Processing math: 100%

yunetive29's blog

By yunetive29, 3 weeks ago, translation, In English

Ciao, Codeforces!

I am glad to offer you to participate in Codeforces Round 1002 (Div. 2), which starts at Feb/02/2025 17:35 (Moscow time).

The tasks were offered by me and Artem shfs. We hope you will find them interesting.

This round will be rated for all participants, with rating below 2100. Participants with higher rating can take part, but the round will be unrated for them.

You will be offered 5 tasks and 2 hours to solve them. The last task will be split in 2 parts.

I want to thank:

We wish you good luch and rating as big as the bottle in the photo!

Point distribution: 500−1000−1500−2000−(1500−1500).

UPD: Editorial

Congratulations to the winners:

  1. potato167
  2. kotatsugame
  3. Mangooste
  4. Rubikun
  5. maspy

Rated only:

  1. RGB_ICPC8
  2. Sanae
  3. ydkm
  4. jtrh
  5. mj1000j

  • Vote: I like it
  • +324
  • Vote: I do not like it

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

Auto comment: topic has been updated by yunetive29 (previous revision, new revision, compare).

»
3 weeks ago, # |
  Vote: I like it +257 Vote: I do not like it

Wow!! 2nd author looks like water bottle.

»
3 weeks ago, # |
Rev. 2   Vote: I like it -26 Vote: I do not like it

GL UPD: first comment after blog has been published

»
3 weeks ago, # |
  Vote: I like it +26 Vote: I do not like it

speedforces?

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

    its actually that go and destroy this contest

»
3 weeks ago, # |
  Vote: I like it +34 Vote: I do not like it

The round turned out to be really cool, hope you'll enjoy)

Tester.

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

I am missing interactive problems hope there is a one in this contest

»
3 weeks ago, # |
Rev. 3   Vote: I like it +18 Vote: I do not like it

it looks like Ciallo, Codeforces!

Div. 2 with 5 problems? Not so common.

»
3 weeks ago, # |
Rev. 3   Vote: I like it -19 Vote: I do not like it

HydroForces!!

»
3 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

As a participant, i will try to empty the water bottle :)

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

    Hope you can success before geting a stomach like a ball.

»
3 weeks ago, # |
  Vote: I like it +55 Vote: I do not like it

As a tester, I have a proof that upvoting this comment will lead to positive delta. But the proof is too long to fit the margin.

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is the water bottle so massive?

»
3 weeks ago, # |
  Vote: I like it -30 Vote: I do not like it

It is going to be a speed force, based on the rating distribution

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

    yeee that's it i hope i can solve the first two of them it will be hard for solving C in think

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

That's a lot of offers!

»
3 weeks ago, # |
  Vote: I like it -28 Vote: I do not like it

I understand the logic, but why is this bottle so big? I've never seen a bottle this big. :P

道理我都懂,可是这个瓶子为啥这么大?我从未见过这么大的瓶子。:P

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

Italian announcement. It will be a fantastic round without a doubt. Ciao!

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

We wish you good luch and rating as big as the bottle in the photo! *luck

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

    I'm a simple man. I see Evelyn, I upvote.

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Would it be advisable to attempt problem E before problem D according to this score distribution?

  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    That's not how it works... E as complete is 3000 and divided into subtasks E1 and E2; that's why individually less than D!! Anyways, me struggling on B itself :(

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

    E1 will be easier than D and have about the same difficulty with C, but E2 will be harder than D (in problems with subtasks, you need to add the scores of the easy and hard version to find the difficulty of the hard version compared to other problems)

    Note that score != problem rating and that the distribution isn't always 100% accurate

    Personally, I'll probably read D, and, if I don't have any ideas, I'll read E1, then decide which one to solve first (if I actually solve the first 3 problems :P)

»
3 weeks ago, # |
  Vote: I like it -11 Vote: I do not like it

It seems that C is more difficult than usual.

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

    yes it will be difficult to solve C

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

    bruuhhhh you will absolutely destroy this contest you've just solved an 2600 rated problemmmm wtfff how did you do that how long did you practice problem solving

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

    you're a legend bro all the respect for you but give me some tips if you see my submissions i have solved 410 problems something like that but i cannot be consistent because the school the exams are destroying me i cannot practice all the time just weekends

    • »
      »
      »
      3 weeks ago, # ^ |
      Rev. 3   Vote: I like it 0 Vote: I do not like it

      No,it's not sloved by myself,i just tested the Deepseek by asking it to slove the problem.

      I can't do this by myslef,if i can,i will not be a Newbie.

      Btw,Deepseek is great.(i have not used it to slove problems in any contest)

      And, Never!!

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

        ohh i see i see i will go and try how much he is smart

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

what is "We wish you good luch"? seems to be a typo

»
3 weeks ago, # |
  Vote: I like it +38 Vote: I do not like it

Is shfs a water bottle?

  • »
    »
    3 weeks ago, # ^ |
      Vote: I like it -6 Vote: I do not like it

    No, a bottle can't make problems, LOL :)

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

cant wait to see how many (will avoid explicitly stating a group of people) will use o3 to win the div2 contest

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

author is taller than water bottle! wow

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

hoping to reach pupil !

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope to tackle C

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

hope you get a positive delta

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

what is your ages guys lets see who's the youngest

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Pupil , I am coming for you!

»
3 weeks ago, # |
  Vote: I like it -29 Vote: I do not like it

As a tester, I have a proof that upvoting this comment will lead to positive delta. But the proof is too long to fit the margin.

»
3 weeks ago, # |
  Vote: I like it -6 Vote: I do not like it

how to increase rating ??

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

This bottle is so big, interesting! good luck!

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

wait, why E1 and E2 have same score as C?

»
3 weeks ago, # |
  Vote: I like it +54 Vote: I do not like it

ahh yes today i compete against deepseek R1.

  • »
    »
    5 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I am trying to register but their is no button for register

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Hello, yunetive29, your contest is my first contest on codeforces, and I got only one problem solved...(yeah,smaller than a bottle)Hope to get more solved in later contests!

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem D got spammed by unrated accounts at the start. (╥_╥)

»
3 weeks ago, # |
  Vote: I like it -18 Vote: I do not like it

Hi @yunetive29,
My username @dp_123 was disabled mid-contest with my latest solution skipped without any system testing mail etc. Please see it.

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

a speechless round. we need special attention.

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

Why is the hack not accessible?

»
3 weeks ago, # |
  Vote: I like it +13 Vote: I do not like it

wtf, how D got over 2k accepted???

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

    cuz it's not hard

  • »
    »
    3 weeks ago, # ^ |
      Vote: I like it -30 Vote: I do not like it

    It's a slightly modified Dijkstra Algorithm!! Would you believe I solved D but not C !!

    • »
      »
      »
      3 weeks ago, # ^ |
      Rev. 2   Vote: I like it +3 Vote: I do not like it

      dude you copy pasted gpt code. lol. Just like i did.

      • »
        »
        »
        »
        3 weeks ago, # ^ |
        Rev. 2   Vote: I like it -19 Vote: I do not like it

        Hehe dumboo.. GPT never writes such codes..

        Edit: Seeing your codes and that too in multiple programming languages ensures that either you are a pro coder or just a cheater who can write such well commented codes in the contests and that gives you no audacity to blame someone else !!

        • »
          »
          »
          »
          »
          3 weeks ago, # ^ |
            Vote: I like it +8 Vote: I do not like it

          hey dumbass. i got the exact same code as you from chatgpt. You just removed comments. u r not fooling anyone. lol.

          • »
            »
            »
            »
            »
            »
            3 weeks ago, # ^ |
            Rev. 3   Vote: I like it -9 Vote: I do not like it

            Dumbass, I always use multiset (easier to remove any element) !! I doubt LLM does that.. Anyways I don't owe you any explanation..

            Prompt to GPT code on Dijkstra's Algorithm : The Link is not working though

            Now, keep comparing the codes and crying over and over !! This platform is just for practice and fun; no one is here to make a statement by rating boosts !!

            #include <iostream>
            #include <vector>
            #include <queue>
            #include <limits>
            
            using namespace std;
            
            const int INF = numeric_limits<int>::max();
            
            struct Edge {
                int to, weight;
            };
            
            void dijkstra(int source, vector<vector<Edge>>& graph, vector<int>& dist) {
                int n = graph.size();
                dist.assign(n, INF);
                priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
                
                dist[source] = 0;
                pq.push({0, source});
                
                while (!pq.empty()) {
                    int d = pq.top().first;
                    int u = pq.top().second;
                    pq.pop();
                    
                    if (d > dist[u]) continue;
                    
                    for (const Edge& edge : graph[u]) {
                        int v = edge.to;
                        int newDist = dist[u] + edge.weight;
                        
                        if (newDist < dist[v]) {
                            dist[v] = newDist;
                            pq.push({newDist, v});
                        }
                    }
                }
            }
            
            int main() {
                int n, m;
                cin >> n >> m;
                vector<vector<Edge>> graph(n);
                
                for (int i = 0; i < m; i++) {
                    int u, v, w;
                    cin >> u >> v >> w;
                    graph[u].push_back({v, w});
                    graph[v].push_back({u, w}); // Remove this line for a directed graph
                }
                
                int source;
                cin >> source;
                
                vector<int> dist;
                dijkstra(source, graph, dist);
                
                for (int i = 0; i < n; i++) {
                    if (dist[i] == INF)
                        cout << "INF" << endl;
                    else
                        cout << dist[i] << endl;
                }
                
                return 0;
            }
            • »
              »
              »
              »
              »
              »
              »
              3 weeks ago, # ^ |
                Vote: I like it +5 Vote: I do not like it

              dude just stop. No one is buying ur bs.

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

              you forgot to remove one remaining ChatGpt comment.

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

                The code I have provided is provided by ChatGPT, not the one authored by me !!

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

        @dummy16661 on CF using chatgpt in contests is not allowed. Dont do it next contest please

  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    It's not that difficult, to be honest. You only need to realize that you will get an answer if both graphs have at least 1 common edge.

    The other part where you find the shortest cost to reach a pair of nodes is quite trivial graph problem.

    Also, this is the type of problem where AI shines.

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

      Modeling this graph, and actually looking that m and n allows you to do that, is not the stuff, that allows 2k ACs

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

        Obviously people cheated. Its nothing new.

        I have seen more surprising ones compared to this.

        Anyways, This problem shouldn't be in a CF Div 2 round. Maybe more suited to Div 3 or Div 4.

        To me this problem is at best 1700-1800 graph problem.

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

      "You only need to realize that you will get an answer if both graphs have at least 1 common edge." — It's not correct criteria, actually. You also need to check if graph is bipartite or not. Because there are cases, when answer would be inf if 2 graphs are bipartite. (Both graphs have single edge (1, 2), s1 = 1, s2 = 2). It is another interesting and easy task about this model. But it can be solved in time O(n + m)

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

        Yeah my bad, forgot about it. But these observations aren't that difficult for a div 2 D.

        After reading the problem, if you think about how to solve this problem in the most brute force way. You will realize soon that you only need to find a common edge such that you can traverse in the same direction at the same time. Also the whole thought process to solve this problem is quite linear.

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    seems plenty of AI generated codes left untouched even after plagiarism check.

    some of the solutions that I think are clearly AI generated: https://mirror.codeforces.com/contest/2059/submission/304082660 https://mirror.codeforces.com/contest/2059/submission/304091441

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

speedforces

»
3 weeks ago, # |
  Vote: I like it +10 Vote: I do not like it

Today, a div2B send me straight to pupils again... damn why I can't even solve B problem properly?

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

    same, i think i am retarded

  • »
    »
    3 weeks ago, # ^ |
    Rev. 3   Vote: I like it 0 Vote: I do not like it

    The moment I saw B, I mentally resigned from the contest. My brain stopped functioning and wouldn't come back :)))

    Edit: it turned out to be quite an easy casework problem in the end, but that's how it always is.

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

    B was overkill,it took me around 1 hour just to get the intuition

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

    If it makes you feel better, I think B was harder than C and D. Although I got B, I completely guessed it out of desperation. I was even a bit surprised I got AC. For C and D I was 100% sure it would pass when I got AC

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

      For now, the next div. 4 is what makes me feel better (I hope)

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

      Today I upsolved C. And it's indeed easy one (compared to B).

      While D is still beyond me, I dunno how people called it standard tho. (I do know dijkstra, but the transition is not really clear to me)

»
3 weeks ago, # |
Rev. 2   Vote: I like it +13 Vote: I do not like it

No way D got 2k solves

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

Please. Somebody explain to me how the 4th test case in problem C is 3? I spent an hour and a half staring at it and could not figure it out

  • »
    »
    3 weeks ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    you might read the i and j wrong. I made the same mistake.

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

      Omg... it doesn't help that all the other examples work with i and j flipped the other way.

  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it +8 Vote: I do not like it

    Switch cols and rows, then it works!

    Actually, I needed like an hour to figure this out. One could say, ok, beginner error. But given that the statements (also of A and B) where kind of hard to read, here in C we are presented with the first three testcases that work with switched col/row also.

    That is bad statement text.

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

    Serve queues 1,4,2,3 in that order

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

    I wasted half an hour in same mistake, you actually have to take transpose of the matrix and then proceed

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

    Saaaaaame. Was staring at this forever and never figured it out. The fact that the 3 explained examples are all the same when transposed is quite bad.

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

B got me like 15 minutes, while C took me around 1 minute to figure out the answer. B >> C

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

    Could not figure out C, I was thinking of some dp approach. How did you solve it?

    • »
      »
      »
      3 weeks ago, # ^ |
      Rev. 4   Vote: I like it +1 Vote: I do not like it

      To maximize the MEX, we need to have 0,1,2,...,x with x is large as possible, that's the whole idea. In fact with any sort of MEX problem, I always think about it at the first thought and it works mostly all the time.

      To get 0, just use the operation on any queue at the nth moment. To get 1, use operation on the queue with the last element is 1 at the n1th moment. To get 2, use operation on the queue with the 2 last elements are 1 at the n2th moment... and so on.

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

      Primary observation is that a[i][j] >= 1. Greatly simplifies the problem and allows you to make the observations that DuongForeverAlone made.

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

        got it, thanks.

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

        Yeah dude, I reduced the problem to “place rooks in a n by n grid” and was like “how the hell can I solve this without using bitmask??”. Then I noticed that and completely changed strategy.

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

How to solve B ? Was stuck on pretest 2 with 3 tries.

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

    Hint: split the problem into 2 case: k==n and k < n. The answer is very simple

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

      Is there a faster approach than O(n^2)? I thought it would be TLE if it was O(n^2), but mine got AC with a pretty solid time.

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

        Rofl just O(n) what are you talking about?

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

          Yea nvm I thought the problem was much more complicated lol

  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    If n=k, calculate the answer naively, otherwise the answer can be either 1 or 2. Calculate it with this piece of code:

    for(int i = 1;i<n;++i) {
        if (a[i] != 1 && n - i + 1 >= k) {
            ans = 1;
            break;
        }
    }
    • »
      »
      »
      3 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      what if n=7, k=6 array a : 1 1 1 2 2 3 3 then the answer would be 4

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

        [1], [1, 1], [2], [2], [3], [3]. so the answer is 2.

      • »
        »
        »
        »
        3 weeks ago, # ^ |
        Rev. 2   Vote: I like it +5 Vote: I do not like it

        Nope, 1|11|2|2|3|3, b = 11230, answer is 2

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

I couldn't understand the statement of problem D until the very end. Why explain two degenerate (test) cases and not the one where you have a finite non-zero value? Reading comprehension got me I guess.

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

    Now I see, I've missed part of the stopping condition in my solution. :sadge:

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

Wtf is C? How to solve C? 4k solves, is it that easy?

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

    There's only one way to get required mex. There's a pattern here. Pattern exists coz he can only serve one row at a time. Take a look at samples again carefully, big hint hidden in them. Or maybe try to construct mex in reverse direction.

»
3 weeks ago, # |
  Vote: I like it +9 Vote: I do not like it

this contest is so trash B is shit C is weird D is free bad contest

»
3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

how did u guys solve C ?? I tried using suffix sum but couldn't solve

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

    same i got the idea of row suffixes but cant able to determine how to avoid overcounting of suffixes from same row ,how to select the better suffix to include from that row.

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

      Yes I tried to maintain that in a priority set but it was a guess so couldnt get accepted

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

        similarly i tried going through every row suffix if it is not present in any other row then take that after saving prefixes in sorted order and break that row . I too dont know about the clear picture if this is correct way or not

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Anyone solved Problem C with prefix sums and DFS?

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

    I thought of that, but the complexity would be too high.

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

    got tle

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

    I used suffix sum for each queue and check if the sum of values equal to the index

    then sort them according to how many checks are correct

    finally, calculate the MEX (if the current value of correct checks is greater than or equal to my MEX, increase it

    solution
    • »
      »
      »
      3 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Interesting, I sorted the suffix sum arrays in reverse lexicographic order and then did greedy.

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

    It actually simplifies to only counting longest suffix of all 1s in each queue.

    304097874

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

      why these values does not matter such as last 2 values are 2 and 1 so this should contribute to suffix 3 as there might be case when there does not exist any other row with suffix value 3?

      • »
        »
        »
        »
        3 weeks ago, # ^ |
        Rev. 2   Vote: I like it 0 Vote: I do not like it

        Intuition comes when you actually begin to maximize the MEX.

        Observe the events from last round to first.

        In last round, we emptied a queue. So 0 would surely be available for our MEX.

        Now we need 1, observe that in each round, each queue increments by atleast 1.

        So the queue that gives us the 1 must look like [.... 1], but since all elements are greater than 0, we must empty the queue before last 1.

        Now we need 2, again an observation here is we used our last 2 operations to get 0 and 1 to our MEX, then the queue which gives us the 2 must look like: [.... 1 1] , there is no other way. And hence we would use an operation here to clear the elements before and get 2 to out MEX.

        Similarly, the queue which gives us 3 will look like: [....1 1 1] and so on.

        Rest is implementaion.

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

          Thanks for your explanation. I missed to observe that MEX can increase only if there are consecutive 1s in suffix, as that's the only way

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

for problem B, is the answer always equals to either 1 or 2 ? (except for the case of n=k, you have to check it manually and it would be higher than 1 or 2)

Spoiler
»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

How is the answer to this sample test case of B equal to 1?

5 4
1 1 1000000000 2 2

Can't we divide array into {1}, {1, 1000000000}, {2}, {2} so that array b becomes {1, 1000000000, 2, 0} and hence first index != i is 2 (a2 != 2). Output should be 2, not 1?

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

    {1, 1}, {1000000000}, {2}, {2}

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

      Oh, I missed the minimum possible part of the statement lol

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

    why not divide it into {1, 1}, {1000000000}, {2}, {2}? now the array b becomes {1000000000}, {2} and first index of b != 1

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

    We can split into {1, 1}, {1000000000}, {2}, {2}. Now array b becomes {1000000000, 2} and hence first index != i is 1.

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

    one can devide in ~~~~~ {1,1}, {10000000}, ... ~~~~~

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

    split it as {1, 1}, {1000000000}, {2}, {2} so that array b becomes {1000000000, 2, 0}

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Wow, it was really hot. I hope that get positive delta.

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I think you both are retarded

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

cooked

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

D is so old :(

»
3 weeks ago, # |
Rev. 3   Vote: I like it +1 Vote: I do not like it

Did anybody think of doing C with O(n^3) DFS with suffix sums ?

Since t*n*n <= 2e5

So , t*n*n*n <= 6e7

seems acceptable to me, but i could not complete that code correctly.

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

    But why? Observe that a[i][j] >= 1. So, the longest suffix of 1s is all that matters.

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

      "the longest suffix of 1s is all that matters."

      Well, I still do not get it. Can you please elaborate?

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

        The problem can be seen as picking suffixes from lines, since that's equivalent to picking prefixes to erase. You'd never choose to pick a suffix that contains an element >1, since that'd mean that you wouldn't get value matching that suffix length in the final xs list. And that would break the mex chain. Just try making up an example where you'd pick >1 element and you'll quickly see it's impossible.

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

    at first i also thought n^3 could pass but at the end i choose n^2*logn

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

    I tried, got TLEed.

»
3 weeks ago, # |
Rev. 3   Vote: I like it +18 Vote: I do not like it

I've not really enjoyed the contest, but at least the problem statements were quite clear.

  • A: pretty clear, decent A problem, definitely good for its place
  • B: I got stuck for some time figuring that it is a corner case problem;
  • C: Here I spent most of the time figuring the condition to stop iterating over the contigious segments of 1s. Definitely a skill issue, the problem is decent.
  • D: No brainer Dijkstra, just typing exercise, very sad that it is div2D. It should reside in div3 or div4.

Overall, not the worst contest but it could've been much better in terms of balance if D was appropriate.

  • »
    »
    3 weeks ago, # ^ |
      Vote: I like it +27 Vote: I do not like it

    Completely agree for D. It's also sad that random grays can type so much faster than lgms. My typing speed is much slower that o3 and r1, unfortunately.

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

      Yeah, I hope plag check will find some of them.

      I've just checked some known cheaters that I added to my friend list. Their solutions all look very similar.

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

        LMAO, 6th place seems very suspicious. The infamous D problem.

        Look at submissions by harsh__h and the spaces between keywords like "for" and parenthesis. In D he has them, whereas in other problems he doesn't have the spaces in between.

        No accusations, just my personal opinion about specific submission. But if you cheated, it will be hilarious.

        • »
          »
          »
          »
          »
          3 weeks ago, # ^ |
            Vote: I like it -10 Vote: I do not like it

          Bro I solved E1, E2 too T_T

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

            You used chatgpt to save time. Which is still cheating.

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

              Don't know man, i didn't use it. Any proof is welcome. I don't even know if gpt can solve this question, any reference will be helpful.

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

                LMAO, "I don't even know if gpt can solve this question,".

                Are you serious? The most streamlined problem I've seen for a while.

                Also, the fact that you ignore the questions about codestyle and still ask for proof, only exarcebates the situation.

                Now, I am sure that you cheated and that's a shame.

            • »
              »
              »
              »
              »
              »
              »
              3 weeks ago, # ^ |
                Vote: I like it -10 Vote: I do not like it

              If he indeed used chatgpt, then it will be another case when relatively high ranked indian problemsetter is found cheating. If you remember Psychotic_D

              • »
                »
                »
                »
                »
                »
                »
                »
                3 weeks ago, # ^ |
                Rev. 2   Vote: I like it +7 Vote: I do not like it

                I thought only low ranked people would be affected by LLM, as they are usually not able to solve past B; but damn, even high ranked people? This is extremely demotivating. So sad I did not reach Master before it got to this point ;( now it seems much harder to achieve and even if I do it won’t be the same cool achievement if I get placed in the same group as a lot of cheaters. Strongly considering just retiring; but let’s do this for another 1/2 months see where I can get :\

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

                  It is sad but I don't think that you should be upset about the cheaters to the point of retiring.

                  Say, if it is a situation about a job, and in this situation the high rank matters, then probably a cheater-high ranked will be exposed easily during the interview. (maybe just my copium, though)

                  So, my point is that if the environment can distinguish high ranked people, then in this environment cheaters will be exposed. And if the environment does not distinguish, then why should I even care about it.

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

            You has AC only in E1, which is easier than E2.

            Also, mind explaining why you have such different codestyle across submissions? In some you use spaces betwee brackets, in some you don't. In some you use endl, in some '\n'. Very suspicious.

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

              Submited E2 just a small modification in the AC submission.

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

              Used '\n' in E1, E2 too ig. Am just saying don't ruin my name without some solid proof, its sad seeing this after doing CP for 4-5 years.

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

                I've not accused you of anything in the beginning. I just noted that the codestyle was suspicious. But you started avoiding the discussion and asked for "some solid proof" (proof of what? I didn't accused you at that point).

                Bro, you are ruining your name yourself by cheating and making attempts to cover it.

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

                  The gap between him submitting B and D is just 3 minutes. He for sure didnt type all that long ass code for D in just 3 minutes.

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

    gpt killed both c and d in just 2 minutes. Hence the high submission rate.

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

    Any hints for C? i was thinking of storing all the suffix sum in min heap, but could not solve it further...

»
3 weeks ago, # |
  Vote: I like it +76 Vote: I do not like it

The worst div2 I have ever seen in this platform.

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

    Coming from a Grandmaster, why you feels so?

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

      D is too easy and gpt-able and E2 is meaningless.

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

        I respectfully do not agree with you. For me problem D was challenging and I tried to solve it this round but I could not come up with an idea. I usually solve Ds though and even right now I am still not sure the ideas I have in my mind work. maybe for a grandmaster it was trivial but for me it was not.

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

    I suppose you haven't ever participated in tracker rounds?

»
3 weeks ago, # |
  Vote: I like it +21 Vote: I do not like it

anuj23334 leaked the solution during the contest on Youtube YT, please ban him MikeMirzayanov, yunetive29

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Not sure if I understood E correctly. Because it seems so simple.

If we see all the i arrays as one long array, then we have to go from left to right, and foreach a[i]!=b[i], insert the b[i] at that position in a[i].

So it looks like a two pointer like construct. Also E2 is simple then.

What did I miss?

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

history.

»
3 weeks ago, # |
  Vote: I like it +28 Vote: I do not like it

The worst div2 I have ever seen in this platform.

»
3 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

Does anyone have a "nice" solution for B? Apart from my horrible implementation, in retrospect it still feels harder than problem D. Maybe I'm just missing something.

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

    It was casework. There are only two cases one where n == k and other where is n < k for the later the answer is either 1 or 2 .

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

    first element has to be in first subarray, then after that the only case when we're restricted by the test case is when n = k, because then we have to choose one element for each subarray. in that case we can easily find the answer. but if k < n then we can always make the 2nd subarray (first subarray of b) have a size > 2. so now our answer depends on that. if all the elements from i=2 to i=n-(k-2) (leaving last k-2 elements for the rest of the k-2 subarrays as 1st subarray was element 1 and 2nd is going to be within this range) are 1 then our answer will be 2 otherwise it will be 1

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

      That is a lot easier than what I've done, thanks!

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

    If n=k, go manual.

    Else, answer is either 1 or 2.

    First try to get 1. (Just try to find an element greater than 1 which is not the first element and if you're able to make K subarrays)

    If not answer is surely 2. To prove this only consider the 2nd and 3rd element of array. If both are equal to 1, you can choose [1,1] (since n > k, atleast one subarray can have greater than one element) as your second sub-array to get 2 as your answer. And actually there is no other case since if it was not the case, you would have found 1 as your answer already.

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

    I had a DP solution that didn’t involve case work

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

The problem ordering for B and C in this round is not very reasonable; perhaps C should come before B.

  • »
    »
    3 weeks ago, # ^ |
    Rev. 3   Vote: I like it 0 Vote: I do not like it

    Not really, I think we're just bad. (Even in my case, if I'm not choke myself in B, I may have the chance to solve til D)

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

    I agree, I do think C is easier to see than B. That being said, they were both very simple. Personally, I'd be more concerned about grays solving D in < 15min. It's so blatant, just look at this guys submissions: https://mirror.codeforces.com/profile/abhi23100.

    Gave up on C++ and started writing GNUC11

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

      A gray player spent 6 minutes solving Problem D, and the code is 160 lines long. WT????

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

        They are obviously just really really fast at typing. And they just happen to be more skilled at speed coding dijkstras in C than C++.

    • »
      »
      »
      3 weeks ago, # ^ |
        Vote: I like it +17 Vote: I do not like it

      And he coded all that in 6 min :D. Codeforces is dying

      • »
        »
        »
        »
        3 weeks ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        It's the GNUC11 that's killing me. Actually, so funny. Can't wait to see this dude's Oscars speech on linkedin for this phenomenal performance.

      • »
        »
        »
        »
        3 weeks ago, # ^ |
        Rev. 2   Vote: I like it 0 Vote: I do not like it

        It's already dead. I just do virtuals of old contests now. No point in competing against LLMs.

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

cat

»
3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Author photo also has A i in it.

Lol that was a signal may be :smile

Author as A and bottle is i shaped

»
3 weeks ago, # |
Rev. 2   Vote: I like it +1 Vote: I do not like it

Did some silly things in C. Tried to calculate suffix sum and transform the problem into picking one number in [1,x] per row and per column, and check if we can pick x numbers. Initially I thought this is a simple Bipartite Matching, but it failed because there are actually three constraints(row, column, value each can appear at most once)

I know this is not the intended solution for C, but I'm still curious about is this triplet(?) flow problem solvable?

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

    I'm curious about this as well, if this is some kind of max flow matching ? I came up with this same idea as well and wrote a brute force DFS which got TLE ofc :(

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

    I did same as you and reduced the problem to place n rooks in a nxn grid. Unfortunately I am pretty confident that using this strategy the problem becomes NP complete, from other “place rook in the chessboard” problems I have seen

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

    As a person who submitted matching without spotting that it can be further optimized, triplet problem is probably unsolvable, but I've spotted a fact that number 1 can only appear if some queue has 1 in the end, so we already know that on the pre-last day some queue will be cleared. Then, 2 can only appear on pre-last day, meaning there need to be two 1's at the end and so on. So we only need edges from number x to row i, if the last x elements of i are 1.

    In order to solve the problem now in O(n3), we run dfs from vertices, that correspond to numbers, in increasing order, and if we didn't find the augmenting path, current number is the answer

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

    I've also failed to notice intended logic for C so I solved the way you described: 304114466

    Hungarian (adapted from e-maxx) does exactly what we need: -v[0] stores current answer for i rows. So I just transposed the matrix and overkilled D2C with hungarian

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

If we have to maximize the score in B what can be possible solution ? Unable to think of one

  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    I think the following solution may work, find a way to mark the indices in the following order, let the index i is marked and the element at index i.e., (v[i] is x), so we have to get index j such that (v[j] = x + 1) and (i + 1 < j <= n) and j should be minimum among all such (v[j] = x + 1), so after getting all these indices, we can say that we marked the indices of elements from [1..p](also ensure index of 1 is > 1) and there are gaps between each pair of elements, so that we can take those subarrays at even positions to get sequence (1, 2, .., p) (assume that indices stored in vector A). Now, all we have to wish is that the condition over k should be satisfied. So, to ensure that binary search on the marked indices, to see that if a answer exists from [1..mid] as the number of subarrays that will be formed here is (2 * mid + n — A[mid])(we can see it is monotonic as A[mid] — A[mid — 1] >= 2). So, see if answer for the current mid exists or not and shift accordingly (if the no of (subarrays >= k), we can select at least mid number of elements).

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

sorry, but, where if solution?

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

are Russian have no access to ChatGPT due to sanction?

seriously, I have no idea how problem like D is acceptable

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

It’s been so long I haven’t seen a Dijkstra on codeforces I was going for a completely different solution. When I finally realized my solution was wrong with 26 min remaining, I stepped back, thought for a while and realized it was literally just a Dijkstra (which funnily enough was much simpler to implement than my previous strategy). I don’t know why so many people complaining, I thought it was a cool question.

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

    I think people complain because you look at the start of the round and you see screenshot.

    Like <2000s solving D in 15m is just terribly egregious. Need to increase verification. Otherwise, virtuals would be more fun.

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

      you can't increase increase it beyond catching the lamest of cheaters who will copypaste the whole solution, only way is trying best to make soln not possible by LLMs and accept they will be used

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

        I mean phone number verification + account lifespan/activity would greatly reduce it, I feel.

        At the very least, we can start removing the brainless copy-pasters who are getting 5 compile time errors across 4 different problems in 10 minutes. But there's not much of a point if they just keep making accounts.

»
3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

For the input in Ques 2

1

8 8

1 1 2 2 8 8 4 4

answer should be 3... But codes outputting 5 get accepted For example this

Submission

Code
  • »
    »
    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Verified. What sorcery is this? Can you hack submissions?

    EDIT: PS: Please post in way that is readable for others. Like put code in Code tag so we can drag in/out etc.

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

      Done..

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

        Did you hack!?

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

          No, I got my pretests passed at first with this wrong code then I changed the solution realising my mistake... But to my surprise my previous wrong code is still getting accepted

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

            I just received a text that hurts more than the breakup I had recently. "Your solution to B got hacked"

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

when will the ratings be updates?? system testing is already finished.

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

can anyone explain what problem D asks us to do ?

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

    it ask us to find the minimum cost.

    the cost can sometime be infinite as we are performing the steps infinite times

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

does E2 soultion depends on E1 soultion?

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Check please this submission. https://codeforces.net/contest/2059/submission/304088287. I think this is ChatGPT and people didn`t delete comments. It is suspiciously. Because Green complete D on 30 minutes.

»
3 weeks ago, # |
  Vote: I like it -83 Vote: I do not like it

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Too many downvotes incoming? :)

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

For the first problem A, I have found a corner case that should give me the output "NO" but is giving me a "YES" output for the editorial solution. The corner test case is: 1 3 1 1 2 1 2 1. Could someone please check if I have misunderstood the problem, or if there is an issue with the editorial?

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

    your corner tc is not clear, it needs to be 2 arrays

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

    None of those arrays ([1,1,2] and [1,2,1]) are good; read the problem carefully

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

    You almost had me XD

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Did codeforces start updating profile ratings immediatly? Gave context after long time. Few years ago it used to take 1-2 days i guess.

»
3 weeks ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

The accepted code of this problem [Codeforces Round 1002 (Div. 2) (A)] for this test case give the answer "YES" but i think it should be "NO". Can anyone define it please let me know .

test case = 1, n = 4, a = 2 3 2 2 and b = 2 3 2 2

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

    "An array is called good if for any element x that appears in this array, it holds that x appears at least twice in this array"

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

C is great!

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

It was a great contest for me. Finally, I became a specialist.

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

Although I became Master in this contest. But I still think using an ad-hoc problem as a distinguishing problem is bad.

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

    Bruh, it's just your personal taste

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

      Sure, but I don't think there's anything wrong with saying that expression (other than English grammar issues).

»
3 weeks ago, # |
  Vote: I like it -9 Vote: I do not like it

Why does the max rating update so slowly? It's important for me because it's my first time to become expert!!!

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

bra

»
12 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Hello, could someone please tell me when Codeforces typically runs the plagiarism check? For example, how many days after a contest does it occur? Also, for this round, have the plagiarism checks (and the process of skipping certain solutions) been completed, or are they still pending?

»
8 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Subject: Clarification Regarding Plagiarism Notice

Hello Codeforces Team,

I recently received a notification stating that my solution (ID: 304086409) for problem 2059A coincides with another user’s solution. I would like to clarify that I did not share my code with anyone, nor did I copy from any source.

Possible reasons for this coincidence might be:

Common Approach: The problem might have a standard or well-known way of solving it, leading to similar implementations. Pre-written Templates: I use a personal template for competitive programming, which may match with others who follow a similar style. Unintentional Leakage: If there is any suspicion of my code being leaked, I assure you that I did not share it publicly. I kindly request a review of my case, and if needed, I can provide further clarification. Please let me know if there’s any way to resolve this issue.

Thank you for your time.

Best regards, tajwone17

»
8 days ago, # |
  Vote: I like it 0 Vote: I do not like it

giant bottle what's the point in that?

»
7 days ago, # |
  Vote: I like it 0 Vote: I do not like it

Subject: Clarification Regarding Code Similarity Flag (Solution 304135837)

Dear Codeforces Team,

I hope you're doing well. I recently received a notification stating that my solution 304135837 for problem 2059A coincides with another participant's solution . I want to clarify that I had no involvement in any form of code sharing, plagiarism, or unintentional leakage. The similarity is purely coincidental, likely due to common logical approaches taken by many programmers when solving such problems.

I take competitive programming seriously and always strive to solve problems independently. I have never shared my code through any public platform like Ideone or elsewhere. If there's any specific aspect that needs clarification, I'd be happy to cooperate and provide any necessary details. I respectfully request a review of my case, as I am committed to fair play and would love to continue growing through Codeforces contests.

Thank you for your time and consideration. I truly appreciate all the effort you put into maintaining a fair and competitive platform for programmers like me.

Best regards,
Tareque101

»
38 hours ago, # |
  Vote: I like it 0 Vote: I do not like it

how did the ratings of tianbincheng and FHQY change although they were master before the contest?

»
17 hours ago, # |
  Vote: I like it 0 Vote: I do not like it

Maybe I'm the only one waiting for this, but when is the problems rating updated?