tibinyte2006's blog

CodeTON Round 3 (Div. 1 + Div. 2, Rated, Prizes!)
By tibinyte2006, history, 2 years ago, In English

Hello, Codeforces! Or, as we like to say in Romania: Nu îmi amintesc să fi adresat întrebări, Codeforces!

I am glad to finally invite you to participate in CodeTON Round 3 (Div. 1 + Div. 2, Rated, Prizes!), which will start on Nov/06/2022 17:35 (Moscow time). You will be given 8 problems and 2 hours and 30 minutes to solve them. It is greatly recommended to read all the problems, statements are short and straight to the point.

I would like to thank:

Scoring Distribution: 500-750-1250-1750-2250-2500-3250-3500

The editorial has been published here!

And here are our winners!

And here is the information from our title sponsor:

Hello, Codeforces!

We, the TON Foundation team, are pleased to support CodeTON Round 3.

The Open Network (TON) is a fully decentralized layer-1 blockchain designed to onboard billions of users to Web3.

Since July, we have been supporting Codeforces as a title sponsor. This round is another way for us to contribute to the development of the community.

The winners of CodeTON Round 3 will receive valuable prizes.

The first 1,023 participants will receive prizes in TON cryptocurrency:

  • 1st place: 1,024 TON
  • 2–3 places: 512 TON each
  • 4–7 places: 256 TON each
  • 8–15 places: 128 TON each
  • …
  • 512–1,023 places: 2 TON each

We wish you good luck at CodeTON Round 3 and hope you enjoy the contest!

  • Vote: I like it
  • +552
  • Vote: I do not like it

Comments Comments (199)
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2 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Good luck in LeafTON round 3.

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2 years ago, # |
  Vote: I like it +83 Vote: I do not like it

omg green round

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

As a tester, I am sure you will find some interesting problems whatever rating you have and I wish you good luck and high rating!

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

First of all, as a tester, I tested. Secondly, I strongly recommend this round!!!

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2 years ago, # |
← Rev. 2   Vote: I like it +12 Vote: I do not like it

As a tester, I'm already yellow, not purple!

Upd: Good luck to all participants! tibinyte2006 orz

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2 years ago, # |
  Vote: I like it +45 Vote: I do not like it

The contest's duration is 2.5 hours in blog but 2 hours in Current or upcoming contests?

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2 years ago, # |
  Vote: I like it +61 Vote: I do not like it

As a tester I suggest everyone to participate in this contest. Even I want to participate in it but I obviously can't :(

I wish everyone good luck and a good experience

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

    As a sad I can't participate in the round myself I'm a tester too!

    good luck everyone! hope you find some interesting problems and have a good time

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

As a tester, Problem A is one of the problems of all time.

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

GREEN>>CYAN

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

statements are short and straight to the point. finally it's in the blog. Thanks

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

As a newbie , it will be more fun to participant in a green author round . Best of luck all.

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

As a tester I tested

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

Leaf forces

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

Excited for a green round :)

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2 years ago, # |
  Vote: I like it -55 Vote: I do not like it

Well, it's obvious that tibinyte2006 is a talented programmer, but he just managed to get negative deltas to become green to pretend to be weak. Also, he is just the blog sender, not one of the problem setters.

So, I wish you guys get positive deltas & have fun!

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    2 years ago, # ^ |
      Vote: I like it +97 Vote: I do not like it

    As a tester, tibinyte2006 is the main author of this round. Please do not spread misinformed opinions.

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    2 years ago, # ^ |
      Vote: I like it +27 Vote: I do not like it

    Capital L

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

    As a Tibinyte's friend I can tell you that he has been developing this round since long time ago and he's the main author of round. Please do not spread misinformed opinions. (and reach master)

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

    As a Leafeon supremacist, I can confirm that this claim is baseless. Also green is the best color because Leafeon is the best Pokemon

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

    Okay I'm sorry for that, I apologize for this.

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    2 years ago, # ^ |
      Vote: I like it +48 Vote: I do not like it

    Personally, a round is the result of everyone's joint efforts. Whoever thinks of the idea, who thinks the solution, who proves the correctness of the algorithm, who perfects the statement, who creates the powerful test data, and who fixes the issue is actually unnecessary so clearly. What matters is our joint efforts and our friendship.

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      2 years ago, # ^ |
        Vote: I like it +13 Vote: I do not like it 
      LLLL
LLLL
LLLL
LLLL
LLLL
LLLL
LLLL
LLLLLLLLLLLLLLLL
LLLLLLLLLLLLLLLL
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    2 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I believe you had a huge surprise...

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2 years ago, # |
  Vote: I like it -32 Vote: I do not like it

Is it rated?

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

How do they'll send cryptocurrency? We have to have crypto wallet?

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

As a tester I have nothing more to add than "Good luck!" and "May the God have mercy on you", because the problem setters surely don't know the definition of this word.

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

Time to become CM now ;)

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

G Gaming

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

Time for Remontad

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

Oooh wow! 58 people contribute to making this round happen.

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

Aiming green in the green round... Wish me luck :)

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

Wish me luck hope i become green after this contest.

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

Hope I will get GM. :)

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    2 years ago, # ^ |
      Vote: I like it +33 Vote: I do not like it

    Oh.. you are close to 2400. Wish you luck for today's contest, hope you become GM..

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

    Same D:

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

      Frate, iti zic io, ajungi si tu (grand)master.

      Marinush

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

        I have just realized that I read statement of problem E incorrectly, I thought that we can shift only whole string D:

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

    Congratulations on reaching GM!!

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2 years ago, # |
← Rev. 2   Vote: I like it -41 Vote: I do not like it

round need to be unrated, because it is round

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

How do they'll send cryptocurrency? We have to have crypto wallet? someone can answer to this question plz!

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

    Create a wallet on this app; this'll give you a wallet address

    https://tonkeeper.com/

    P.S. Checkout the TON username market on telegram : https://fragment.com/ where you buy and sell in TONs

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

      I've created my wallet address and update my profile but did not receive the prize yet. Would you like to tell me the reason or what should I do to receive the prize?

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

    Yes, you'll have to provide your wallet address to receive prize.
    Here is the list of TON wallet applications: https://ton.app/wallets

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

      Who should I provide my wallet address to?

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

        In CodeTON Round 2 there was a Codeforces message from System, containing link to form. So I think you should wait for that message (it will be visible in notifications panel).

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        18 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        me

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

Will score distribution be announced briefly before the beginning of the round?

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

    It will be posted along with the editorial as to not spoil anything about the round

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

Hope that I will be Pupil after this round

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

Score distribution?

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

In the previous CodeTON rounds, I lost my rating. Hope Today I will gain plus point.

Love to see short statements and straight to the point.

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

let's go

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

time to + 100

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

Good Luck!!!!!!!!!

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

Last few contests gave me cancer

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2 years ago, # |
  Vote: I like it -16 Vote: I do not like it

What is the conversion ration TON to INR ??

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

How to optimize E from O(n^2) to O(n) or O(nlogn) ?

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

Good problems, very good D and E was easy

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

How to calculate the number of numbers from 1 to M coprime to N ?

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

Took me 40 minutes to search and read about Mobius function(Problem D). Not sure if there is any solution that doesn't utilize it.

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

    Use the Inclusion-Exclusion Principle.

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

      Yeah, I thought about that one too. But I believe that I won't be able to implement in time.

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

      I thought that would maybe time out. Took me an eternity to get the Mobius solution to work.

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

        It won't be time out. Think about this, this principle works on a[i]/a[i+1], and the product of them is a[1]/a[n] which isn't more than 1e9.

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

        It won't time out because, you will apply factorization and inclusion exclusion stuff only when $$$a[i] != a[i-1]$$$. The prefix gcd can only change $$$O(log(m))$$$ times, because everytime the gcd changes, it has to reduce at least by a factor of $$$2$$$.

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

What was logic for C , Solved D but was not able to process C's logic.

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

    Consider the xor of the two arrays, you will notice something ...

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

    Some useful observations are:

    1. If every index is equal/unequal in both strings, then it's possible.
    2. When an operation is performed, either bit in string a, or bit in string b changes, so if all of the index were equal in both, then it becomes unequal, and vice-versa.
    3. You just have to make all index 0 in string a, let's say. So, string b can be all 0's or all 1's. If all 0's then we're done.
    4. If not, then just add 3 extra operations: (1,1) (2,N) (1,N)
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2 years ago, # |
  Vote: I like it -8 Vote: I do not like it

How to round number in D?

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

    If you think of the number I think of, you have to round it down :)

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2 years ago, # |
← Rev. 2   Vote: I like it +25 Vote: I do not like it

Upvoted the contest for the story behind Problem E. I couldn't focus on the Problem, because I had to show the story to my fiancée and my brother and needed to hype it. Beautiful Daemon Targaryen. I would upvote more if I could.

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

Did anyone solve D using Euler's totient function?

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2 years ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

I used about 10 minutes to think about the conclusion of problem E but used more than an hour on coding. Anyway, I think this round is the best round I've ever participated.

UPD: after I solved problem D I got rk20. :) (though my final rank is 200 :(

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

HAHAHAHAHAHAHAHA

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

Literally dying atm. Could get a grasp on D (Yes, I did think of inclusion-exclusion here and there but did not find a way to preprocess the values of the mobius function. Did I want to copy paste an overkill $$$O(n^\frac{2}{3})$$$ method to preprocess prefix sums of the mobius function? Hell no.) but couldn't finish. C felt hard, like, very hard. Might be even convinced that C > D if C was after D.

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2 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it 
is it the approach for problem c ?

If A^B == 0000 or 11111 then answer exists 

Then we shift all ones to left side and zeros to right for A and do same operations for B also.
which makes A as 00001111 and B as 11110000 something like this
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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The answer is only possible if for all i either a[i]=b[i] or a[i]!=b[i]. Just make all a[i]=1,you'll observe that all b[i] become either 0 or 1.If they are all 0 just do 1,n so all a will be 0 and b will remain 0,else if all b are 1 then do (1,1) (2,n)

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

    Yes. I did operations on $$$[i,n]$$$ for all $$$i$$$ such that $$$b_i$$$ and $$$b_{i-1}$$$ are different. Then $$$b$$$ consists only of $$$1$$$s or only of $$$0$$$s. The same goes for $$$a$$$. Then you just need to look at $$$a_1$$$ and $$$b_n$$$ (they did not get changed yet) an accordingly make everything $$$0$$$.

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

I always lose rating in div 1+ div2 contest any suggestions…?

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

    I guess, in div 1 + div 2 contests it is more important to solve the earliest problems fast enough (since there will be more people solving the same amount of problems, relatively speaking). Interestingly, for me it is the other way round: I always win rating in the joint contests...

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

How to solve E ?

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

    Take a segment. Let's call $$$C$$$ the number of closing brackets and $$$O$$$ the number of opening brackets. Let's call $$$M$$$ the minimum value the prefix sum of the segment reaches (Opening bracket is $$$+1$$$, closing is $$$-1$$$). The cost of a segment then is $$$max(0, O-C)+min(0, M)=max(0, O-C)+M$$$.

    Explanation

    You can calculate both parts of this sum for all segments idependently by iterating and doing confusing calculations. See 179633595 the two blocks after long long ans=0;.

    For me it was easier to visualise the bracket sequence as a graph of the prefix sum and then look at area on the graph which both of the terms contribute.

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

      Very clean code and solution, thank you! I knew that the solution is very clean, but did not have the skill to find it tho.

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2 years ago, # |
  Vote: I like it +65 Vote: I do not like it

E was nice. Until the last 40 minutes, I thought that the first operation can be only applied to the entire string. That led to a much more complicated problem. :(

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

    Wow... I was thinking and thinking with that misunderstanding of the first operation for the whole contest. I got to read the statement a bit more carefully.

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

    Me too, I have written two brute-forces to understand that it's not the case...

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

    And I realized only after reading your comments that we can apply to any substring :(

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

How to solve problem E? The most I can do is a brute force O(n^3) solution.

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

Great contest!Congrats to the authors!Cool problem D, unfortunately I didn't solve it.I'm waiting for editoiral!Thanks!

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

Me After Solving A & B :')

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

Can anyone please explain how to solve problem C with intution?

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

    Consider A^B. Every time we conduct an operation A^B would be XORed by 111...1. So for a YES case, A^B would either be 000...0 or 111...1 after whatever operations. Now you just need to make A to 000...0. In the end, B is either 000...0 or 111...1. Then it's easy to make a few more operations to get job done.

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

Why the time limit of G is 3s... My FFT solution passes. It does not seem to be the intended solution.

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2 years ago, # |
  Vote: I like it +149 Vote: I do not like it

Congratulations to conqueror_of_tourist for not only living up to their username, but also becoming LGM with Python!

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

    you also achieved the goal of your username with python only. Screenshot-2022-11-06-200424

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

Countforces

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

I wasn't able to solve A, with just looking at the problem and understanding it, I saw that in the correct cases, position 0 had an 1 or position 1 had a 2, so intuitively I went on and did an algorithm that checks that case. Too bad for me it was just when it started with 1.

But can someone explain why is it when it starts with 1? I did it so if it's 1 or 2 at the first 2 positions, i.e. arr[0] == 1 or arr[1] == 2 then yes, else then "No".

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

    before u completely look at my answer read the question once for clarity..... constraint : i < j < k 1st position element has to smaller than kth position element element at 1st position cannot be changed.... but using 1st position you can swap any positions of j and k .

    so the smallest possible element that can take 1st position must be one .

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

Problem D is similar to this problem which I proposed to codechef before.

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2 years ago, # |
  Vote: I like it -25 Vote: I do not like it

Sorry, but I didn't like this round at all:
A: just one simple observation
B: just one simple observation
C: just one simple observation
D: just find number of coprime with n not greater than m (testers in comments, is it interesting? Really?)
E: closed this problem as soon as I've read. Yet another "count some *** for all subsegments"
Can say nothing about F+, because didn't read
No, don't misunderstand me, each problem is good. But as set of problems in one contest...

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    2 years ago, # ^ |
    ← Rev. 5   Vote: I like it +38 Vote: I do not like it

    I'm sorry that you found the first 2 problems of the contest(which are supposed to be easy) too easy for a candidate master / master, I will talk to tibinyte2006 and kindly ask him to give out some smart and well hidden minimum cost maximum flow with a lot of observations and implementation details for problems A and B in his future rounds. As of C, consider it a simple observation for you, judging by the counter of solves it was just as hard as it had to be and I also needed some time to figure it out(but I might also have some skill issues so ignore me if you want). I really liked problem D because of its "smooth" solution that included some simple math and observations, so I found it interesting and I am sorry that this does not fit your view. Your argument about problem E is probably by far the dumbest thing I read on the internet today and this really means something as I've seen a lot of dumb memes today. Anyway, I am sorry that you didn't enjoy the round and wish you good luck in the next ones.

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

      First and foremost, L

      Second, this is not how you argue, this is how you correctly argue this is satire, by all means the purpose is not to offend anyone

      Dear contestant,

      Please don't worry if you had to little to no observations to pass some problems that were set to not cater to you. In a Codeforces round, the mechanics employed by some problem do not by any means determine the quality of any problem*. Apologies if this does not fit your view.

      Thanks for the feedback,

      Tester of Codeton Round 3

      *: (although, admittedly, it can determine the quality of the round as you stated-- I won't argue with that because I am not in the mood for jejune disputes)

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

      you found the first 2 problems of the contest too easy for a candidate master / master

      Where have I said this?

      consider it a simple observation for you

      It's my point: this is "one observation problem". You either find this observation->"ez ac" or you don't find this observation at all->you can't solve problem. This argument is applied to A and B

      I really liked problem D

      It's your personal opinion and I've shared mine.

      Your argument about problem E

      It's not argument, it's my personal opinion. I don't find problems which require some counting on all subsegments interesting. I didn't solve D incontest because I'm dumb, but if I solved D, I would skip E.

      The dumbest thing is "This contest is the most bad/good/... contest ever" comments without any details. I tried to share my opinion about whole problemset why I didn't like this contest. And seems that you misunderstood my initial comment. Once more: every problem is good. But you don't find problemset good if every problem is about, for example, geometry, do you?

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        2 years ago, # ^ |
        ← Rev. 5   Vote: I like it +8 Vote: I do not like it

        "you found the first 2 problems of the contest too easy for a candidate master / master" — I just wanted to point out that it is a normal fact that you found them easy because you have a higher rating than the people that should find them challenging, I didn't want to say that you mentioned this, because you didn't.

        "It's my point: this is "one observation problem". You either find this observation->"ez ac" or you don't find this observation at all->you can't solve problem. This argument is applied to A and B" — I find this quite common in the modern day competitive programming where problems are more and more based on tricky observations, I wouldn't say that I love it, but we should adapt to these kind of problems if we want to do CP, and if not, as we say in Romania, "ayaye".

        "It's your personal opinion and I've shared mine." — I respect that.

        "I don't find problems which require some counting on all subsegments interesting." — I get that, but in a contest(especially in a serious one with a higher stake like OI or ICPC) you can't just say that you don't like a problem, you just have to try to solve it because you wanted to take part in that competition(I will assume that no one can force you to take a contest if you do not want that).

        "I didn't solve D in contest because I'm dumb" — I know this is a bad feeling, but you probably just missed out on something, it happened to all of us, no need to be so harsh about it.

        "The dumbest thing is "This contest is the most bad/good/... contest ever" comments without any details" — I agree that this is usually toxic and people tend to say that after a round that got them positive delta, but my opinion as a tester was just that the round was good and recommended others to try it.

        "But you don't find problemset good if every problem is about, for example, geometry, do you" — No, I do not. I also recommend that you spend some time analysing the problems with editorials and realize that they are not just some combinatorics / dp problems, but also require some clever observations. Also, A-C are not counting at all and E and H are not combinatorics / dp(as I know).

        Anyway, I think we discussed more about this than it was necessary. The contest is done and others will come.

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

          Thank you for discussion!

          Just one correction: in ICPC you have wider set of topics to choose from and what is more important you have teammates. This allow you to skip/delegate some problems if you don't like it more often, than in personal contests.

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

Why is my output on the code I wrote when run on my compiler different from the one used in the contest ?

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

Great contest

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

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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

    Hope updated ratings would changed my rating by 2 points

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

      Hope updated ratings would change my rating by 54 points

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

problem : https://codeforces.net/problemset/problem/1750/D

input : 4 1000000000 60 30 1 1

output: 595458194

how? can anyone explain?

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

Personally, I think this E is very good. I can use BIT or segment tree, or even deduce the expression O(n) directly. I really like this problem

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

in problem D

if m was <= 1e6 , could we solve the problem using Möbius Function?

something similar to this problem
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2 years ago, # |
  Vote: I like it -12 Vote: I do not like it

My friend rating is 1110 and rank in contest is 9000 by solving only problem and by rating is 926 and got 8200 rank by solving 2 questions...But the worst things my rating decreases only by 6 and my rating decreases by 28....How is it even possible.....Please help anyone

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

    You get more rating points than you should first 5 or 6 contests you participate in, which is most likely to be the situation with your friend.

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2 years ago, # |
← Rev. 3   Vote: I like it +11 Vote: I do not like it

Alsalam Alykom,My solution for A,B is skipped and i swear i didn't use ideone or another account or anything its my clear solution ! I swear to god i didn't cheat why something like that happen to me ? how to avoid this MikeMirzayanov tibinyte2006

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

How do I receive my money?

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2 years ago, # |
  Vote: I like it +65 Vote: I do not like it

Where is my ton please?

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2 years ago, # |
  Vote: I like it +68 Vote: I do not like it

How can I receive my money?

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

Has anyone received the prize yet?

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

Has anyone received their TON yet? The deadline was supposed to be on Nov 26.

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

    Hi. The wallets were collected and sent to TON. Of course, they need several days to process the data.

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

Is the team still in the process of sending prizes? Or is it already finished and I didn't receive anything due to some mistake from my side?

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2 years ago, # |
← Rev. 2   Vote: I like it +31 Vote: I do not like it

I just received my prize a minute ago. Maybe you guys will receive it soon

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

Will this round send NFT trophies? Like the kind of NFT trophy in round 1 and round 2.

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