yes.. only Kotlin will be available for the contest... watch this contest you may understand about the contest rules and language Kotlin Heroes: Practice 7

The practice problems are from past Codeforces contests, you can search them online to find the solutions.

Kotlin was taking a while to compile when I tried it in VSCode. I switched to IntelliJ, and that sped up the compiling significantly.

Most of the owners of these unrated accounts might be android developers, who never heard of codeforces before. Like me, they just might have come across Roman Elizarov's tweet or one of the Jetbrains' announcements about this contest and decided to register to participate.

It'll never get fixed sadly. People comment on every Kotlin Heroes post to do this but MikeMirzayanov and BledDest seem to enjoy ghosting them. As long as they get their sponsorship money, they don't really care.

As far as I understand, JetBrains want to use these contests to try promoting both Kotlin in the world of competitive programming and vice versa (competitive programming among people who just use Kotlin to write code and are not familiar with ICPC or anything like that), that's why they expect people who didn't know about CF to participate and offer them prizes. Unfortunately, I don't know how to limit the usage of alternative accounts without limiting the ability to win a T-shirt if someone didn't participate in a CF round earlier.

A better idea might be a requirement of 1 previous non-T-shirt-based round (obviously they should have solved at least 1 problem). The round itself can be unrated, but it might make people who make multis at the last moment realize they can't participate. Then again, people who have long-standing multis really can't be stopped.

Some multis I found, and why I feel like they are multis:

submitpurposework submitjavak submitkotlin

You found this too, just here are all the associated multis I could find. Each was created at the same time, and each left at the same time.

If anyone cares, this issue has disappeared (or somebody fixed it) and now it's possible to submit solutions for Kotlin Heroes: Practice 7. The Problem A page now has "choose file" and "submit" buttons as one would expect.

Was this a one time glitch? Or does Kotlin judge always become dysfunctional roughly half an hour before Kotlin contests on the codeforces platform?

Because there's no need to promote C++. Without this, most people wouldn't even have heard of Kotlin, but almost every programmer has heard of C++.

Every contest is C++ Heroes

On E, I got 2 WA and wasted 30 minutes because apparently using for (x in a) { print(x); print(" ") } is 50x slower than println(a.joinToString(" ")) (roughly tested on my computer)

I think your guess is correct. I also got TLE with stack and got AC by replacing it with IntArray.

After also getting TLE in H using a Stack, I replaced it with an ArrayDeque (which provides the same methods and some more) and got AC in ~1.2 seconds. Seems like ArrayDeque should just always be preferred over Stack.

$$$O(2^knm\log{n})$$$ passed in 1.5 seconds

120901124

Yes, Stack was the culprit. I implemented Stack using array in the last 2 min and managed to get AC in 780ms. Almost cost me a t-shirt today.

Have a look at my AC submission if you wish to.

Yeah basically don't ever use java.util.Stack. An ArrayList / MutableList would do the job, using .removeLast() as the pop operation

I use following code to find nearest less element on right in array $$$a$$$ :

vector<int> next_l(n);
for (int i = n - 1; i >= 0; --i) {
    next_l[i] = i + 1;
    while(next_l[i] < n && a[i] <= a[next_l[i]])
        next_l[i] = next_l[next_l[i]];
}

It's short, fast and without stack.

One observation that you can make is that greedily extending the string as far as possible is a correct solution. Because of this, the sum of answers is at most $$$\mathcal{O}(n\log{}n)$$$, so you just need to be able to calculate the answer for some $$$k$$$ in $$$\mathcal{O}(ans\log{}n)$$$ time. You can do this by constantly binary searching for the longest length that the next string could have. You can see my submission here: 120888512

Notice that each part will have length at least $$$k$$$, so the answer for fixed $$$k$$$ is no more than $$$\frac{n}{k}$$$, so the sum of all answers if $$$O(n \log n)$$$, which means that you can explicitly build all partitions, you just need to find longest allowed segment starting at $$$i$$$ in $$$O(1)$$$ time. Notice that you can find first position $$$j_0$$$ such that there are $$$k+1$$$ zeros at $$$[i; j_0]$$$ and first $$$j_1$$$ such that there are $$$k+1$$$ ones at $$$[i; j_1]$$$. Then the longest segment is $$$[i; \max(j_0, j_1) - 1]$$$. Let's find $$$j_0$$$, for example. If you calculate number of zeros on each prefix, you want first prefix with $$$sum_0[j_0] = sum_0[i - 1] + k + 1$$$, and you can just precompute such $$$j_0$$$ for each sum on the right from $$$1$$$ to $$$n$$$.

It was exactly the same for me. I surely heard about Kotlin before and wanted to try it some day, but never was motivated enough to actually do it. This contest gave me the so much needed push. And Kotlin syntax turned out to be very easy to learn. It's time to try some Android apps development now :-)

I'm also a bit surprised that so few people participated. But who cares, it's their loss.

There is an important detail in the problem statement. "There is exactly one character (and exactly one character) in this sequence" Your counterexample does not satisfy the condition.

:catThink:

Bump

It would be weird if they announce winners for this Kotlin Heroes round before they announce the winners for the previous Kotlin Heroes round :)

Everyone is so excited about random t-shirt prices knowing the fact that winning one of those has a probability of about .0000000000001% lol.

:catThink: