. The fact that I know number of solutions of B must be of the count 0, 1 and 2 actually makes the problem harder.

        → Reply
»
2 years ago, # |
  Vote: I like it -11 Vote: I do not like it

The worst contest i had seen!

→ Reply
»
2 years ago, # |
  Vote: I like it +44 Vote: I do not like it

How to solve Div.1 C?

→ Reply
»
2 years ago, # |
  Vote: I like it +314 Vote: I do not like it

:)

→ Reply
»
2 years ago, # |
  Vote: I like it +8 Vote: I do not like it

i devoted more than 1 and half hout to problem B and ended up getting wrong answer . It just screwed up . But still enjoyed it.

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +70 Vote: I do not like it

I can't blame authors for my mistake but I think it wasn't obvious from the statement of problem div2C that queries must be performed in given order

→ Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +14 Vote: I do not like it

    What? I sorted the queries by length and wasted 1.5 hours and about 10 wrong submissions. I debugged on almost 50 test cases.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    I spent about an hour trying to solve and I still didn't know this was the case until I read your comment

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Glad to see I wasn't the only one. Took a long time to figure out, though I place all blame on me and none on the author. I have a bad habit of rushing the reading of the longer questions.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    I think many have done this, and I feel coloring according to our own order is slightly more difficult question, I think I correctly solved that version, but the real question is very much easier and straightforward!

    → Reply
»
2 years ago, # |
  Vote: I like it +24 Vote: I do not like it

Anyone has an idea for pretest 6 of div2C?

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    It contains large values for each $$$l_i$$$. Using ints and adding them up will cause overflow, leading to WA.

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

It's not as crowded as recent contests but main page still can't load properly :(

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

i don't get it, seems like most struggled in this round , why would there be such a huge difference in div 2 rounds ?

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

was it a normal DIV2 contest?

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

The moment when you get runtime error in Div. 1 C in the last 5 mins....

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

How to solve Div1B?

→ Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +11 Vote: I do not like it
    Spoiler
    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    The main observation is that the numbers in your array must have strictly increasing MSBs (you can prove this by induction). From there, you can iteratively calculate the number of arrays with MSB at most $$$2^i$$$ for each $$$i$$$, and it's a slight modification once you have to do the MSB of $$$d$$$.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    Sequence is ok iff msb($$$a_i$$$)<msb($$$a_{i+1}$$$). It can be calculated with simple dp.

    → Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +42 Vote: I do not like it

    An array that satisfies the given constraints mandates that the MSB of an element is greater than it's previous. No two elements can have the same MSB.

    Let $$$A[i]$$$ be the number of elements that MSB at $$$i$$$th position that are less than $$$d$$$.

    The answer is : $$$ ( (1+A[0])*(1+A[1]). . .*(1+A[32])) - 1 $$$.

    We choose one number from each MSB bucket. We subtract the case where we choose 0 elements. We also take modulo $$$m$$$ at each multiplication step.

    → Reply
»
2 years ago, # |
  Vote: I like it +109 Vote: I do not like it

Problem B(Div1) D(Div2):

You can't find this in OEIS

Me: 1 Hour trying to find it in OEIS

→ Reply
»
2 years ago, # |
  Vote: I like it +15 Vote: I do not like it

VeryLongCodeWithoutBugForces

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

Anyone got an idea for Div2D ? Thanks :)

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    Main idea: We can find an array b for a given array a iff, for all i, a[i] has a larger leading bit than a[i-1].

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

Super balanced contest. Apart from that, ABD seem pretty nice, C could be but getting to right implementation can be tough and the constraint $$$2^20$$$ seems too tight.

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

    After being unable to correctly implement my idea for a while, I realized it seems equivalent to a greedy which subsequently passed.

    Apply f on the topmost node such that after applying f there is no 0 in depth <= g. Very easy $$$O(nlogn)$$$ implementation for $$$n=2^h$$$.

    Hopefully correct.

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

      Probably correct. I simulated that in a much more complicated way (and forgot that $$$G$$$ isn't $$$H-1$$$ because the sample doesn't contain that), but your example supports my point that a lot of this problem's difficulty is getting bogged down in ugly implementation.

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Yeah, my code got extremely messy exactly due to $$$G$$$ not being $$$H-1$$$ as well. I didn't enjoy the problem either, but I guess it's still nice that a short correct code exists.

        → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      Proof: Consider the first time we make an operation not on the root, but on the path that making an operation on the root would modify. But making an operation on the root would be strictly better, as then all vertices would contain a smaller number. Also, applying that operation immediately after the last operation on the root instead of where it was originally done doesn't change anything.

      Thus, we can first greedily apply operations to the root, then independently solve all branches of the path operating on the root would modify.

      → Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 4   Vote: I like it -8 Vote: I do not like it

    Actually my solution was super simple to code. I guess there are many ways to think about this problem, because mine solution is significantly different than one described by Enchom

    Its code is here: https://ideone.com/1XiSlu

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +55 Vote: I do not like it

    Personally I don't think that it's ok when D1B level problem becomes D1D level because you have to restore the answer.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      Well, restoration of the answer in my solution is

      for (int i = n; i >= 1; i--) {
  if (!taken[i]) { cout<<i<<" "; }
}

      where taken[i] means that I take number a[i] in the final answer :) (which I guess everybody needed to determine in order to get the first number right)

      Or were you talking about Div1D and somehow messed up which comment you reply to :P?

      EDIT: Ok, I think it is me who messed up xD. I kinda thought discussion under this post is about Div1C only, based on comments above. Don't mind me :<

      → Reply
»
2 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Div1 D solution? Seems really hard. Seen a ton of similar problems but none of the usual techniques worked. Probably missing an observation of why some greedy part is correct.

→ Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it +10 Vote: I do not like it

    Number of operations we have to make is

    $$$ 1 + \max(\lceil\frac{\text{# of adjacent pairs of same character}}{2}\rceil, \max_{c \in \Sigma} (\text{# of adjacent pairs of character c})) $$$

    As each operation decreases both quantities by at most one.

    Greedy that works is always choosing the color with most adjacent pairs, then selecting some interval starting or ending with an adjacent pair of that color, and not both starting and ending with an adjacent pair of that color.

    Implementing this greedy in $$$O(n^2)$$$ is easy, but I couldn't implement it fast enough to not TLE :(

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    find all occurrences 'aa' for all characters. Lets say if these type of occurrences is maximum for character a currently, then delete substrings of type aa....bb or bb....aa till it is possible to do so. This process ends when there are occurrences 'aa' for only one type of a character. Though the implementation was a bit too frustrating for me.

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    You can reduce it to the following problem: You are given a string and in every step you can remove two consecutive letters if they are different or an arbitrary single letter and you want to make it empty in the smallest possible number of steps.

    Answer to that problem is max((length+1)/2, max frequency of a single letter)

    Reduction from original problem is that for every pair of consecutive equal letter, you put that letter into that string in reduced problem. It can be easily seen that operations of removing beautiful string in original problem is allowed operation in that reduced problem, so they are equivalent.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks. I made the reduction but for some reason the answer wasn't obvious to me. Quite disappointed that this is the solution to be honest.

      → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      And how to solve the problem after reduction? I found $$$\mathcal{O}(n^2)$$$ solution, but it shouldn't fit in TL if $$$n \approx 10^5$$$.

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

        Oh well, that is a bit tedious task. In each step you determine the letter with biggest frequency (in O(26) cause I keep them explicitly) and we want to remove one of its occurences with some other one. Then you iterate over all other letters (another additive O(26), but beware, it's amortized!, because of reasons that will be clear later) and check whether there is any place where these two letter currently touch. In order to perform a single check you keep two-dimensional array (with size 26x26) of vectors of pairs that tell you where are currently neighbouring occurences of these pairs of letters. When you decide to remove some pair you need to remove 3 pairs and add 1. However removing is a bit inconvenient, so I do that lazily. That is, I do nothing in that moment of removing that pair and whenever I see some candidate for a neighbouring pair in one of these 26x26 vectors, I check whether this pair is still valid (that is, both letter are still alive) — if it is then I use it, if it isn't then I remove it. In order to know what new pairs of neighbouring letters are created I use my own bidirectional list. In order to restore indices in current string (based on global ones that I use everywhere) I use another bidirectional list for alive original positions and Fenwick tree.

        I hope my code can be of some help: https://codeforces.com/contest/1329/submission/75411250

        → Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

How to solve Div 2 D ?

I tried to do kind of cheating using Berlekamp-Massey algorithm to extrapolate the sequence manually but didn't work out.

Also if anyone got AC using BM algorithm please post your solution link.

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

How to solve div2 B? My submission gives TL on pretest 2.

→ Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    maybe you used memset in each test case

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

      yeah, is it wrong?

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

        you have 10^4 test in each test case and you used memset on a array size 2e5 so it become n^2

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it -6 Vote: I do not like it

          Using memset for size n won't give tle as sum of n over all testcase always less than 200000. And all a_i is also less than n.

          → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          for clearing the arrays in independent testcases its better to use fill function . this will take the time of size n given in the input . It doesn't lead to tle since sum of n over all testcases is <=2e5.

          → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I also used memset in each test case... 4 times (whoops? :)) )... and I passed all the pretests, so that shouldn't be the problem.

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          you just pass the pretests :))) but if you pass the protests that means memset is not the problem

          → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          If you use memset for size n then will never give tle but if you use memset for size 20000 in each case then you will surely get tle.

          → Reply
          • »
            »
            »
            »
            »
            »
            2 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            I used size 200000 each time. So... I guess a sweet TLE is awaiting me? :)): ...

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              2 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              I hope as 10^9 iteration is very large to complete in 2 seconds.

              → Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 8   Vote: I like it +1 Vote: I do not like it

    You can use a multiset and a set
    Create a multiset and a set that will be used as the prefix
    Create another multiset and a set that will be used as the suffix
    Iterate through the input array add the current value to the multiset prefix and the set prefix too
    And remove the value from the multiset suffix and from the set suffix if the multiset doesn't contain the current value again
    Now to check if a state is valid the multiset prefix size should be the same with the set prefix size and the largest element in the multiset prefix should be equal to the size of the multiset prefix check same for the suffix(multiset and set)
    If valid add i and n — i — 1 to the result

    Code
    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Thanks a lot, I had a same algorithm but with memset, so I got TL(

      → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      you could use a vector and sort it then go throw it and check if each element is equal to its index+1

      → Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 4   Vote: I like it +2 Vote: I do not like it

    Traverse the array until you get 1st repeated element. When you get that break the array into two array at the point where you get 1st repeated value. Then simply check if two array is permutation or not.

    Repeat the process starting from back.

    → Reply
»
2 years ago, # |
  Vote: I like it -13 Vote: I do not like it

In problem B, it was clearly written that two permutations are of non zero length, but this surely wasnt the case.

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

    No, both permutaion length was >0

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    are you sure about it?

    → Reply
  • »
    »
    2 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    But they are of non zero length. a problem with the statement was the constraints saying both l1 and l2 <= n , not only their sums.

    still , they are of non zero length , maybe you adding that to your code only fixed another bug.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Possibly, but adding an empty permutation somehow passed that testcase.

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        then unfortunately you probably will fail the main tests. this test is clearly wrong for you:

        1

        2

        1 2

        it should output: 0

        while yours probably from what your saying will output:

        1

        2 0

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          1 2 1 2 answer for this should be 1 2 2 why it is wrong?

          → Reply
          • »
            »
            »
            »
            »
            »
            2 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            my bad , i am not sure why the comments here don't end lines like intended , fixed now.

            → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    And Poor me, couldn't figure out if l1=l2 they are considered as same permutation in an hour

    → Reply
»
2 years ago, # |
  Vote: I like it +38 Vote: I do not like it

I read div1A statement and for some reason, I thought "well, it's optimal to sort the values in decreasing order", but i never read that I wasn't allowed to change the order in which I use the array values. I realized about this when there were 10 minutes remaining of the contest. So sad to lose rating for that silly mistake. Goodbye div1, it was nice to meet you :(

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    If you're truly worthy you'll get it back soon :)

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I did the same, and then after 30 minutes I realized that the queries should be done in order :/

    → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Same here

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

How to do problem B div2?

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    First try checking valid sequence like all elements are 1 to n-1 , atleast there is one element with count 2. After that you can check till particular index if max=index and min=1 and sum=sum of permutation till index than insert in ans Same way for reverse array Take common elements

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks!

      → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Hey can you tell me why this fails ? link

      Thank you!

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Because you are checking validity of permutation using sum of first x numbers. According to your code, both the below testcase will be considered valid permutation because their sum is equal to the sum of the first 5 elements. Sum({1 2 3 4 5}) = 15 [Valid] Sum({1 3 3 4 4}) = 15 [Not Valid]

        → Reply
        • »
          »
          »
          »
          »
          2 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          I submit 3 wrong submissions while contest due to this, but finally Igot the point where I am stucking

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

Never noticed that the input is not a permutation in problem C, so sad......

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

What are the solutions for Div2C and Div2D?

→ Reply
»
2 years ago, # |
  Vote: I like it +7 Vote: I do not like it

How to solve Div2 C ?

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

    You are given m segments of different colors and you can slide these segments in a row of size n.
    The maximum length we can get by spreading these segments is equal to the lengths of all the segments(i.e without any overlapping), let denote it by sum.
    So if sum is less than n, it is not possible to cover the whole row.
    Now if sum >= n we have to compress it to fit in the row size of n. We have to overlap some units of the cells(possibly zero) such that all the colors are visible.
    The number of units cells to be overlapped is simply sum-n.
    So start placing the segments from start of the row and after leaving sufficient units of cell place next segment(trying to overlap as much as possible but less than required length), and keep doing this for all the segments maintaining the required number of cells to be overlapped.
    And if at any iteration the end of segment goes out of the window of size n, in that case answer is not possible.
    My solution with above approach

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      thanks brother it really helped...

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

Does it take some time for problems to appear in the Problem Set after a contest is over? I think I finally found out what I was doing wrong in my Div2B code after an hour smh.

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

1A(2C):hacker's paradise

→ Reply
»
2 years ago, # |
← Rev. 2   Vote: I like it +42 Vote: I do not like it

Tried hacking O(M^2) solution on A (http://codeforces.com/contest/1329/submission/75404901) with ~2.5B operations. Passed in 1.45s (facepalm).

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

    It looks like the optimizer changes while(ans[i] + a[i] - 1 < lst) ans[i]++; to ans[i] = lst - a[i] + 1 ?

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

      No, since if it did, the code would run much faster than 1.45 seconds. It's just that the operations are very simple and the processor manages to execute them in time.

      → Reply
»
2 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

In problem c (div2) i tried following approach : first sort in descending order and then choose 'p' value such that first position of the current coloured segment does not coincides with first position of last coloured segment.my submission : link .can someone tell the mistake in approach ?

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

    You can't change the order

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      I sorted back according to the indices so that order remains .

      → Reply
      • »
        »
        »
        »
        2 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Colors are painted in the order of input, so you can't firstly paint one with largest $$$l_i$$$, then second largest etc.

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

          Thanks , you are right . Because changing the order will cause some colour disappear when we reorder them to print . I removed sort lines from my code and it got accepted . submission

          → Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    the order of coloring will remain as given.

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

How to solve Div1B/Div2D?

→ Reply
  • »
    »
    2 years ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    Use the observation that the required sequence is essentially a sequence of numbers such that the most significant bit is in increasing order. So, the max length of this sequence is at most 30.

    → Reply
    • »
      »
      »
      2 years ago, # ^ |
        Vote: I like it -35 Vote: I do not like it

      Oh god, I got that observation before, but I just stopped thinking and doubting the problem statement which says it can't be found in OEIS, lol. I started spend too much time seeing similar pattern in OEIS and looking for how to code it. I should've just trusted the problem statement, what a nice lesson learned.

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

        Or you know, don't use OEIS to solve problems.

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

Wow, that's a pretty bad contest if I've ever seen one. Speedforces for a lot of blues, who will now undeservingly be candidate masters.

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

Can anyone tell what could be the pretest 2 for DIV2 B...

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

WOW div2D/div1B is quite tricky. You can find the sequence on OEIS this sequence but only those few numbers in front of the sequence is right for this problem. So tricky/QAQ

→ Reply
»