AlFlen's blog

Codeforces Round #750 (Div. 2)
By AlFlen, 3 years ago, translation, In English

Hello, Codeforces!

RedMachine-74 and I are glad to invite you to our Codeforces Round 750 (Div. 2), which will be held at Oct/24/2021 13:05 (Moscow time). Notice the unusual time of the round. The round will be rated for all the participants with rating strictly less than 2100. At the time of the round the CheReKOSH olympiad will be held, where the problems from this round will be used!

We would like to thank everyone who helped us a lot with round preparation.

On the round you will be asked to help characters from animated series Luntik and his friends. You will be given 7 problems, one of which has two subtasks. You will have 2 hours 30 minutes to solve them.

UPD: Score distribution: $$$500-750-1500-1750-2500-(2000+1500)-3250$$$.

UPD2: Editorial

UPD3: Congratulations to the winners!

Div. 2:

  1. int65536

  2. hehezhouyyds

  3. trunkty

  4. m3owp1mp

  5. Chtholly-Nota-Seniorious

Div. 1 + Div. 2:

  1. SSRS_

  2. emthrm

  3. int65536

  4. turmax

  5. hehezhouyyds

We wish everyone good luck!

  • Vote: I like it
  • +480
  • Vote: I do not like it

Comments Comments (134)
| Write comment?
»
3 years ago, # |
  Vote: I like it -34 Vote: I do not like it

hoping for a good contest :)

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

    You wrote the exact same comment in the announcement of rounds #748 and #749, but didn't participate in both of them. Will you participate in this one? XD

    → Reply
    • »
      »
      »
      3 years ago, # ^ |
      ← Rev. 5   Vote: I like it +1 Vote: I do not like it

      Bro actually i participated in div3 round but my solution of D1 concides with 1 participant , i dont have any idea how this happens to me and in round 749 i did not participated due to issuse on codeforces like site is not opening properly . I am new to codeforces and if u want then i will not post such comment from next contest onwards.sorry for my comment .

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

        I never said there's anything wrong with your comment, I just thought will you participate this time. Chill dude, you can post whatever you want. :)

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

      if he participate he cheats, so better he shouldn't!

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

    will see !!

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

    why are tou skiped?? you cheated again ?!?!?!

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

      The moment I saw 3 compilation errors, I knew something was fishy.

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

      No boy, he isn't a cheater! I know him personally, and he is a really hard-working (just see his graph) and honest guy.

      As far as skipping is a concern, it's because he didn't knew that we can't use 2 ids during a contest.

      → Reply
»
3 years ago, # |
  Vote: I like it +51 Vote: I do not like it

As a tester, good luck anyone who participates will have a positive delta!

-QuangBui(YT/CP)

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

    sum of delta = 0 intensifies

    → Reply
»
3 years ago, # |
  Vote: I like it -60 Vote: I do not like it

Anxious for the contest to arrive

PS: I take this opportunity to leave a link to a blog with a collection of segment tree problems that I hope will be helpful to someone. https://codeforces.net/blog/entry/22616

Thanks to AminAnvari

→ Reply
»
3 years ago, # |
← Rev. 4   Vote: I like it -6 Vote: I do not like it

why cf is organising contest at unusual time these day? By the way hoping for big delta +ve

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

    At the time of the round the CheReKOSH olympiad will be held, where the problems from this round will be used!

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

    At this point I dont know which is the usual time anymore :))

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

traktor. Me when I read A

→ Reply
»
3 years ago, # |
  Vote: I like it +50 Vote: I do not like it

As a tester, GOOD LUCK!

→ Reply
»
3 years ago, # |
← Rev. 6   Vote: I like it -44 Vote: I do not like it
nice meme
→ Reply
  • »
    »
    3 years ago, # ^ |
    ← Rev. 5   Vote: I like it +9 Vote: I do not like it

    It's better to put the meme inside a spoiler so it doesn't take so much space :)

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

does one subtask means that one problem will be divided to easy and hard versions?

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it -28 Vote: I do not like it

    Probably.

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

    Yes, subtask means same problem will have different constrains and one of those can be passed with a easier solution.

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

i wish i could creat a contest like this one day :3

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

Unusual Time

→ Reply
»
3 years ago, # |
  Vote: I like it +52 Vote: I do not like it

I hope you surprise us with A1 and A2

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

I hope to become MASTER again through this contest.

Good luck!

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

I want green

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

Good time for Chinese participants!

I wonder why recent contests take place earlier.

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

    Maybe today's contest is early because otherwise it would have clash with CodeChef's Cook-Off !!

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

I don't know whether it is appropriate to post this comment under this blog.

If I registered for round #751(Div-1) and I lost candidate master after today's contest, will I be able to participate in tomorrow's Div-1?.

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

ppl say the unusual time of this round is because its mirroring an olympiad, but we all know the real reason is that it doesn't clash with the T20 World Cup: India vs Pakistan. Thanks Codeforces, appreciate it.

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

At least the first question can be easy for newbies!!!

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

    Boruto you are annoying as ever !! First cheating in chunin exams and now this

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

      A new chunin exams starts today (3:30 IST)! Hope you are still in the game Kakashi!

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

        I am Kakashi hatake the copy ninja .. I know a 1000 Jutsu and I will use them all and live up to my name !!

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

      Kakashi sense(or else I can say stupidgoat) can you teach me the Rasengan! For the current chunin exams so that I won't cheat, please...

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

        Bro I am a copy ninja just stop kidding

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

          I wonder about that because you don't have the dear Sharingan anymore Mr.Copyninja!!

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

            Bro codeforces round at 3.30 ... I have sharingan ... Itachi is alive what do you think is happening .... It's INFINITE TSUKUYOMI !!

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

              Those Who Do Not Understand True Pain Can Never Understand True Peace.

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

      Interesting !

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

is there gonna be score distribution?

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

Yay! Perfect Timing for me. Today is IND vs. PAK World T20 from 7:30 pm (UTC+5.5).

→ Reply
»
3 years ago, # |
  Vote: I like it -25 Vote: I do not like it

Is it rated?

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

    What part of the sentence, "The round will be rated for all the participants with rating strictly less than 2100," do you not get?

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

      he's trolling obviously or testing how many downvotes he will get

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

      What part of sarcasm you don't understand? Oh, all of it.

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

        Eh, I thought he was being deadass lol

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

Todays Contest is Awesome..... I am only able to solve one question but B and C questions solution i know but there time complexity is way too much........ Today gonna learn something new...... Very nice contest Again!!!

→ Reply
»
3 years ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it
  • Problem 1: Pretest passed in 7 mins.
  • Problem 2: Pretest passed in 3 mins.
  • Problem 3: Pretest passed in 12 mins.
  • Problem 4: WA on pretest 2 :(
→ Reply
»
3 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Guys does Codeforces really use Adobe Flash Player? I feel like it is a prank, I can't do hacks.

Open
→ Reply
»
3 years ago, # |
← Rev. 2   Vote: I like it +12 Vote: I do not like it

WTF is happening with me I am solving A in more than 1 hr and B,C in 30 mins is it just me or nowadays problem A is becoming trickier. I suck

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

    Not you, todays problem A was tricky af. Not hard, just tricky. Seemed trivial at first but then...

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

Thanks for such a good contest!

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

WHY THE CONSTRAIN sum <= 1e9 is in D there are solutions that cut because of this what the hell is point? you could make it 2e9

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

    I feel that was the whole point of that qn.. There was a way("we had to think that") to get it within the limits. I couldn't pass my code btw..

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

      I don't think we should think to reach the idea of the problemsetter for specific problem

      especially constructive problems like this has many many solutions

      this is hole point of creativity

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

    That made it interesting for odd sized array.

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

How to solve E? I tried a O(n^1.5) dp approach but didn't work.

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

    Most probably you must have taken the limit as sqrt(n) but it should be sqrt(2*n)

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

      No, it took sort(2*n). Got runtime error in pretest 8

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

        I also got RE in pretest 8. After declaring the array globally, my code passed.

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

          Yeah, mine passes when I use vector instead of array. Any idea why?

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

    O(n*sqrt(n)) is well enough.try constant optimization.

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

    dp[k][n], and kmax = 450

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

please tell me if i have submitted two pretest passed solutions for the same problem, which one will be judged for main tests??? please some one tell meee pleaseeeee

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

Are pretests for F1 really weak?

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

    Unfortunately, it's true. That problem didn't have a multitest testing so it didn't work out to cut off as much solutions as possible using a little number of pretests. A little number of tests was needed to avoid queuing.

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

    unfortunately yes :(

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

Someone solved the problem F2 with a Gaussian basis?

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

    I don't think that it is possible to solve this problem with gaussian basis. The expected solution uses the fact that you have at most 5000 different values for $$$a_i$$$. You just need to solve $$$dp[k][xor] =$$$ minimum position you can reach value xor, using only $$$a_i <= k$$$

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

Can someone give a hint for A? I was trying it case by case but I felt like there were way too many cases with my method. In the end, I was really short on time and couldn't even enumerate all the cases.

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

    for Problem A : the answer is either 0 or 1 proof : let A , B , C be durations ( 1 min each ) of the 3 musics A + 2B + 3C is the total duration now consider 2 groups : ( B + C ) and ( B + C ) so we remain with A + C duration now if A + C is odd => 1 will be extra or else A + C will be equally distributed

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

      How do you split 2B + 3C into 2 groups of B+C? A 3 minute song cant be split and neither can a 2 minute one.

      And for A=1, B=0, C=1 the answer will be 2. Is my understanding of the problem flawed ot something?

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

        B can't be zero

        a,b,c >= 1

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

          Oh I missed that. I still don't understand the explanation above though :(

          → Reply
  • »
    »
    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it
    1. First of all, it's obvious that answer can't be more than 3.
    2. Answer can't be 3, because in that case we can move from the bigger subset 1 or 2 to smaller one (or, if bigger subset consists only from 3's, that means all 1's in 2's in smaller, and we can exchange 3 with 2 or 1 and make answer smaller).
    3. Answer can't be 2 using the same logic. Obviously, if bigger subset has at least one 1, we can make the answer 0 easily. If answer is 2, then all 1's in smaller subset. Bigger subset consists only of 3's? Exchange 3's from that subset with 2 and 1 from smaller, afterwards move 1 to smaller. Answer is zero. Bigger subset consists only of 2's? Exchange 2 from bigger with 1 in smaller. Answer is zero. 3's and 2's? Exchange 2 with 1 from smaller.

    4.Answer is obviously can be 1, because a + 2b + 3c can be odd.

    Any natural (and > 0 depending on definiton) numbers a, b, c, such that a + 2b + 3c is even can be divided to equal groups.

    Proof: answer can't be bigger than 1, proved that in 1. — 3. If cannot be divided, then answer is 1, and that means that sum of bigger subset and smaller subset is odd, which contradicts with what we prove. So the answer is (a + 2b + 3c) % 2.

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

D was nothing but x(+y) + y(-x) = 0 and x(-y) + y(x+z) + z(-y) = 0. One can easily use 1st relation to solve for even arrays but tricky part comes for odd arrays where u need to find x and z such that there sum is not 0.

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

    the problem guarantees $$$a_i \ne 0$$$ for all $$$i$$$, so at least one of $$$a_1+a_2$$$, $$$a_1+a_3$$$, $$$a_2+a_3$$$ is not zero. Just find it out and solve $$$a_4$$$ to $$$a_n$$$.

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

How to solve A quickly?

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

Great problems!

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

I don't understand the answer for A from editorial:

cout << (a + c) % 2 << '\n';

If a=0, b=1, c=0, shouldn't the answer be 2 minutes? (not zero as the editorial suggest)

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

    a,b,c (1≤a,b,c≤10^9) — I didn't see this limitation myself at the beginning and went to solve C.

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

    a,b,c >= 1

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

    a and c cannot be 0 from the constraints.

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

      This problem would be much more interesting if 0 <= a, b, c. Writing clean and not bugged solution in this case is a skill. Disrespect to authors or coordinators.

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

        The problem is an A problem for a reason.

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

        The solution was already a guess, adding more corner cases is dumb, specially on the first problem.

        → Reply
        • »
          »
          »
          »
          »
          3 years ago, # ^ |
          ← Rev. 2   Vote: I like it -36 Vote: I do not like it

          It can be solved without any corner cases even if 0 <= a, b, c, you just can't see it

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

            Sure, say your solution then. Me and probably many others would just add ifs and get ac faster then trying to think of a way to implement it without corner cases

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

The question for the experts is, is there really not a single test in A that would reject a solution with an overflow of the int variable? https://codeforces.net/contest/1582/submission/132905404

I spent half an hour looking for a hack, for this solution, for me it became a kind of challenge...

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

Did anyone else solve F2 with bitset in $$$O(\dfrac{a_i^3+na_i}{w})$$$? I think that I will FST(

→ Reply
  • »
    »
    3 years ago, # ^ |
    ← Rev. 4   Vote: I like it 0 Vote: I do not like it

    I solved by assuming that only the first and last occurrence of a number will matter (Not 100% sure that it is correct)

    Upd: Not Correct

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

    $$$O(N + A^2 logN)$$$ passed in 400 ms, god bless g++!

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

      If I am not wrong, your solution is $$$O(N+A^2log(\frac{N}{A}))$$$.

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

I needed more 5 minutes to fix my bug for D if I just didn't be like a moron in A and waste 20 min on it pfff I hope my solution is WA after fixing that bug that would actually make me feel less worst lol

Edit: Accepted but wasn't gonna make it in time without my stupidity in A anyway that actually feels even more less worst my idea is correct and I wasn't gonna make on time better than the solution is completely wrong or i just needed more 5 min to get Accepted

→ Reply
»
3 years ago, # |
← Rev. 2   Vote: I like it -18 Vote: I do not like it

Problem — B

Left shift
Logarithimc power method
Pow

Is'nt left shift and logarithimic power method better than pow??
anyone can explain why?? thanks in advance.

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

    (1L<<zc); returns an int, so if zc > 30 you will get 0 in the answer. Change it to ll ans = (1LL<<zc); and you should get ac.

    You are also taking the modulo in the logarithm power, which makes 0 sense in this problem and this is why youre getting wa

    → Reply
»
3 years ago, # |
← Rev. 3   Vote: I like it -59 Vote: I do not like it

Nice problems! If problem A explanation was correct this contest was one of the great contest in cf

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

Problem D has shaved years off my lifespan.

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

FSTforces

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

2 minutes silence for those who got FST on F1.

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

I don't understand one thing

Why does xor of empty subsequence equal to zero?

Another interesting thing is that there was no such pretest in the tests.

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

    0 only as empty sequence was in first sample

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

      But there was a subsequence with xor 0

      There is no pretest with only empty sequence xor 0

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

        wasn't, only three correct subsequences: {2}, {4} and {2,4}

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

          I think he means that if you consider all non-decreasing subsequences (not just increasing), and you don't consider xor 0 as always possible, you will fail only with a test where there is no non-decreasing subsequence with xor 0.

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

To get F1 accepted it was more important to keep in mind that xor 0 is always possible than the fact that increasing subsequence is different from non-decreasing.

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

.

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

    132901810

    so you submitted your first code in c++ just by studing that link?????

    haha. another shameless indian lying after being caught.

    just like the other indians

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

After the round 750, I received the message:

" Your solution 132867227 for the problem 1582C significantly coincides with solutions vpike/132864849, RefreshedCoder/132867227 "

The two solutions can be found here:

132867227 132864849

It's a simple greedy approach and I wrote every character in my IDE. It's a wrong judgement and may I ask the admin of Codeforces to examine the two submissions manually?

I personally don't know vpike and it's just a coincidence that our solutions are similar in this simple problem. I solved in total 6 problems in the round and there's no motivation for me to cheat in problem C.

Please, could anybody help me to get my rating back?

UPD: Ping MikeMirzayanov here

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

    I'm sure too, it's a rare coincidence. Why two submissions for simple problem with small amount of code cannot be close to each other? Please examine submissions manually.

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

    There is no any feedback on this issue for a long time. Dear AlFlen and RedMachine-74, could you comment it?

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

Can somebody hack my D? I think my submission is wrong, because I dont check this condition the sum of their absolute values cannot exceed 10^9. 132943758 upd: it is correct

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

Is F2 solvable with java?

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

Thanks for fast Editorial!

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

Can anyone tell me why me solution https://codeforces.net/contest/1582/submission/149071151 gives TLE on F2 If I use while loop which I have commented instead of for loop it gives AC.

→ Reply