shiven's blog

A summary of what passes for an Indian ICPC Contest, apparently
By shiven, history, 7 weeks ago, In English

The online round for the India’s ICPC was held on 23rd. And also a week before that. The first time around, the organisers tried debuting a self-hosted DOMjudge instance. That didn’t go too well, so a recontest was scheduled on CodeChef.

If you’re not too familiar with the Indian ICPC system — this online prelims is used to select teams that can participate in the offline regionals. From a single university, usually approx. 2 teams (for the top unis) or 1 (for almost all unis that aren’t in top-15) qualify.

As you could guess, it’s quite easy for good teams to miss a spot. Looking at the first contest’s rankings, teams from rank 2 to 100+ all had 4 solves. Owing to severe system issues in the first contest, the time penalties weren’t too reliable differentiators, leading to the recontest this Saturday.

In hindsight, we forgot to be grateful for the problemset and reliable testcases the first time. I’m sure everyone agrees that the basic requirements (ignoring all other stuff like problem quality, balance, …) from an online contest would be:

  1. A functioning judge
  2. Correct, validated testcases
  3. Testcases sufficient enough to weed out obviously wrong solutions. (Notice the obviously: as we’ll soon see, the official solution shared by the contest admin isn’t even in polynomial time. Of course, it passes all cases with $$$N=3000$$$)
  4. Problems that can’t be easily Googled or insta-GPTed.

The first contest lacked (1), while the re-contest failed at (2), (3), and (4).

However, the organisers seem to believe that the re-contest was a fair metric just because CodeChef didn’t break, so here’s a compilation of what went wrong. Maybe they re-evaluate their decisions, maybe they realise that we deserve better. Regardless, even as part of a team that qualifies for both regionals from either version of the prelims, the experience was too bad to pass as a standard for ICPC.

Disclaimer: This is just a list of issues that I sourced by talking to the handful of teams at my university. Since the official Telegram group was blocked by the organisers following the re-contest (with a final message requesting to ‘trust the process’), there wasn’t an avenue to discuss the details. Hopefully this blog becomes one, because I’m sure that I’ve just discovered a small subset of issues. Additionally, all submissions are still hidden so we can’t figure out any more issues by looking at accepted code.

The re-contest

It had 5 problems, available here.

4, for all practical purposes, since the first was a cakewalk (and insta-GPTable). And also easily Googleable, because it is the exact same as another CodeChef problem. Also, as will be a recurring theme here, the official solution is easily hacked. Just to point out the obvious, whenever this happens, it means that the in-contest testcases don’t account for this, because the solution written by the contest admin would agree with / be the model solution.

Now for the remaining problems.

Small Indices

Here’s a summary of (a subset of) what ACs for this problem. A reminder that $$$N=3000$$$.

  • GPT writes an obvious $$$O(N^3)$$$ DP. Works.
  • $$$O(N)$$$ greedy. Wrong, ofc. But works.
  • $$$O(N)$$$ DP. Wrong, ofc. Also works.
  • $$$O(2 ^ {\log^2 N}) \approx 2^{121}$$$ brute force, written by the contest admin, finalised as the official solution.

Let’s talk more about this last one.

This video shows the original problem had $$$N=10^5$$$, presumably with some incorrect model solution. The admin then ACed with a recursive brute-force, with the above non-polynomial complexity. They probably discussed this and realised that the earlier faster solution was wrong, and also (wrongly) agreed that the brute-force works in $$$O(N^2)$$$. So they reduced the problem constraints to intend $$$O(N^2)$$$ solutions to pass.

Somehow, the end result → neither does the model brute-force solution work in time, nor do the test-cases enforce any time complexity (given that the model solution was itself guilty of being the slowest).

To add to that, the solution is presented in a way that pretends that the complexity is obviously correct.

The claim basically is: ‘if the recursion depth is $$$N$$$ and any single chain branches off atmost $$$K$$$ times, the tree has atmost $$$N2^K$$$ nodes’. There is no attempt to justify this. In fact, it is not even stated anywhere explicitly. Just assumed. The video goes: ‘maybe there are not too many states, let’s try’. It passes their own bad testcases (remember, still for $$$N=10^5$$$), and it was thus finalised as a problem. I’m surprised that this was not a huge red flag for the setting team.

We quite easily hacked it, and Igen reported it in detail. The response? A band-aid of caching on top, and a prompt ‘can you now find a counter-case’? Come on :\ Can we now get an acknowledgement of how much was messed up in this problem? Can setters now admit that a contest needs a provably correct solution? (On a side-note, our team ACed with a DP with $$$N^2 \log N$$$ states. It seems correct to us, but given that the test-cases apparently let anything slide, our word is as good as anyone’s. Can detail our solution later if needed.)

Yet Another GCD Problem

Constraints mention $$$K \geq 1$$$. Testcases don’t agree with this :(

No clarification regarding this was made during the contest. Now in a 2.5h ICPC-style penalty contest… you get the idea of all that can go wrong due to this. In all the contests that I’ve heard or been a part of, the organising team monitors for any such obvious validator mess-ups that lead to failing submissions. They announce the issue ASAP (and fix it or call it off, depending on the severity). This didn’t happen here. This was addressed after being pointed out by participants in CF comments of the official editorial.

Equations

After the contest, we pasted the statement into ChatGPT. Without any prompting, it quickly wrote a DSU which maintains a bi-colouring with other required info. Great.

This is not surprising at all. All of our immedate reactions on reading this problem — why does this library-checker problem exist? Just a thin veil of ‘observation’ (which is itself very standard — make a graph when pairwise constraints on sum, XOR, or whatever). As is well known, LLMs excel at these.

Okay, maybe I should have been more careful while calling the ‘observations’ lacking. Because among the only novelties in the problem was realising the $$$w=0$$$ case, and it was apparently enough to mess up the testcases. The official solution gets hacked... again.

As a sidenote, I appreciate the commitment from the organisers to hold a re-contest after the judge broke. Unfortunately, even after the re-contest, it doesn’t look like we’re in a better situation. You don’t expect the most awaited annual competition [with guesstimated upwards of 3M INR collected just as registration fees, along with sponsorships and stuff] to be far worse then online contests held every week throughout the year on every OJ.

  • Vote: I like it
  • +925
  • Vote: I do not like it

»
7 weeks ago, # |
  Vote: I like it +19 Vote: I do not like it

Is your n^2 logn dp based on there being 20 Fibonacci numbers under 6000?

→ Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +26 Vote: I do not like it

    Yes, that’s the $$$\log$$$ multiplier in the state.

    → Reply
»
7 weeks ago, # |
  Vote: I like it +84 Vote: I do not like it

I really want whatever the problem setters are smoking.

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

    Contest admin Mr. Anup Kalbalia stated on the telegram group that they won't rejudge any AC solution. The only implication this has is that 3/5 problems were wrong and didn't adhere by the global ICPC rules of 10^8 operations per second which is totally unacceptable.

    → Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +68 Vote: I do not like it

    I would assume the re-contest was prepared in a rush.

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

      but sir atleast they could have added extremes as test cases such that n^3 solutions couldn't pass for n = 3000 even some wrong greedy solutions were passing

      → Reply
    • »
      »
      »
      7 weeks ago, # ^ |
        Vote: I like it +35 Vote: I do not like it

      What do rules in other ICPC regions (throughout the world) say about a contest whose 4 out of 5 problems have issues?

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

      Yes, it looks like the organizers were under pressure to conduct the online round ASAP so the regionals schedule wouldn't get affected, and the online round would happen while most colleges still had an ongoing semester.

      Sadly, they failed to do even the most basic sanity checks and the situation has gotten very messy now :(

      → Reply
      • »
        »
        »
        »
        7 weeks ago, # ^ |
          Vote: I like it +44 Vote: I do not like it

        Based on everything I’ve seen, I believe a regular Codeforces round is far better than the ICPC India prelims.

        Firstly, they should have prepared an additional problem set. Secondly, considering their budget ( INR 1100 per team for one regional), they could have collaborated with experienced problem setters from Codeforces. These setters could have provided original problems, valuable feedback, and assistance with testing. Instead, what they delivered was disappointing, and their current inaction and silence on this issue make it even worse.

        I am requesting a recontest. Even some of the teams who qualified are asking for a recontest (apart from those who cheated their way to qualification) .

        If they choose not to do recontest, it will decrease the credibility of ICPC.

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

          what makes it worse is that the contest was a re-contest

          → Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it -174 Vote: I do not like it

I understand but you have to understand doing a recontest in a week was a very big challenge.

→ Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +101 Vote: I do not like it

    Unless they proposed the statements in a week, it is more than enough to:

    1. Prepare strong testcases
    2. Validate the testcases
    3. Check if the problems can be ACed using GPT

    Given the funds ICPC Prelims have from the registration, these are the very basic requirements.

    → Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +46 Vote: I do not like it

    The decision to conduct a recontest was in the hands of the problemsetting team. It is a hard task, but it is also something that thousands of teams care about — and have paid good money for. Many teams have put in well over a year of consistent effort to do well in the online round.

    The contest's fairness is not a "good-to-have", and it is the problemsetting team's responsibility to conduct the online round properly. If it was not feasible to do a contest in a week, they should have taken more time to verify everything. Negligence is not okay.

    → Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +19 Vote: I do not like it

I believe everyone is still trying to make sense out of the problem Small Indices, if you could take some time for the sake of it, please write an editorial that clears out this problem once and for all addressing why caching works (if it does) for the $$$N^{3}$$$ dp and stating the correct $$$O(N^{2} \log N)$$$ approach. Thanks in advance!

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

    Let dp[i][j] be a tuple (number of maximum small indices of subarray from [1 .. i], maximum element in [1...i], second maximum element in [1..i]). Now from dp[i][j] you have some cases on how large/small A[i+1], B[i+1] are compared to largest, second largest elements in [1..i](stored in the tuple dp[i][j]). Accounting all the transitions gives you a N^2 solution.

    → Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +58 Vote: I do not like it

    An $$$O(n^2)$$$ solution, that is correct, exists. We just need to maintain for each sum

    1. The number of smaller indices encountered so far.
    2. The largest element corresponding to that sum, and number of indices.

    Now, Why would a sum not prefer fewer smaller indices with a larger element?

    Consider the following scenario:
    Let $$$ a + b = c + d $$$, where $$$ c > a \geq b > d$$$.

    Given

    1. $$$a + b = { ( \text{mx}, a })$$$
    2. $$$c + d = {( \text{mx} - 1, c})$$$

    You can achieve a greater sum by forming $$$(c + a)$$$ or $$$(c + b)$$$, which will include at least $$$(\text{mx} - 1)$$$ smaller numbers. This, in turn, is better than $$$(\text{mx} - 1, c)$$$ with the sum $$$(c + d)$$$.

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

      During the contest, we did come up with a similar idea but dismissed it quickly, assuming a simpler one existed due to the high number of accepted submissions. Unfortunately, this decision led to us not solving the problem. Looking back, it would have been wiser to proceed with this approach regardless.

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

        Unfortunate, I was in the same boat. However, I didn't notice the number of accepted solutions and ended up solving in n**2 time

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

      I used the same approach. I don't think your proof is correct, tho that was my reasoning when solving the problem.

      → Reply
»
7 weeks ago, # |
  Vote: I like it +96 Vote: I do not like it

also why not conduct separate prelims for every site? good teams won't be bothered as much by the stupid preparation of one contest if they get another chance. we are anyways paying for all the sites separately.

→ Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +89 Vote: I do not like it

Just to mention, every team paid 2200 INR (assuming they participated for two regionals, 1100 if one) for this "ICPC prelims". I could have used it to buy whatever they were high on.

→ Reply
»
7 weeks ago, # |
  Vote: I like it +57 Vote: I do not like it

Gym contest problems with their statements altered are less GPTable than these "original" problems

→ Reply
»
7 weeks ago, # |
← Rev. 5   Vote: I like it +85 Vote: I do not like it

I don't think people realize how bad the situation with small indices is, there were over 200 solves on the problem, ignoring the obviously wrong greedy and $$$O(N^3)$$$ solutions, Most of the teams that didn't use these sols, Used a $$$O(N^2)$$$ one. I have looked at 2 solutions of 2 top 15 teams and multiple others from higher teams that are $$$O(N^2)$$$ and found an edgecase for all of them, currently there is only one $$$O(N^2)$$$ sol that i have found to be correct with a rough proof. We were rank 3 and used an $$$O(N^2log)$$$ sol ourselves, i bet if they open the submissions and people tried to look for edgecases, then there would be very less teams that actually solved the problem.

Edit: the correct $$$O(N^2)$$$ sol i mentioned has been mentioned here.

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

    I think this $$$O(N^2)$$$ solution is also correct.

    • $$$dp1[i][j]$$$ -> {sum, mx} pair for maximum sum(of top2 elements) with $$$j$$$ small indices in $$$prefix[i]$$$

    • $$$dp2[i][j]$$$ -> {mx, sum} pair for maximum max-element with $$$j$$$ small indices in $$$prefix[i]$$$

    (here, $$$sum$$$ is sum of maximum & 2nd-maximum element)

    now, for a fixed no. of small indices, either we will get better top-2 sum by using greater max-element or by using greater sum of top2 elements,

    so we will compute out trasition for $$$dp1[i]$$$ using $$$dp1[i-1]$$$ and $$$dp2[i-1]$$$, and for $$$dp2[i]$$$ using $$$dp1[i-1]$$$ and $$$dp2[i-1]$$$.

    Code

    P.S : It is passing all the edge cases I've encountered so far, and logically seems correct. Do tell me if I am missing any edge case.

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

      you can reduce memory used to order of $$$n$$$ instead of $$$n^2$$$ by basic knapsack concept. The solution is mathematically provable, I have the proof in mind.

      → Reply
    • »
      »
      »
      7 weeks ago, # ^ |
      ← Rev. 2   Vote: I like it +9 Vote: I do not like it

      Your solution will only fail if at the end of $$$ith$$$ prefix, it possible to have a three different configuration of choices with same number of small indices having (sum,mx) pairs as $$$(a,b),(c,d),(e,f)$$$ at the end of $$$ith$$$ index such that $$$a<c<e$$$ and $$$b>d>f$$$. I have a bit of a rough proof that this will not happen.

      Actually, if the above sol is correct, then you can do it in $$$O(nlogn)$$$. The main observation is that the optimal solution will not have more than $$$O(logn)$$$ indices that are not small indices.

      Instead of $$$dp[i][j]$$$ representing $$$j$$$ small indices in $$$prefix[i]$$$, $$$dp[i][j]$$$ will represent $$$j$$$ non-small indices in $$$prefix[i]$$$, $$$j$$$ will have an upper bound of around $$$2logn$$$ and we will make the same dps, as you have mentioned.

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

        Agree with the part that it could be actually reduced to $$$O(nlogn)$$$ by storing non-small indices.

        But, can you elaborate further, on why my solution will fail on the given case you mentioned..?

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

          Made a small mistake in the sol, it should be $$$a<c<e$$$ and $$$b>d>f$$$.

          After the $$$ith$$$ prefix, we can add the following $$$(a_j,b_j)$$$ pairs, $$$(c,c),(c+d,c+d)$$$ and then $$$(1,1)$$$ 100 times. In the following case only the pair $$$(c,d)$$$ will be optimal.

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

        https://codeforces.me/blog/entry/136522?#comment-1222131
        is an incomplete proof.

        "The main observation is that the optimal solution will not have more than O(logn) indices that are not small indices." is there any proof for the same?

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

    how do you see others' solutions I am unable to see them by going on the submissions tab.

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

Ideally, there should be a re-contest again.

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

And they think rejudging the submissions will solve all the issues as mentioned here

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

    They only rejudged the submissions for "Yet Another GCD Problem" since it contained invalid test cases, that is, cases containing k = 0. They've decided not to change the verdict of any submissions for the "Small Indices" problem. The submissions for the "Small Indices" problem should also be rejudged with much stronger test cases. There are many brute-force $$$O ( n^{3} )$$$ solutions and many other incorrect greedy solutions that have allegedly been accepted. Many deserving participants are directly affected by this alleged inaccurate judging of solutions.

    → Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Our N^2LogN solution

A trivial solution would be to make a n^3 dp where dp [i][j][k] would represent the max number of small indices you have if the max value and second max value till the ith index is j and k respectively.

But think about number of NON small indices, those can only be around logn.

You can just swap the states of the dp matrix to something like dp[i][j][k] would represent the second maximum value till ith index if j the the maximum value and you have k NON small indices up till now.

Note:- Max value and Second Max value is Cj and Ck respectively.

As 1<=i<=n & 1<=j<=2*n & 1<=k<=25 and for each state there only some constant transitions.

Memory limit was kinda strict for us so we mem optimized it.

→ Reply
»
7 weeks ago, # |
  Vote: I like it +38 Vote: I do not like it

The thought of a solution that runs in 2^121 is quite hilarious, does the editorialist even know what he's supposed to be doing? How can the organizers be this incompetent

→ Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +16 Vote: I do not like it

I don't think they will do anything now they muted the telegram group also. It is very shameful that at such a big level this things are happening.

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

Now I know why CM team from our college only able to solve 2 problems

→ Reply
»
7 weeks ago, # |
  Vote: I like it +13 Vote: I do not like it

We r demanding for a recontest whereas they dont even care to add a strong test case for Small indices where n^3 solutions fails. According to them it will be injustice to them who submitted n^3 solution in a problem where n=2000.

→ Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 3   Vote: I like it +16 Vote: I do not like it

    I am not against of re-contest but If they can fix the k = 0 case for the unsatisfied array. Why can't they fix the cases for which n ^ 3 solutions passes. They should do something otherwise it will be a total mess for this year ICPC. The testcases are so weak that greedy solutions are also passing.

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

Imagine the organizers saying, "We can't penalize any team right now because they weren't informed during the contest," while the author themselves had no clue how to solve the problems during the contest. What even is this logic? :p Only 1 correct problem out of 5, that too exists on the internet.

→ Reply
»
7 weeks ago, # |
  Vote: I like it +77 Vote: I do not like it

Looks like I graduated at the right time. Sad

→ Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

How does someone even try writing a n^3 solution for a n=3000 and gets accepted ! (knowing that there is 20 min penalty on wrong submission)

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

    yeah man no one in there right mind will write o(n^3) solution for that problem but more than 80% ac solutions on that problem is o(n^3) that shows how much cheating has affected deserving people and weak test cases too

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

      Many have also written wrong $$$O(n^2)$$$ that have ACed including mine.

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

Each year, the quality of online prelims is degrading. Can anyone from the management answer, why with utmost funding, there is no improvement? What's the point of holding even a single contest for all regionals, if the quality is not up to the mark. It's understandable somehow that holding different online rounds requires alot of time, but since it is reduced to a single contest since 1 or 2 years, is the expectation too high to have a normal experience? It's better if this gets highlighted before it gets worse.

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

    The respected organiser is busy doing podcast (check here).

    Unfortunately, nothing is going to change because they know students will register again next year anyway. It has essentially become just a way to make money, nothing more.

    Also his reply on the first failed contest of icpc. Click.

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

      The simplest solution is to have people in the management who respects these issues and the sport of this competition. I wonder why they don't have people who genuinely contribute for the uplifting of this competition.

      → Reply
»
7 weeks ago, # |
  Vote: I like it +61 Vote: I do not like it

The problem-setters and/or admins have stated that it is unfortunate that weaker solutions have passed the Small Indices problem. However, many believe it would be unfair to participants to modify the problem now—by removing it, adding more test cases, or otherwise—because teams that solved it during the contest might have come up with better solutions if they had received a WA verdict at the time.

But how is this justification valid without considering the teams that couldn’t solve the problem? Many teams spent hours analyzing it, watching the number of solves climb, while knowing that an O(n³) solution should not pass the time limit. These teams, understanding proper runtime analysis, refrained from submitting such solutions to avoid incurring a WA penalty. What about these teams? They likely discarded incorrect greedy solutions because they recognized they were flawed.

The claim that teams could have fixed their solutions if they had received a WA verdict during the contest is equally unconvincing. Many teams spent the entire duration of the contest working on correct solutions. It’s highly likely they wouldn’t have had enough time to debug or come up with an alternate approach either.

How can this contest be considered valid? Teams dedicate an entire year preparing for ICPC, sacrificing so much in their college life to perform well. For many, this is their final year of college and their last chance to qualify. And now, that opportunity has been taken away—perhaps unfairly—by teams that either didn’t understand the correctness of their solutions or weren’t even aware of runtime constraints.

→ Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +28 Vote: I do not like it

    Enigma27 Arpa is there any chances of recontest?

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

    I would like to add that most of the accepted submissions for this problem were in the last 5 minutes(approx 100 out of 300) so there logic doesn't apply on many submissions.

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

What about the participants who had their last attempt this year and couldn't perform well due to all this mess

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

To add, Yet Another GCD Problem is almost copied from here .

→ Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +25 Vote: I do not like it

    Yo what the hell, it may be possible they actually copied it as this question allows $$$k = 0$$$ which they did while copying the problem on their initial testcases.

    → Reply
»
7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

So who actually suffered in all this? The final year teams comprising of Masters and CMs? NO, they would qualify anyway. The noobs and the cheaters? they don't either coz they never had the chance to begin with, but rather they would be happy coz they might have submitted the O(N^3) solution or the greedy one(without caring about the penalties), and they might as well be going to the regionals. The people who suffered are those enthusiastic second or third years that fall between these two categories, many of whom participated for the first time. If given the fair chance, I'm sure people from this category can replace the people from the second category. These are also the people who'd most likely be in the regionals next year, the experience from this year's qualification would've helped them a lot. And ultimately would help India's position at the global stage. But thanks to the dumb organizers.

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

    yeah sir we are in that last category I am in my 2nd year participated for first time and it was worst experience for me, although there was skill issue too with me but our team could have done better than what we did in the contest

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

    Seems like I am the perfect example of a Type 3 participant :(

    → Reply
»
7 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Only one thing I want to say $$$-$$$

There exists KARMA. One day, Arpa, Enigma27, Th3_M3nt0r, cosmic_vector, anupkal, or their children, close relatives, or even themselves might participate in the ICPC prelims. Then, they will face similar issues: poor problems, weak tests, cheating, etc. ...

→ Reply
»
7 weeks ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

Can someone give few testcases for unsatisfying array since mine fails on tc 5.

My approach is to put the minimum value possible at any given instance. If 1,2,3 and 6 cannot be used at a index, use 4 and remember 1,2,3 to not use them till their respective ending index and ignore condition for 6 since 4 is already smaller than 6.

or please give any general edge cases here is the code:


int main()
{
    DRAMA;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

   int t;
   cin>>t;

    for(int o=1;o<=t;o++){

        ll n,k;
        cin>>n>>k;

        vector<pair<pair<ll,ll>,ll>> v;

        for(int i=0;i<k;i++){
            ll a,b,c;
            cin>>a>>b>>c;
            v.push_back({{a,b},c});
        }

        sort(all(v));

        ll j=0;

        vector<ll> ans(n);

        int r=0;


        vector<ll> temp(n+2,0);

        priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> pq;

        for(int i=0;i<n;i++){

            vector<pair<ll,pair<ll,ll>>> temp1;

            while(j<v.size()&&v[j].first.first-1==i){
                temp[v[j].second]++;
                temp1.push_back({v[j].second,v[j].first});
                j++;
            }

            while(!pq.empty()&&pq.top().first-1<i){
                temp[pq.top().second]--;
                pq.pop();
            }

            ll res=mex(temp);

            if(res>n){
                r=1;
                break;
            }

            ans[i]=res;

            for(int i=0;i<temp1.size();i++){

                if(temp1[i].first<res){
                    pq.push({temp1[i].second.second,temp1[i].first});
                }
                else temp[temp1[i].first]--;
            }
        }

        if(r==1)cout<<-1<<endl;
        else {
            for(auto el: ans)cout<<el<<" ";
            cout<<endl;
        }

    }
}
→ Reply
»
7 weeks ago, # |
  Vote: I like it +41 Vote: I do not like it

I'm sorry about what happened in the ICPC prelims. The problems passed to me just 10 hours before the contest. As you can see in the video almost all the time I was just engaged with solving them. The time was such tight that I was occupied with just solving the problems to finish them before the contest started. I didn't have any time to test.

I know that the organizers are making a decision that will make everyone happy.

→ Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +94 Vote: I do not like it

    I know that the organizers are making a decision that will make everyone happy.

    What exactly they planning to do

    → Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    If organizing a recontest is not possible, then increase the number of seats in the regional competition as much as accommodation permits, and allow more teams from each college. Top university teams are the ones most affected due to close rankings with their counterparts. Increasing the number of teams would satisfy most participants.

    → Reply
  • »
    »
    7 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +33 Vote: I do not like it

    One week and there hasn't been any official statement.

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

      I am 5th in my college and my solution was an incorrect greedy :)

      → Reply
  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    They have updated the selection criteria here

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

      What is the logic to prepare Ranklist 2, they should remove the wrong solutions and not the complete problem. There would hardly be any differences in both the ranklists.

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

        I already mentioned this before. Both the ranklists are practically the same. They only have 19/220 extra teams in the Amritapuri region with their new scheme. The number of teams affected are significantly higher, probably upto 300.

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

Given the amount of effort teams put in throughout the year to prepare for such contests, it's disappointing how the organizing team handled the situation. At the very least, a re-evaluation of the submissions with stricter test cases for 'Small Indices' would be a minimal step toward fairness.

Alternatively, increasing the number of qualifying slots for this year’s regionals to compensate for the errors could be a solution that benefits deserving teams without compromising the fairness of the competition.

Hoping this gets addressed before the credibility of such a prestigious competition takes further damage!

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

    I am not sure though but after everything they have have said(also considering the time that they are taking), I think increasing the number of qualifying slots is something they are looking forward to. Which also would be a good enough solution for the error that they made.

    [JUST A WILD GUESSS!!!]

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

      I don't really think that would be a great idea, just because trying to increase the capacity of regional sites in such a short time is likely to cause issues in regional rounds. Also, selecting the extra students based on final performance is itself a contradiction. There still might be teams not making to regionals but are competent enough in a well executed contest.

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

        I think they will choose this option. After all, regionals are the only thing left to be spoiled now by them.

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

        Well I only said that because I think they are probably not thinking about a recontest. And as they said they won't give wa verdict to those wrong solutions that were AC during the contest

        [So I thought the only option left is this] or they won't do anything and leave the standings as it is.

        → Reply
»
7 weeks ago, # |
  Vote: I like it +45 Vote: I do not like it

From a single university, usually approx. 2 teams (for the top unis) or 1 (for almost all unis that aren’t in top-15) qualify. As you could guess, it’s quite easy for good teams to miss a spot.

This is a kinda unfortunate feature of ICPC contests — it's much easier for a good contestant to reach further from a weak school since the number of spots is mostly a small constant like that. With regionals there's already a serious effort to balance that by giving more advancing spots to strong regions, but with hard caps on number of advancing teams from any school (including 1 for finals), the optimal strategy seems to be forming a team ahead of time and picking a school that won't give you rivals and it's only balanced out by the fact that ICPC is a secondary concern after actual education.

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

Selection Process

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

    Not expected this weird thing from ICPC

    However, we are doing our best to accommodate most of the affected teams.

    Why most? and not all

    "Yet Another GCD Problem" Adjustments: We rejudged all non-accepted submissions after removing the K=0 case. Penalties were adjusted based on the first accepted submission.

    And what about the teams that wasted their time trying to debug incorrect cases?

    "Small Indices" Problem Removal: After the "Yet Another GCD Problem" rejudge, we then removed the "Small Indices" problem completely and recalculated the ranklist.

    What about the teams that solved this problem with the correct time complexity?

    Doing all this doesn't compensate for the time wastage of contestants, frustration level during contest.

    Both the released ranklists are nearly identical, showing that the adjustments made little to no impact. The core issue is that these adjustments fail to account for the time wasted during the contest i.e. debugging the K=0 case or solving the problem in better time complexity.

    Some teams could have utilized this wasted time to solve other problems, but our precious ICPC doesn’t seem to consider the time factor. Their approach assumes that if a team solved 3 problems in 1.5 hours, they could have solved the same problems in 2.5 hours, completely ignoring the competitive dynamics and time pressure of the contest.

    Imagine if such mistakes happened in a Codeforces round. Would Codeforces handle it with adjustments like this? Obviously not. There would be a recontest.

    This wasn’t an offline round where organizing a recontest would be difficult. Instead, it feels like the organizers were more concerned about keeping the onsite round on schedule. They seem to be rushing to wrap things up(unfairly) without giving much thought to the teams who have been preparing for this competition for years.

    There was/is a simple and more effective solution to this- recontest

    → Reply
»
7 weeks ago, # |
  Vote: I like it -9 Vote: I do not like it

They should conduct recontest, altering selection criteria is not going to make this contest fair.

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

We rank 2nd in our college and the first team filled both regionals same as us i.e. Kanpur & Chennai.If they decided to go to only one of the regionals say Kanpur, did we get the chance to go to Chennai regionals or the seat from our college left vacant.

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

Man, I feel like the original ranklist was better than this fked up merged Ranklist. How does this even make sense?

Considering penalties, let's say one team didn’t solve the problem small indices, while another team first attempted that problem but accumulated greater penalties for other problems. How do you merge such lists? Now let's say while merging the lists the team that didn't solve small indices(means one problem less) is placed higher than that who solved it (and also solved all the problems that the other team solved), how can you just randomly remove a problem and compare it to another ranklist?

How do you compare apples to oranges and decide which is the better fruit?

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

Refund our money back or do a proper recontest , please don't degrade the quality of Indian Teams and what standards or benchmark are you setting for future of ICPC in India , Is this the way how Competitive Programming will prevail in India , why don't you think as an individual team like they all put their efforts throughout the year for such an unfair contest , DIV4 is really better than this in terms of satisfaction.

Why are you thinking of just managing everything while making a bigger mess . Many of us have affected our endsem exams just to perform well with a hope of our future .

Why are we deserving this . Why you don't want powerful teams from India . What you think of world finals . Why good teams don't deserve world finals .

Why are you so helpless . "**IS THIS THE LEVEL WE ARE AT , INDIA CAN'T EVEN CONDUCT A FAIR ICPC SINGLE ROUND . WHY ???** "

→ Reply