fangahawk's blog

By fangahawk, 6 weeks ago, In English

Hello Codeforces!

We're glad to invite everyone to participate in Codeforces Round 940 (Div. 2) and CodeCraft-23, which will start on Apr/21/2024 17:35 (Moscow time). The contest will be rated for all the Div. 2 participants (Rating < 2100).

The Programming Club at IIIT-H organizes this event as a part of our techno-cultural fest Felicity @ IIIT Hyderabad.

Programming Club at IIIT-H

You will have 6 problems to solve in the duration of 2 hours 2 hours and 15 minutes. One of the problems is divided into 2 subtasks. Do ensure you read all of the problems!

There have been a bunch of people that have helped us in making this contest a (hopefully) great one!

Score Distribution: $$$500 - 1000 - 1500 - 1750 - 2250 - (2250 + 1250)$$$

Hope you enjoy the problems!

UPD: The editorial can be found here.

UPD: The Winners

Official

  1. Alphaqwq_
  2. misfits
  3. Monkey1ng
  4. Nanani_fan
  5. lmq3z

Unofficial

  1. BurnedChicken
  2. maspy
  3. kotatsugame
  4. zdc123456
  5. zjy2008
  • Vote: I like it
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  • Vote: I do not like it

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6 weeks ago, # |
  Vote: I like it +135 Vote: I do not like it

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6 weeks ago, # |
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Hmmm cool

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6 weeks ago, # |
  Vote: I like it +12 Vote: I do not like it

I hope I will not reach specialist in this round

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6 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

As a participant,hoping for clear and short statements

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6 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

Damn, never hoped that after Indian ICPC, IIITH Programming club would go international! Really looking forward to attend it!

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6 weeks ago, # |
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Hope to the positive delta after this round

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

bet

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

Will the score distribution of the problems be published? Or will the contest be held on ICPC rules?

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

    It will have a score distribution. We will publish it soon.

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5 weeks ago, # |
  Vote: I like it +33 Vote: I do not like it
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5 weeks ago, # |
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5 weeks ago, # |
  Vote: I like it -14 Vote: I do not like it

Hope so there is no moss for this :)

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5 weeks ago, # |
Rev. 2   Vote: I like it +51 Vote: I do not like it

As a tester I really liked the problemset and encourage everyone to participate in the round!

akcube fangahawk orz

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    5 weeks ago, # ^ |
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    Great to hear you enjoyed the problem set! Your encouragement really makes me want to dive in and give it a go. Thanks for the motivation!

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5 weeks ago, # |
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Can you post the score distribution as well?

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5 weeks ago, # |
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Hoping for pupil

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5 weeks ago, # |
  Vote: I like it -8 Vote: I do not like it

I will upvote if I become CM

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5 weeks ago, # |
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Wishing everyone a good codeforces round after a week.

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5 weeks ago, # |
  Vote: I like it +8 Vote: I do not like it

akcube round, orzzz!!!

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5 weeks ago, # |
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what algorithms book and resources gennady korotkevich (tourist) read?

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5 weeks ago, # |
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Will the score distribution of the problems be published? Or will the contest be held on ICPC rules?

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

Best of luck so that you can get good response for your questions, from IIIT Hyderabad.

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5 weeks ago, # |
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clear and short description we like it

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5 weeks ago, # |
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Minecraft !!

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5 weeks ago, # |
  Vote: I like it +83 Vote: I do not like it

IIITH Programming Club orz

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5 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

As a participant, I would like to get upvotes as minus looks ugly in my profile!

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5 weeks ago, # |
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1K stalkers and I will write this one.

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5 weeks ago, # |
  Vote: I like it +93 Vote: I do not like it

As a monkey tester, I was offered $$$6+9+6 \times 9$$$ bananas to write this comment.

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

I too wanted to join IIIT bcoz of the coding culture there, my clg sucks

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    5 weeks ago, # ^ |
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    Yeah I hear a lot about IIIT too....but then everything can't be helped I guess....at this point you should be contributing to building the culture there.

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

Hoping to reach pupi.

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5 weeks ago, # |
  Vote: I like it +30 Vote: I do not like it

Why codecraft-23 in 2024, isn't it codecraft-24?

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5 weeks ago, # |
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Best wishes for everyone's A, B, C, D

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5 weeks ago, # |
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When will the score distribution be announced?

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    5 weeks ago, # ^ |
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    what meant by score distribution

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      5 weeks ago, # ^ |
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      Usually contest happen in either icpc style, i.e. every question has equal weightage, i.e. whoever does any question first gets higher rank. Whereas normal contests on codeforces (non educational rounds) have ratings attached to them so that they are indicable that it is 50% chance that it can be done by someone having rating about equal to what's shown, and the if you do a higher rated question its more rank than doing first few questions.

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        5 weeks ago, # ^ |
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        Score distribution $$$\neq$$$ problem rating, those are two separate things.

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          5 weeks ago, # ^ |
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          But scores are proportional to the earned points ig

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            5 weeks ago, # ^ |
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            Usually, yes, but his explanation was still wrong, he mixed up the two systems. The score distribution does not tell you how much a problem is rated, only its value in the contest.

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              5 weeks ago, # ^ |
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              Yeah agreed problem ratings depends on the problems itself there is no way one can just guess without opening the problem and not even the order justifies that I have C labelled 800 rated problem

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

            To clarify, scoring distribution is different from the ratings attached to the problems post-contest. Ratings are given to problems after the contest is over based on number of solves, etc. and are indicative of the rating required to solve the problem in contest. Scoring distribution is just the points you as a contestant would gain if you solved it at the 0th minute. It's more or less only meant to be indicative of what the setters + testers approximate the relative difficulty of the problems to be.

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              5 weeks ago, # ^ |
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              Got it — "Ratings are given to problems after the contest is over based on number of solves"

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5 weeks ago, # |
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Will try to solve 3 questions !! Wish me luck :)

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5 weeks ago, # |
  Vote: I like it -7 Vote: I do not like it

ee sala cup namde

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5 weeks ago, # |
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Good luck everyone

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

I taught this was a programming contest,looks like a math test to me

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5 weeks ago, # |
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so much math i loved it, knew solutions within minutes of every question, but while implementation so many mistakes :(

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5 weeks ago, # |
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can someone pls enlighten me on how to solve D

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    5 weeks ago, # ^ |
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    it is clear from the question that we need for each y

    ax ^ ax+1 ^ ax+2 ... ^ ... ^ az-1 ^ az > ax ^ ax+1 ^ ax+2 ... ay ^ ... ^ az-1 ^ az.

    note in lhs, ay is missing so basically what we need is all those ranges(x, y) in which sum of MSB of ay is even.

    we need x, y, z. traverse through y, now for each y calculate no. good x and z pairs.

    suffix and prefix dps can be maintained for each bit for odd and even counts.

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      5 weeks ago, # ^ |
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      Is MSB = max set bit? That makes sense but I'm kinda confused about how you would implement the prefix and suffix dp. If you don't mind, can you please give me the gist? Appreciate it.

      Edit: oh wait nvm I see how to do it. Thanks broski

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        5 weeks ago, # ^ |
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        okay, both will have 3 states -> suffix/prefix, bit, odd/even. i will tell for prefix, suffix is similar. pref[n+1][30][2] (n+1, for simple implementation, 30 bits and 2 odd and evens)

        dp state pref[i][j][0] represents count of bit j till ith elements such that parity/zor of that bit is 0(even)

        basically

        a1, a2, a3, a4, a5, a6, a7 a8 ....

        if we are at a8, and we are taking a look at 4th bit (i.e., bit at this position -> (1<<4)) then if we take all suffixes ending at a8 and xor each suffix, the state represents no. of such suffixes that have THE 4th bit off.

        think for transitions it's not that difficult/look at my code., if you don't understand then ill write it. (if anything is unclear don't hesitate to ask or even dm me.)

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      5 weeks ago, # ^ |
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      why msb ?

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        5 weeks ago, # ^ |
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        because in a range for a particular bit there can either odd or even no of that bits. now, we are taking a range (x, z) and we have to remove any element from that range new range -> (x, y-1) ^ (y+1, z), if in new range there are odd bits, then old range will have even bits of MSB.

        basically i am structuring my solution such that it is dependent upon MSB.

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          5 weeks ago, # ^ |
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          ok,i know what you mean, just because remove aj will happen some change, and MSB will be make this change simple. thx

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          5 weeks ago, # ^ |
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          can you explain this part please, I get the pO*aE + pE*aO part but why did you add aO+ pO, can you please explain

          Spoiler
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5 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

I was way too close to solving problem C! Great problemset :D

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5 weeks ago, # |
  Vote: I like it -14 Vote: I do not like it

Thanks for the math contest authors. Always appreciate a -100 delta.

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5 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

C was way too hard for me... After an hour of intense math i found a formula, which gives much bigger answer for the second sample, but i just can't find any single thing wrong with it... If we won't place any rook on main diagonal there are $$$30$$$ possible positions for the first rook, after that there will be $$$12$$$ possible positions for the second rook and $$$2$$$ positions for the last one. In total it is $$$30 * 12 * 2 = 720$$$, which is already bigger, than $$$331$$$... What's wrong?

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    5 weeks ago, # ^ |
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    I was able to find a recurrence for C but was not able to submit in time because of runtime errors.

    F(n) = F(n-1) + (2*n-2)*F(n-2)

    where n is the total number of row and column numbers which were not used. Not sure if it is correct.

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      5 weeks ago, # ^ |
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      I also came up with same relation but I got WA 257630731

      I forgot to add dp[0]=1; Oh god.

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      5 weeks ago, # ^ |
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      Can you provide the intuition for this?

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        5 weeks ago, # ^ |
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        Let us assume some n pairs (r,c) where the rooks have already been kept. Now we cannot keep the rook in any of those (r,c) or (c,r). So, that leaves us (n — r -c) rows and (n — r -c) columns.

        Now, let us list down these x = (n-r-c) viable rows and columns as a1, a2, a3 .. ax.

        So, we can keep the rook at a1,a1 and a1 will be removed from the above list and we will have x-1 elements in the list. (so, f(x-1) part)

        If we take any other ai, we will have x-2 elements left, and we can arrange a1 and ai in 2 ways multiplied by numbers of ways to arrange those x-2 elements. And there are x-1 ways to chose some ai. Hence 2*(x-1)*f(x-2).

        So, the recurrence becomes : f(x) = f(x-1) + 2*(x-1)*f(x-2).

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        5 weeks ago, # ^ |
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        I'm not great at explaining, so forgive me if this isn't clear.

        Initially, I noticed that placing a rook at coordinates (ri, ci), where ri ≠ ci, eliminates 2 rows and 2 columns, reducing the chessboard from n*n to (n-2)*(n-2). Similarly, if ri = ci, it removes 1 row and 1 column, making it (n-1)*(n-1). Let's say after K moves, we've removed some rows and columns, leaving a chessboard of size m (m ≤ n). Now, this smaller board becomes a blank canvas for determining how many boards we can build by arranging the rooks. The challenge lies in determining the number of possible boards we can build. Initially, I considered a brute force approach where I place a rook on a cell, compute the resulting reduced matrix after the computer's move, and repeat.

        $$$\displaystyle f(n)= \sum_{i=1}^n \sum_{j=1}^n

        \begin{cases} f(n-1),& \text{if } i=j\\ f(n-2), & \text{otherwise} \end{cases}

        \text{ where cell } i,j

        \text{ has a white rook}$$$
        $$$ f(1)=1,f(2)=3; base cases $$$

        However, this solution is inefficient with a time complexity of O(N^2) and suffers from overlapping solutions.

        For instance, consider a 3*3 matrix where a white rook is fixed at (1,1) you will get 3 possible boards.

        W _ _	W _ _    W _ _
        _ W _	_ _ W    _ _ W
        _ _ W	_ B _    _ W _
        

        Also when you fix rook at (2,2) you will get 3 boards.

        W _ _	_ _ W    _ _ B
        _ W _	_ W _    _ W _
        _ _ W	B _ _    W _ _
        

        As you can see we have one overlapping solution.

        By examples I observed that we can calculate all possible boards by fixing the rooks in just one row. So we take a row and see all the variations of that row and calculate the number of boards for each variation and this would give us the unq count.

        W _ _	_ W _    _ _ W
        _ _ _	_ _ _    _ _ _
        _ _ _	_ _ _    _ _ _
        
        _ B _    _ _ B
        _ _ _    _ _ _
        _ _ _    _ _ _
        

        Below is the formula of our revised approach ( the base condition remains same)

        $$$\displaystyle f(n)= \sum_{j=1}^n

        \begin{cases} f(n-1),& \text{if } j=1\\ f(n-2), & \text{otherwise} \end{cases}

        \text{ cell } 1,j

        \text{ has a white rook} + \sum_{j=2}^n f(n-2)

        \text{ cell } 1,j

        \text{ has a black rook} $$$

        If you simplify this you will get

        $$$ f(n)= 2*(n-1)f(n-2)+f(n-1)$$$
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      5 weeks ago, # ^ |
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      .

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    5 weeks ago, # ^ |
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    I think the problem is that operation order doesn't matter, so it should be divided by $$$(n/2)!$$$

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      5 weeks ago, # ^ |
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      Maan, I am so bad at combinatorics)

      Thank you)

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    5 weeks ago, # ^ |
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    Thought the same. Someone enlighten me.

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    5 weeks ago, # ^ |
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    The 720 possibilities are overcounted. Eg. These two configs are same;

    - W(1,3), B(3,1), W(2,4), B(4,2)
    - W(2,4), B(4,2), W(1,3), B(3,1)
    

    The final rook positions are same in above 2 cases.

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    5 weeks ago, # ^ |
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    yeah, thought the same, i could figure out only that 1 is the way we put all in diagonal, so i need to come up with 330, maybe 720/2 gives 330 lol

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    5 weeks ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    I just solved it once the contest was over:(((((((((((( Unfortunately I couldn't submit my solution which I am sure that it is correct :((

    #include <bits/stdc++.h>
    using namespace std;
    
    #define ull unsigned long long
    #define ll long long
    #define ld long double
    #define int ll
    
    #define endl '\n'
    #define EPS .0000001
    #define MOD 1000000007
    
    void calculate();
    
    int dp[300010];
    signed main() {
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2; i <= 300000; i++) {
            dp[i] = dp[i-2]*(i-1)*2+dp[i-1];
            dp[i] %= MOD;
        }
        int t = 1;
        cin >> t;
        for(int i = 1; i <= t; i++) {
            calculate();
        }
    }
    
    void calculate() {
        int n, k;
        cin >> n >> k;
        int x, y;
        set<int> st;
        while(k--) {
            cin >> x >> y;
            st.insert(x);
            st.insert(y);
        }
        n -= st.size();
        cout << dp[n] << endl;
    }
    
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      5 weeks ago, # ^ |
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      I dunno what got into me, but i have found a recurrent formula, similar to yours, but for some reason I was trying to make it linear...

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    5 weeks ago, # ^ |
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    My way of thinking was similar. There are $$$10$$$ ways for the second rook, because $$$30-(4*8-4*3)=10$$$. And because by filling up the remaining diagonals, all these could be considered as final positions. So after placeing 2 rooks, there are $$$30*10=300$$$ possible positions, which again could be considered as final ones. And then $$$1+30+300$$$ gives $$$331$$$ (the $$$1$$$ is needed because the initial position can also be considered as a final one). But what I don't get is that we overcount the positions after 2 rooks, so it should be $$$150+30+1=181$$$. Can anyone please explain it to me?

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    5 weeks ago, # ^ |
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    Btw, my A got WA on system testing, wow... $$$a_i \le 100$$$ got me there... Definitely, not a good day for me...

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    5 weeks ago, # ^ |
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    this solution would have been correct if two rooks position swapped yields a different setting. but according to problem, they are same. and if you notice carefully, its 6! = 720 and there is a reason for that.

    you can look at my solution i did not do recurrence i used combinatorics only.

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    5 weeks ago, # ^ |
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    did the same!

    what you need to do is to divide this big number i.e. 720 with 3!.

    It is because there are repeated groups in counting. and the number of repetitions for each group is 3!.

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5 weeks ago, # |
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Does anyone have any advice for improving on problems like C in today's contest?

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5 weeks ago, # |
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OMG, I SUBMITTED D 8 SECONDS BEFORE THE END

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5 weeks ago, # |
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I think D is more intuitive than C.

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5 weeks ago, # |
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Problem B in a Nutshell, huge skill issue

Spoiler
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    5 weeks ago, # ^ |
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    basically you can just check whats the biggest number for Math.pow(2,x)-1 <= k because Math.pow(2,x)-1 is only 1's, the rest of the number doesnt really matter and you can sum it up with the difference and the rest 0's

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5 weeks ago, # |
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How to solve D?

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    5 weeks ago, # ^ |
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    a ^ b < a implies that msb of b should be present in a 
    which means for given A[i] ans = no. of pairs l, r such that msb of 
    A[i] should be present in that xor(l...r).
    i couldn't implement in time but i'm sure it's correct 
    

    UPD :- i was correct 257663607

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5 weeks ago, # |
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can someone please explain b? I am getting wa.

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    5 weeks ago, # ^ |
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    Think about how any number that looks like this $$$2^{n}-1$$$ has all ones in it's binary representation. Also if $$$n=1$$$ just print the value of $$$k$$$. In all other cases where $$$n\geq1$$$ you only really need to print two numbers that sum up to $$$k$$$.

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    5 weeks ago, # ^ |
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    Let $$$b$$$ be highest bit in k. Then you can get $$$b-1$$$ bitcount by just printing $$$s = 2^{b} - 1$$$. Then print out $$$k - s$$$ and fill the rest with zeros.
    Why does this work?
    Either $$$k - s$$$ will be greater equal than $$$2^b$$$, then you've got all bits in answer and you can't make it bigger (note that it means k has no 0s in its binary representation,
    or $$$k - s$$$ will be smaller, but that means that you "spent" 1 from b-th position to cover up all 0s in lowest bits.

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5 weeks ago, # |
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Hello, I want to report a cheater "Sputnik123". No need to long text lol, he just cheated.

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5 weeks ago, # |
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I Solved C during the contest but once I fixed some minor bugs in my code I found that the contest is over :(

Here is my (should be) Accepted code for problem C:

#include <bits/stdc++.h>
using namespace std;

#define ull unsigned long long
#define ll long long
#define ld long double
#define int ll

#define endl '\n'
#define EPS .0000001
#define MOD 1000000007

void calculate();

int dp[300010];
signed main() {
    dp[0] = 1;
    dp[1] = 1;
    for(int i = 2; i <= 300000; i++) {
        dp[i] = dp[i-2]*(i-1)*2+dp[i-1];
        dp[i] %= MOD;
    }
    int t = 1;
    cin >> t;
    for(int i = 1; i <= t; i++) {
        calculate();
    }
}

void calculate() {
    int n, k;
    cin >> n >> k;
    int x, y;
    set<int> st;
    while(k--) {
        cin >> x >> y;
        st.insert(x);
        st.insert(y);
    }
    n -= st.size();
    cout << dp[n] << endl;
}
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5 weeks ago, # |
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I didnt had dream last night for maths formula of c. so i was not able to solve it( btw one of worse c i have seen). (If author love maths then give olympiads)

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5 weeks ago, # |
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Found my bug in C five seconds after contest ended. That sucks.

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5 weeks ago, # |
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    5 weeks ago, # ^ |
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    Hint: Brute force a solution for cases in which n = 2. Generate random testcases keeping n = 2 and see where your code gets a different answer from the brute force solution.

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5 weeks ago, # |
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Solved D but at what cost.. nice D tho 🙂
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road to pupil :pray: thank you

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how to solve F?

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Anyone explain why this fails? 257627963

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    5 weeks ago, # ^ |
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    Try cases with low value of $$$n$$$.

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    5 weeks ago, # ^ |
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    Your submission fails for cases like n=2 and k=100. One answer is [63,37], with 7 1s, however your code returns [1,99] which has only 4 1s. The optimal way is to find such m, that 2^m-1 <= k

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5 weeks ago, # |
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Any hints for problem F? I thought of doing merge sort segment tree with heavy-light decomposition, then binary search to find the lowest value that the count is not equal, but I realized there some edge cases.

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5 weeks ago, # |
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My B problem got Judgement failed veredict, what that means?

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5 weeks ago, # |
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Mathforces

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5 weeks ago, # |
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problem C last minute submission got very much raised after 2:05 due to this solution being leaked online in a telegram group please MikeMirzayanov look into this behaviour of participants : screenshot of the solution , i will later reveal those solutions which are being cheated from here!(https://ibb.co/TTQGHV3) Vladosiya othmaine also look into this once please

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    5 weeks ago, # ^ |
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    one submission which was a random person in my friend list: link

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    5 weeks ago, # ^ |
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    found three cheaters they tried very much to keep their submission for not being detected 257640596 useless variables and deleting those we will get exact same as the solution even the line space etc is also exact same (spacing is same) ! 257641772 exact same , 257640016 exact same again . Please look mostlikely they may be undetected due to some minor changes but the format and spacing is exact same . MikeMirzayanov Vladosiya

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      5 weeks ago, # ^ |
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      I also checked out the previous contests which those three gave, Infact there too they cheated from a telegram group . Infact now the first user upsolved the B ,C problems which he/she already did in contest time . That also leads us to why he/she cheated the same code exactly which i shared earlier just because of some useless variable declarations like int p1=INT_MIN; he/she will not get caught at all.

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5 weeks ago, # |
Rev. 4   Vote: I like it +16 Vote: I do not like it

Could anyone please tell me why the answer for n=6 in E is 24? I am getting 23 by hand and code, I was able to simplify the formula for C(i,j) mod j as follows:

1) if j is prime then ((i/j) mod j)*(j-1)

2) if j is not prime then we only need to do this for j=4 because (j-1)! mod j is 0 for all other composite numbers

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5 weeks ago, # |
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I guess D can be solved using dp

Screenshot-2024-04-21-223826

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    5 weeks ago, # ^ |
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    It's more observation i would say, if you can think of the final equation the dp is strightforward to use to calculate it

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    5 weeks ago, # ^ |
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    It's not combinatorics which has exponential time complexity in brute force. It's just a very classical "change your perspective to group quantities together" type of problem.

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      5 weeks ago, # ^ |
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      Exponential? You mean cubic?

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        5 weeks ago, # ^ |
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        Please read carefully of what I wrote before commenting. That's basic respect.

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          5 weeks ago, # ^ |
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          Oh sorry, I thought the combinatorics had something to do with the problem

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Could anyone please explain how to do problem C? I always seem to struggle in these types of problems related to combinatorics :(

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    5 weeks ago, # ^ |
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    It is possible to solve using combinatorics but the solution would be way harder. (You need some techniques to reduce time complexity).

    I guess the official solution would be DP: DP[n] = DP[n-1] + DP[n-2] * (n-1) * 2.

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      5 weeks ago, # ^ |
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      Would you kindly explain how you figured it out? Like explain why this works? I haven't solved many dp problems too.

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        5 weeks ago, # ^ |
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        Assume there is an N X N chessboard, and you know how many different configurations are there for 1 ~ N-1. Think N X N chessboard as an extension of an N-1 X N-1 chessboard.

        1. It is possible for you to put a rook at (N, N). The number of cases equals to DP[n-1].
        2. If not, a rook should be added somewhere between (N, 1) ~ (N, N-1). The number of cases equals to DP[n-2] * (n-1) * 2.
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      5 weeks ago, # ^ |
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      yes

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mathForces

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5 weeks ago, # |
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Can't believe I'm this bad. Can't even solve 3 problems

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5 weeks ago, # |
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Problem C was quite to my liking and I was able to sprint to rank 50 after solving C, although temporarily(^_^)

But D was too difficult and I fell to rank 900+ for not solving any problem after C._(:3]<)_So how to solve it? I was confused by my prefix and suffix array, so tried brute force, but it got WA rather than TLE.

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    5 weeks ago, # ^ |
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    For any number, $$$a_i$$$ you can find that it satisfies the condition only when the xor of subarray in which $$$a_i$$$ is present has the MSB of $$$a_i$$$ set to zero, now you just need to find these subarrays for each element which can be done using dp for each bit j.

    To find the first observation, try to think of an answer for $$$n^2 logn$$$ where you iterate on each subarray and find the number of possible y's in that subarray. What will be the condition for y?

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      5 weeks ago, # ^ |
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      why only MSB? If any bit is set in ai, but not set in xor of the whole subarray, then it would lead to a valid answer, right? It doesn't necessarily have to be MSB.

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        5 weeks ago, # ^ |
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        yes,i mean same of you, but i think twice MSB just make this question be simple. MSB just a "point"

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        5 weeks ago, # ^ |
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        because if the msb is set in the subarray then taking xor will make it zero and then for any bit< msb even if u set it you can't satisfy the conditon since $$$sum_{i=1}^{n-1} 2^i < 2^n$$$.

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5 weeks ago, # |
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Can someone explain to me what this test result means? Also it's not even shown like -1 in results, just empty square, as if I havn't submitted this problem

https://codeforces.com/contest/1957/submission/257589755

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5 weeks ago, # |
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combinatorics is the most no skill thing you can ask a programer to do

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My solution for q1 has been given the verdict as judgement failed, what does it mean?

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everyone solved C using dp but I only did some n choose r math thingy 257639189

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How to solve E, F?

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It says judgement failed for problem B https://codeforces.com/contest/1957/submission/257590765

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E problem was just observations i think.. i didnt prove anything concrete, just saw first few values and came to conclusion that n = 4 and n = prime are kind of special cases to handle.

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5 weeks ago, # |
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I am getting judgement failed verdict on this submission.

https://codeforces.com/contest/1957/submission/257590736

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does anyone know what the verdict "Judgement failed" means. Muhammad-Saram got it and I feel bad for him. His logic was on point. Here's his submission

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5 weeks ago, # |
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my solution for B is still not judged https://codeforces.com/contest/1957/submission/257594447 even when system testing is finished :/

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 3 pages of judgement failed, strange.

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    5 weeks ago, # ^ |
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    And all of them were made almost at the same time

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What is judgement failed??? Is it wrong answer?

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https://codeforces.com/contest/1957/submission/257616209 I faced judgement failed in this submission?

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https://codeforces.com/contest/1957/submission/257627565 https://codeforces.com/contest/1957/submission/257651698 submitted the exact code again and it passed now but gave runtime error during system testing. What's going on? Codeforces high or wat?

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For Problem B, can someone please explain why my submission 257616558 is incorrect? I agree I overkilled it a little bit, but I cannot see how the Jury's answer is better.

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5 weeks ago, # |
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I would like to report a suspicious contestant: F9O1OvO

I think he cheated because his template has changed massively:

  • His main function looks completely different
  • He switched from typedef to define
  • He suddenly uses "namespace wshb"

His submission for problem A in today's contest: 257575498

His submission for problem A in his previous contest: 255636465

His performance is also very unusual — I find it very weird that a specialist who has participated in more than 50 contests suddenly gets 5th in a div.2 round.

I believe he might have given his user to another person.

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    5 weeks ago, # ^ |
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    And they said: "If you solve ABCD, cheaters won't affect you", lol

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      5 weeks ago, # ^ |
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      Yeah lmao. When I previously reported some cheater, some user told sth similar also,

      Debunked
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    5 weeks ago, # ^ |
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    Yeah a sudden Rk 5 and his template changing is definitely an indicator it wasn't him doing the contest.

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can anyone tell me whats wrong in my code. I thought if sum of array is k and number of 1 bit is maximum it work. but it is not why? sub :https://codeforces.com/contest/1957/submission/257631018

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    5 weeks ago, # ^ |
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    same here

    t = int(input())

    for _ in range(t): n, k = map(int, input().split()) q = k ans = [0]*n l =0 while k>=2 and n>0: for i in range(0,k+1): if pow(2, i) >k: ans[l]= 2**(i-1)-1 l+=1 n-=1 k=k-(2**(i-1)-1) break

    if sum(ans)<q:
        ans[-1]+=q-sum(ans)
    print(*ans)
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    5 weeks ago, # ^ |
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    I think there is this set of data: n = 14, k = 720729. The number of 1s actually equals to 19, and you give 17. I think your solution is inaccurate for the case where s is not equal to k.

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Nice round, but why problem E doesn't have a bolded notification that sum of n over subtests is unbounded? It messed me a bit up.

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✋✋✋**help wanted** ✋✋✋✋✋✋ i know i am dumb but can anyone tell me where the hell am i wrong here! it works like magic bitcount are correct+ sum of the nums are correct then what?

t = int(input())

for _ in range(t): n, k = map(int, input().split()) q = k ans = [0]*n l =0 while k>=2 and n>0: for i in range(0,k+1): if pow(2, i) >k: ans[l]= 2**(i-1)-1 l+=1 n-=1 k=k-(2**(i-1)-1) break

if sum(ans)<q:
    ans[-1]+=q-sum(ans)
print(*ans)
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    5 weeks ago, # ^ |
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    atleast format the code before asking for help.

    cout << "Please!!" << endl;
    
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what an impeccable contest and i hope that there will be more contests like that

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Mathforces (More like combiforces tbh).

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5 weeks ago, # |
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This is one of the best rounds I have had in recent times, with clever ideas in every question! Hope to see more rounds like this~

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What is the time complexity of question E?

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Cool

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In the next codecraft we'll have "How does the Knight move??"

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    5 weeks ago, # ^ |
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    I'm glad someone got that reference :D

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Nice Round !

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5 weeks ago, # |
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Can anyone explain what's wrong with my Solution

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4 weeks ago, # |
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Well

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4 weeks ago, # |
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I hope to reach 1200 points after this compete

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4 weeks ago, # |
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Why I'm not in the winners list??? I'm 2nd!!!

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13 days ago, # |
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i wish you all reach what u want :)