When I enter the link, it says "No such contests"

have some sense of humor, ts pmo 🥀 🥀

Improve your concentration, it might help

As a participant, I will enjoy a wondANDful time during the contest!

Easy to make it overflow:

1

20 15

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5

The code will return 20, though the answer is 19. Also, put your code in spoiler when you are presenting it, it is inconvenient to make a long scroll to view other messages.

Hoping that this contest contains only very easy and doable problems)

Added, thx!

As a CM, I was cooked!

Thanks, included!

Included, thanks!

Depends. There are some tasks that require some data structures like Trie, or Segment Tree. But in some times those might even appear in 1400-1500 rated problems. There might be greedy based problems in 1600-1800 rating threshold, constructive, interactive, anything, though there will not be complex topics such as FFT or flows

Jokes aside, I think you should aim not for slightly above 1400, but for 1600-1700. If you will be able to solve this rated questions, you will be able to easily solve 1400. Sometimes it will require you a lot of time, but it is worth it, because it is a new experience. You can try to integrate easier questions as a warm up. Or, when approaching question, trying to think about easier version of the same question. Maybe the solution comes from it. Additionally, as the other guy mentioned, make observations and try to combine them.

Hoping this will help you!

Why do you need to be string at problem solving? Be an array at solving questions above 1400 rating. It will pay off istg

There is a clash, ABC ends at 13.40 UTC

Thanks, added!

Clashes with Atcoder Beginner Contest

The problems are extremely hard((

Added, thanks!

Hoping that problems will not be hard

Could you please mention the participants' handles for me to add them to the table?

Bro congrats. Problems were very easy, but not for me(

Problems are extremely hard ;(

You forgot USA my friend

Yea thats a true point

Thanks, added!

E-European

G-Girls

O-Olympiad

I-Informatics

Hoping that problems will be very easy so that I can reach Master)

They are hard(

Thanks, added!

Thanks, added!

Added, thanks. Good luck to yall!

Thanks, added! Gl to your team!

Hoping that problems will be very easy

The reduction definitely isn't easy, as you need to make a couple of observations to bring this task to the graph one. This greatly fits for the position of E

Hoping that the problems will be easy.)

I don't think that there's a way to make compressions online. But there is a solution which will make all the queries online while there's no need for compression: using Implicit Segment Tree. It can support operations on segments with r <= 1e9.

You may consider using coordinate compression for this problem. You should compress the values of a[i], b[i], p[i], x[i]. Now, the values are up to 8 * 10^5 instead of 1e9. You can do a normal BIT there instead of that one which is implemented with map

Maybe it is because D was firstly proposed as C, and C1 was proposed as D1. That is why some people might consider D easier than C

Added, thanks!

.

I am sorry, is UK Ukraine or United Kingdom?

Added, thanks!

Added, thanks!

Thanks, added! glhf

Added, thanks! Wish you good luck!

added, thanks!

Thanks, added!

Added, thanks!

Azerbaijan team:

Aykhan Damirli (dmraykhan) — 2nd time at IOI, 0 attempts left

Hasan Valiyev (Hasanv) — 1st time at IOI, 1 attempt left

Ali Aliyev (Nxxlt) — 1st time at IOI, 0 attempts left

Elvin Imanli (Captain_Georgia) — 1st time at IOI, 0 attempts left

problems are hard ;(

A person who thinks all the time

Hoping that problems will be easy

They are from Iran...

How do you estimate the difficulty of a problem?

Can someone tell the proof of problem A?

Mirzoyonov

From which country will the T-shirts be sended? BledDest

Give this comment negative contribution too

bro sold

The number of distinct f(x)'s for x <= 10^9 — 2 is small enough to precompute them.

Idk why this blog is downvoted. If you disliked this blog, please, leave a review.

Not yet, BledDest, when the results are going to be announced? 2 weeks passed already

I mean that the participant's solution may be wrong, but it generates the same output as author's solution.

Yes, that is the one of explanations. But I wonder, can't wrong solutions pass with these manually created tests? Although there is only answer for the task, the algorithm for reaching it may be flawed. I wonder did authors consider this?

I wonder how did you verify your tests for problem D? Is the condition for validity just checking that the number of groups of (x + y) with odd number of elements is 1, and the number of groups of x with odd number of elements is 1? I think that there should be other conditions satisfied.

Proof_by_QED

Your aggression to the blog author indicates that your solutions are all AI-generated. Ngl ts pmo

ngl idgaf ts pmo

1) This may only prevent the low percentage of cheaters in the best case, as one can just take a screenshot of the problem statement and submit to LLM. Alternatively, one can use image to text converters to grab the text of the statement. It could be better to add screenshot restrictions, but it is really difficult to implement this perfectly as this is bypassed easily.

2) Too problem-specific.

3) There's a blog section where you can report cheaters. Not guaranteed that it will grab attention of organizers, but cheater report system doesn't guarantee it either, as there may be fallacious reports.

Jokes aside, it is because of the api problems and this is not related to carrot. Other rating predictors face the same problem

If it's broke, just give it money, and it will work anew

F is insane

There's an observation that if you have $$$a_i \lt = a_j$$$ and $$$b_i \lt = b_j$$$, then it is optimal to put $$$i$$$ and $$$j$$$ in one group. Therefore, $$$i$$$-th cat will not influence the result. Therefore, we can remove cats, values of which are "nested" into other cat. As a result, if you sort $$$a$$$ in increasing order, $$$b$$$ will be sorted in decreasing order. Now, you can do CHT.

When the winners of prizes will be announced?

Firstly, you need to do some transformations of the statement.

Define $$$mex(a)$$$ as the number of $$$k$$$'s such that every number in the range $$$[0, k]$$$ is present in $$$a$$$. This number actually equals to $$$mex$$$, because by the definition, all numbers from $$$0$$$ to $$$mex - 1$$$ are present and $$$mex$$$ is not present.

Also, let's count the contribution of each subarray independently. Now, you can sum up the contributions of each $$$k$$$ separately. For this purpose, you can brute force the value of $$$k$$$. It is possible to make all integers from $$$0$$$ to $$$k$$$ appear in the subarray if the subarray contains all the fixed positions where $$$a_i \le k$$$, because if this doesn't satisfy, i.e. there's a value $$$\le k$$$ in the whole array, but not in the subarray, then you should repeat that value $$$2$$$ times to make this value appear in the subarray.

Now, let's keep track of the minimum/maximum position of a fixed element that is $$$\le k$$$, let's denote them as $$$[lx, rx]$$$ correspondingly. Then, for the subarray $$$[l, r]$$$ to have a positive contribution, it should contain the segment $$$[lx, rx]$$$.

To calculate the contribution of segment $$$[l, r]$$$, let's denote $$$cnt$$$ as the number of elements that are $$$-1$$$ in the current segment, and let's denote $$$miss$$$ as the number of elements in the array that are $$$-1$$$, and let's denote $$$x$$$ as the number of elements that are $$$\le k$$$ that aren't present in the array. The contribution becomes $$$P(cnt, x) * (miss - x)!$$$ . This is because you must insert those $$$x$$$ missing elements $$$\le k$$$ in the subarray to make all of them appear, and you can insert them in any order and in any position that is $$$-1$$$ in the subarray, so the number of such arrangements is $$$C(cnt, x) * x!$$$, or $$$P(cnt, x)$$$. And all other $$$miss - x$$$ elements can be inserted arbitarily into $$$-1$$$ positions, so the number of those arrangements is $$$(miss - x)!$$$ . Together, the number of ways equals to $$$P(cnt, x) * (miss - x)!$$$ .

To calculate this value fast, you need to fix the $$$cnt$$$, because $$$miss$$$ and $$$x$$$ are the same for all the subarrays. For each $$$cnt$$$, you should save the number of segments that contain $$$[lx, rx]$$$ and have the number of missing elements equal to $$$cnt$$$. Now you can notice that the number of $$$[lx, rx]$$$s for which the answer should be calculated equal to $$$n$$$, $$$k$$$ can take the maximum value of $$$n - 1$$$, so you can fix $$$cnt$$$ for each of those segments independently.

Let's iterate $$$rx$$$-s in decreasing order and add segments which have $$$r \ge rx$$$ and take a fenwick tree for each $$$cnt$$$. For each of those segments $$$[l, r]$$$, you need to add $$$1$$$ to the $$$l$$$-th position of the $$$cnt$$$'th fenwick tree. Now, the number of segments that contain $$$[lx, rx]$$$ and have the number of missing elements equal to $$$cnt$$$ is given by the $$$sum(1, lx)$$$ in the $$$cnt$$$-th fenwick tree. Basically, among segments with $$$r \ge rx$$$, you calculate the number of those with $$$l \le lx$$$. For more implementation details, you can look at my code.

D messed up the whole contest, but E was saver

B's complexity is $$$O(n + \log mn)$$$ I guess, since $$$\gcd$$$ reduces $$$\log$$$ times, or it does not change. After $$$\gcd$$$ becomes $$$mn$$$, $$$\gcd$$$ function is calculated in $$$O(1)$$$.

We can just leave codeforces and migrate to other platforms ig

Div2D editorial:

Case 2: s=0 or 2∤s It's not difficult to prove that the answer in this case is l. (Why?)

Come on, if it isn't difficult, why do we need editorials? Can someone explain the solution properly?

Had the similar solution, which makes [-1, 0, 1] and [-1, 1] operations on difference array instead. Then, the complexity of making [-1, 0, 1] operation becomes O(1) with prefix sums, and [-1, 1] operation has O(n) because of repeating [-1, 0, 1] n times. The idea is to make an array full of 0's by having [+x, 0, -x] and [+x, -x] (which is more time-consuming, so we need to to it less). The sum of array is always 0, because it is a cyclic difference array. If you need more details, you are free to ask. 308089138

virtaul

G1<F maybe, but E is definitely easier

:crying_emoji:

Let's wait for a while, there will be 100+ contribution on your comment ;)

Yes it is. More rating, more respect

You are right, but you wrote about problem setting, so I commented about that.

Anyways, he is right. Problem setting is a hard task

C is harder than usual, anyways nice contest.

Agree

because it's wrong

Thanks for the organizers, hated this contest

C is crazy

Nxxlt

I solved C with going till $$$max(min(n,d),k)$$$ and it passed. You can try to hack, submission id: 238742923

$$$suf[i]$$$ equals to first such $$$(j \gt = i)$$$ that satisfies the following expression : $$$a[j] \gt = mid - (j - i)$$$. To build this array we need slightly modify our expression, $$$a[j] + j - mid \gt = i$$$. So, the possible $$$i$$$ for each $$$j$$$ lies in segment $$$[1, a[j] + j - mid]$$$. Thus, it will be correct that $$$suf[i]$$$ is applicable for all indices before $$$i$$$ also. It means that $$$suf[i]$$$ will be either: $$$suf[i + 1]$$$, if $$$a[j] + j - mid \gt i$$$, or minimal $$$(j \gt = i)$$$ for which satisfies equation $$$a[j] + j - mid = i$$$ (it can be easily handled with pre-building suffix array). If I was unclear, please tell.

It's possible to speed up this solution to $$$O(N\log N)$$$ by using suffix minimum. My submission: 217428923

Use fast input. Passed in 202 ms. Code: 215151784

The second proof is that if the added value is greater than $$$avg$$$ then $$$avg$$$, increases, otherwise it decreases. So if the average will be a maximum element, then the addition of new element will be $$$ \lt = avg$$$, because all elements are less than or equal to $$$max$$$.

Suppose that there is answer that is larger than maximum element, then there is subsequence $$$\frac{a_1 + a_2 + .. + a_k}{k} \gt max$$$, so $$$a_1 + a_2 + .. + a_k \gt max * k$$$, and it's clearly impossible because upper bound of sum $$$a_1 + a_2 + .. + a_k$$$ is that each element is maximum and it's exactly $$$max * k$$$, so $$$a_1 + a_2 + .. + a_k \lt = max * k$$$, contradaction.

Oh, ok. Thanks for help