dreamoon_love_AA's blog

Codeforces Round #631 (Div. 1 and Div. 2) — Thanks, Denis aramis Shitov!
By dreamoon_love_AA, history, 14 months 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

Comments Comments (492)
»
14 months ago, # |
  Vote: I like it +13 Vote: I do not like it

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

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

    But, be careful of test 1

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

        Joke has gone above your head.

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

          Indeed you are quite right

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

    I see you know what meme is good

    → Reply
  • »
    »
    14 months 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
    • »
      »
      »
      14 months ago, # ^ |
        Vote: I like it +4 Vote: I do not like it

      But not all of them show up in contest.

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

    ohh really?

    → Reply
  • »
    »
    14 months 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
    • »
      »
      »
      14 months 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
»
14 months ago, # |
  Vote: I like it +15 Vote: I do not like it

How many problems?

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

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

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

    I think there will be 5 problems for both division.

    → Reply
    • »
      »
      »
      14 months 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
      • »
        »
        »
        »
        14 months 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
        • »
          »
          »
          »
          »
          14 months 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
          • »
            »
            »
            »
            »
            »
            14 months ago, # ^ |
              Vote: I like it -12 Vote: I do not like it

            Maybe it's just a round without div.1 A So it will be extremely hard.

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

        .

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

      You mean for each division?

      → Reply
      • »
        »
        »
        »
        14 months 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
»
14 months ago, # |
  Vote: I like it +21 Vote: I do not like it

wow Setting Problem After such a long time

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

Didnt invited sorry_dreamoon?

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

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

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

it better be virus themed questions

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

    there is no secret to it. just solve more tasks

    → Reply
  • »
    »
    14 months 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
    • »
      »
      »
      14 months 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
»
14 months 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
  • »
    »
    14 months 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
  • »
    »
    14 months 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
»
14 months 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
»
14 months ago, # |
  Vote: I like it +86 Vote: I do not like it

dreamoon_love_AA here I come!!!

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

to understand the meme for none one piece fans

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

tourist vs Retired_MiFaFaOvO. Who will win ?

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Excited! Looking forward to your problems.

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

.

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

Who is Denis Aramis Shitov????

→ Reply
»
14 months 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
  • »
    »
    14 months 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
»
14 months 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
»
14 months ago, # |
  Vote: I like it +31 Vote: I do not like it

Let us see if this round can get 25000 registrations

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

Thank You Denis Aramis Shitov

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

tourist vs. Retired_MiFaFaOvO It will be exiting!

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

Thank you mr. Denis Aramis Shitov

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

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

I hope I can do well in this round

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

is it rated?

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

Time is not good for me!

→ Reply
»
14 months 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
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Wow！It seems good！

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Will it be a Mathforces Round?

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

I hope its going to be an interesting round

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

Looking forward to seeing concise and crap free problem statements.

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

Thank you Mr.Denis Aramis Shitov.

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

.

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

thank you Denis aramis Shitov and to all other contributors.

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

Friendly reminder: watch out for FST.

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

What's the score for each problem?

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

.

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

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

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

Hope I will turn blue after this contest.

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

When the problem scroring will be posted?

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

I think , this contest will be very intersting

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

good luck for every one)

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

Why hasn't the score distribution been published yet?

→ Reply
»
14 months 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
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Please post the score distribution of the problems.

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

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

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

Is it rated ?

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Enjoy the contest

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

site seems fast today

→ Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

lets do well guys ..best of luck..

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

noExplanantionsForSamplesForces

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

Is this the second april fools round?

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

→ Reply
  • »
    »
    14 months 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
    • »
      »
      »
      14 months 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
      • »
        »
        »
        »
        14 months 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
        • »
          »
          »
          »
          »
          14 months 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
  • »
    »
    14 months 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
»
14 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Is this round bit hard!!!

OR only i feel so?

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

Great contest, I hate it.

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

        C made my heart go up and down.

        → Reply
      • »
        »
        »
        »
        14 months 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
»
14 months ago, # |
  Vote: I like it -11 Vote: I do not like it

The worst contest i had seen!

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

How to solve Div.1 C?

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

:)

→ Reply
»
14 months 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
»
14 months 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
  • »
    »
    14 months 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
  • »
    »
    14 months 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
  • »
    »
    14 months 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
  • »
    »
    14 months 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
»
14 months ago, # |
  Vote: I like it +24 Vote: I do not like it

Anyone has an idea for pretest 6 of div2C?

→ Reply
  • »
    »
    14 months 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
»
14 months 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
»
14 months 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
»
14 months ago, # |
  Vote: I like it +3 Vote: I do not like it

was it a normal DIV2 contest?

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

How to solve Div1B?

→ Reply
  • »
    »
    14 months ago, # ^ |
    ← Rev. 2   Vote: I like it +11 Vote: I do not like it
    Spoiler
    → Reply
  • »
    »
    14 months 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
  • »
    »
    14 months 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
  • »
    »
    14 months 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
»
14 months 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
»
14 months ago, # |
  Vote: I like it +15 Vote: I do not like it

VeryLongCodeWithoutBugForces

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

Anyone got an idea for Div2D ? Thanks :)

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

    maybe you used memset in each test case

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

      yeah, is it wrong?

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

    No, both permutaion length was >0

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

    are you sure about it?

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

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

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

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

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

    Same here

    → Reply
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How to do problem B div2?

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

      Thanks!

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

      Hey can you tell me why this fails ? link

      Thank you!

      → Reply
      • »
        »
        »
        »
        14 months 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
        • »
          »
          »
          »
          »
          14 months 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
»
14 months 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
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What are the solutions for Div2C and Div2D?

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

How to solve Div2 C ?

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

      thanks brother it really helped...

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

1A(2C):hacker's paradise

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

    You can't change the order

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

    the order of coloring will remain as given.

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

How to solve Div1B/Div2D?

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

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

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

    Try your code for this input:

    input
    → Reply
    • »
      »
      »
      14 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I am getting

      1

      4 6

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

        It failed on mine :( I got:

        2

        2 8

        4 6

        which is definitely wrong because I forgot checking if all numbers from 1-m exist. Took me an hour to notice it RIP

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

    both l1 and l2 permutation of same number

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

    1 1 1 Answer is 1 1 1

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

      it was written in the problem statement that a[i]<n, hence this cannot be possible

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

    Try this input to your code

    Input
    → Reply
»
14 months 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
»
14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I spent an entire hour on C not realizing that all lengths must be FULLY painted (when at length 3, you cannot start painting at the next to last position for example). Even though it was very clearly mentioned in the statement. I'm mad at myself.

→ Reply
»
14 months ago, # |