By awoo, history, 3 years ago, translation, In English

Hello Codeforces!

On Feb/22/2022 17:35 (Moscow time) Educational Codeforces Round 123 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

Harbour.Space

Hey, Codeforces!

Once again, it is time for another exciting scholarship opportunity from Harbour.Space!

We have partnered with various tech companies to offer Bachelor’s or Master’s degree scholarships in Computer Science, Data Science, Cyber Security, and Front-end Development and work experience in the partnered companies.

We are looking for various junior to mid-level positions to fill in different fields such as:

  • Java Spring / Node.js Back-End Developer
  • DevOps Engineer
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  • React / React Native Front-End Developer
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Requirements:

  1. High School Diploma for Bachelor degree applicants or Bachelor’s degree for Master degree applicants
  2. Professional fluency in English
  3. Previous experience is a must for Master student applicants and a plus for Bachelor student applicants

Make sure to apply before Mar 13, 2022, to be eligible for the scholarship and reduced application fee.

APPLY NOW →

Keep in touch and follow us on LinkedIn for more scholarship opportunities. And follow us on Instagram to evidence student life, events, and success stories from students.

Good luck on your round, and see you next time!

Harbour.Space University

UPD: Editorial is out

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3 years ago, # |
  Vote: I like it -16 Vote: I do not like it

Hoping for a great contest. Good luck and high ratings for everyone.

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3 years ago, # |
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Its great that cf is arranging two contest in two continuos days ;-;

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    3 years ago, # ^ |
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    Then there is google hashcode on 24. Man, this week is packed with excitement for programmers.

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3 years ago, # |
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Glad to see MVL switching careers to CP and leaving Chess

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3 years ago, # |
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Spoiler
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3 years ago, # |
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May all who deserve, gain ++ ∆

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3 years ago, # |
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I'm looking forward to this contest.

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3 years ago, # |
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Monsters Incoming !

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3 years ago, # |
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looks like the gap between E and F is too big...

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3 years ago, # |
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How to solve E?

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    3 years ago, # ^ |
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    count the cells which must be missed.

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    3 years ago, # ^ |
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    After processing the first $$$i$$$ characters, lets suppose we are at some position $$${x, y}$$$ .

    Assume wlog we have $$$s_{i} = D$$$.

    We clearly can't even reach points with $$$x' \lt x$$$ or $$$y' \lt y$$$, so let's assume we've marked them as unreachable in the previous $$$i - 1$$$ steps.

    If this is the last character, we can clearly reach all points with $$$x' \geq x$$$ and $$$y' \geq y$$$, so we don't need to subtract anything. Otherwise we don't need to worry about points with $$$x' \gt x$$$, since the next position (which will be on row $$$x + 1$$$ since $$$s_{i} = D$$$) is better suited to that task since any repetition from that point will use the same sequence of repetitions, except for one less downward step.

    So that leaves the points to the right on the same row. If the character $$$R$$$, appears $$$k$$$ more times after position $$$i$$$ of the string, then we will go out of the grid trying to access the last $$$k$$$ columns of this row, so they are unreachable. The rest can clearly be reached by repeating the last occurrence of $$$R$$$ before position $$$i$$$.

    So we just need to subtract $$$k$$$ from the answer at this position. Any further position is in a lower row so we won't overcount.

    We can see due to symmetry the same holds for $$$s_i = R$$$ with $$$D$$$ occuring $$$k$$$ more times after position $$$i$$$.

    One last thing to be careful about, is that for the first substring of equal characters, there is no character to move us downward / right to reach those cells perpendicular to our direction of motion, so we always have to subtract $$$n - 1$$$ in that case.

    Solution — 147333480

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3 years ago, # |
Rev. 2   Vote: I like it +37 Vote: I do not like it

I'm curious how many others just kept randomly shuffling the permutation till they got one that was anti-fibonnaci in B lol.

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    3 years ago, # ^ |
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    Did one of them actually passed the system tests ?

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      3 years ago, # ^ |
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      Yeah mine did.

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      3 years ago, # ^ |
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      using shuffle() instead of next_permutation() did pass

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      3 years ago, # ^ |
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      It reasonably should, I intuitively feel that the probability of selecting an anti-fibonnaci is either $$$\frac{1}{c}$$$ for some small constant $$$c$$$ or at worst $$$\frac{1}{n}$$$, both of which will easily pass.

      I still ran the max case ($$$T = 48$$$, $$$n = 50$$$) and it passed in $$$15$$$ ms.

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    3 years ago, # ^ |
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    I just reversed the first half and second half for every partition at index (1,n)

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3 years ago, # |
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C harder than D and E

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    3 years ago, # ^ |
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    Once you know the solution, you will facepalm. I figured it out after spending more then an hour and ha half and it's soo simple I was overcomplicated it. Wasted D tho :(

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      3 years ago, # ^ |
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      Same thing happened with me..C is basically dp with kadane's algo i have wasted all my time doing C

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      3 years ago, # ^ |
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      i'm so frustrated. why did i stop extending left/right when there's a negative number... certified facepalm, didn't make it in time

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        3 years ago, # ^ |
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        I did something similar but in D. Instead of multiplying k I was adding k to the answer... certified facepalm.

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    3 years ago, # ^ |
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    Once you figure out its DP, it becomes easy.

    DP[i][j] — max continous subarray sum upto the ith index in array and j addition of X.

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3 years ago, # |
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speeeeed forces. didn't expect this :(

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3 years ago, # |
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How to solve F?

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3 years ago, # |
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Could someone take a look at why my O(N) solution for D is failing with a TLE? Code here

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    3 years ago, # ^ |
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    Look carefully at the constraints on $$$n$$$ and $$$m$$$

    Spoiler
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    3 years ago, # ^ |
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    n & m can be large per testcase you can use set instead.

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Logic for C?

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    3 years ago, # ^ |
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    First, we will calculate the max sum for each subarray for length i=0,n, We can do this just by 2 simple FOR loops. So now we have an array SubArraySum[], and subArraySum[i] represents the max sum among the ith length subArray. Now to find the best answer when we can add some k elements of value x, we can do this by iterating over the subArraySum array, and for a SubArray of length =i we can increase its value by x*min(k,i).

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3 years ago, # |
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Thanks! Enjoyed the problems, however, don't you guys feel like the difficulty gap between CF div2 rounds and edus has become large? Or is it just me who's good at such problems and it's meant to be this way? Somehow my performance in edus (well, last two) is exponentially better than in normal div2s and i dont remember edus being so friendly even a little while back. i feel one could exploit edus for large rating gains <doublestrike> like i did </doublestrike>.

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    3 years ago, # ^ |
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    I've observed the same thing recently, with EDUs being ridiculously free :thinking:

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    3 years ago, # ^ |
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    I agree, although I feel like this contest in particular was pretty easy even for an EDU.

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3 years ago, # |
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finally a good contest :)

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3 years ago, # |
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Could someone tell me why my solution is giving TLE for question D link-https://codeforces.net/contest/1644/submission/147348233

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    3 years ago, # ^ |
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    you are creating vectors inside the test case loop, n,m can be upto 2e5. doing this vector creation of this much size t(testcases) time, leads to TLE. To solve this you need to create the vectors outside the test case loop and set and unset it

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Can anyone please tell why my code is getting TLE on testcase 6 in ques D.. Acc to me is O(N) solution.As I couldn't find the reason in contest and after it also. My code https://codeforces.net/contest/1644/submission/147349933

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    3 years ago, # ^ |
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    It does not guarantee that the sum of M and N is within the 1e5 range.

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    3 years ago, # ^ |
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    The problem is declaring again and again those 3 vectors, because of n and m being up to 2e5 every time, it takes a lot of time to always fill the vectors. A simple replacement of those with a map and declaring one of them global will do the trick (Of course, with .clear() before each testcase). Here is your solution but modified as said above : https://codeforces.net/contest/1644/submission/147351646

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3 years ago, # |
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got the logic of d, but not able to implement how to check two cell visited previous or not?

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    3 years ago, # ^ |
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    Process queries in reverse order. Then for any query q, check if either you have seen the row and the column of this query before (in which case this query shouldn't be counted), or if all the rows or all the columns have been painted already (again, in which case this query shouldn't be counted).

    Now we know all queries that should be counted (call them valid_queries), the answer is simply $$$k^{validqueries}$$$.

    https://codeforces.net/contest/1644/submission/147343490

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      3 years ago, # ^ |
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      can u please elaborate the logic behind processing queries in reverse order?

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        3 years ago, # ^ |
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        For any element r,c, we need to know whether it was painted over completely later (in which case its not a valid query and we don't consider it). It's easier to find this out if we reverse the queries, since at any point we have state of all painted cells so far, and can check if current query paints any new cells or not.

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      3 years ago, # ^ |
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      Why is my solution 147366362, which is conceptually identical to your solution, TLE'ing? I've tried replacing long long with int to no avail.

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        3 years ago, # ^ |
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        Issue was using find(set.begin(), set.end(), value) rather than set.find(value). Apparently this makes a difference.

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          3 years ago, # ^ |
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          Yeah , first one is O(N) , where as second one is O(logN) , where N is number of elements in set.

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3 years ago, # |
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I guessed the algorithm, but I can't calculate it. :(

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3 years ago, # |
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A — D, were great (didn't read E or F yet).

It's a pity I didn't read D thoroughly and was thinking about a way harder problem (one in which we're not given coordinates of operations — just q)

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    3 years ago, # ^ |
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    wait, what problem were you thinking about exactly lol? cus the answer changes depending on the coordinates xD

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      3 years ago, # ^ |
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      Version in which no coordinates are given.

      Just n, m, q and k in input, nothing more.

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        3 years ago, # ^ |
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        But I couldn't really figure it out.

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        3 years ago, # ^ |
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        oh I think I understand, you're trying to maximize the number of colorings among all possible queries?

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          3 years ago, # ^ |
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          Yes.

          What is the number of possible distinct colorings if you know grid is n x m, there are k colors and you know that q coloring operations were performed (but you don't know anything about those operations).

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            3 years ago, # ^ |
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            oh my bad LOL, I just realized that what you originally said made sense, I sort of forgot the problem tbh xD

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    3 years ago, # ^ |
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    At the beginning I thought that I can change the order of q operations

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3 years ago, # |
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I solved 2 Questions. I see no rating increase in my profile.Reason being???

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3 years ago, # |
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Thoroughly enjoyed A through D. Wish I had more time to read the rest

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3 years ago, # |
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any hints on how to do B ques?

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

    The first permutation is n,n-1,...,1. Let's say this is (1).

    Then just swap every adjacent pair of (1). There are n-1 adjacent pairs so you will get total n pairs.

    Let's understand this with an example of n=4.

    1: 4 3 2 1

    2: 3 4 2 1

    3: 4 2 3 1

    4: 4 3 1 2

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    3 years ago, # ^ |
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    you may think about how to construct an answer for $$$n$$$ when an answer for $$$n-1$$$ is known.

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    3 years ago, # ^ |
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    Sort the permutation in descending order and shift the smallest element (i.e 1) one place at a time from right to left. For example if n=4:

    4->3->2->1, 4->3->1->2, 4->1->3->2, 1->4->3->2

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3 years ago, # |
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Can Someone mention a testcase where my submission for C is failing. Submission: Link to Code

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To users who get Time Limit exceeded at test 6 of problem D (like me, spend almost half an hour to figure it out)

You should not initiate the rows = [-1]*n and cols = [-1]*m, because it is not guaranteed that the sum of m or n do not exceed 2*10^5. You should using dictionary or hashmap instead.

And thanks for testers for setting test case 6, or I will definitely get hacked after the contest.

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If you are/were getting a WA/RE verdict on any of the problems from this contest, you can get a small counter example for your submission on cfstress.com

Problems added: "A, B, C, D, E, F".

If you are not able to find a counter example even after changing the parameters, reply to this thread, mentioning the contest_id, problem_index and submission_id.

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3 years ago, # |
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who solved C without dp?

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    3 years ago, # ^ |
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    I used a segment tree.

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    3 years ago, # ^ |
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    i think i did without dp, though after the contest

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    3 years ago, # ^ |
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    You can just find the max sum for each subsegment for a fixed size for all sizes from 1 to n. and then just try all values of max with given k

    my implementation.

    Code
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    3 years ago, # ^ |
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    Me, well don't exactly know if it would be called as DP or not, but there was a little bit of pre-calculation for calculating the maximum possible sums of each of subarrays and then finally iterating over them to get optimal answer for each k from 0 to n.

    vector mxsum(n+1,INT_MIN);

    mxsum[0]=0;
    for(int i=0;i<n;i++)
    {
        int sum=0;
        for(int j=i;j<n;j++)
        {
            sum+=a[j];
            mxsum[j-i+1]=max(mxsum[j-i+1],sum);
        }
    }
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3 years ago, # |
  Vote: I like it +76 Vote: I do not like it

I can't understand why in Problem F the constraint is $$$1 \le n,k \le 2*10^5$$$.

Is the key point of the problem how to calculate Stirling numbers using NTT?

Absolutely this problem should focus on how to find the solution, not how to use polynomial.

This wastes me too much time so I can't solve it in contest.

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    3 years ago, # ^ |
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    I was not able to solve it in time either(cause stupid me thought it was impossible to calculate the sum of Stirling numbers fast enough). But since I hate NTT and I like this problem I have upsolved it without NTT in O(n * sqrt(n))147355537
    After some optimizations(147356152) it is now running in 1,5 seconds, wich is 1/4 of TL, so maybe this solution is actually intendent

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      3 years ago, # ^ |
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      Will you please explain it ?

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        3 years ago, # ^ |
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        1) I've just realized I am damn and solved this problem in O(nlog)(147475186)
        2) My solution is almost same as in the editorial, but I compute ans = sum S(i,1)+S(i,2)..+S(i,k) in O(min(i,k)) with no fft or other techniques. Since S(i,j) = sum(1<=t<=j) C(j,t) * t^i * (-1)^(j+t) / j! we have:
        S(i,1)+S(i,2)+..+S(i,k) = sum(1<=j<=k,1<=t<=j)( t^i * (-1)^(j-t)/((j-t)! * t!) ). Let d = j-t, then
        ans = sum(1<=t<=k,0<=d<=k-t)( (t^i/t!)*((-1)^d / d!) ) iterate over t and sum(0<=d<=k-t) ((-1)^d / d!) can be precomputed

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3 years ago, # |
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i cant even understand the question D, i solved the answer for no of uniques after all operations, my brain found this sub problem easier and got attached to it, soo much more to learn

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    3 years ago, # ^ |
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    Spoiler
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      3 years ago, # ^ |
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      thanks but i knew that i just was confused about the question itself

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Implementation forces

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3 years ago, # |
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anyone else who did random shuffle in B :)

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3 years ago, # |
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Can Someone give small hints for Problem E.

Make sure to not spoil it for other people.

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3 years ago, # |
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Could someone take a look at why my O(N) solution for D is failing with a TLE? Neon https://codeforces.net/contest/1644/submission/147331915

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    3 years ago, # ^ |
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    Your solution is exactly $$$O(nT)$$$, and there isn't any constrant on $$$\sum n$$$.

    The hacker generated a testdata with $$$t = 10000, n = m = k = 2\times 10^5, q = 20$$$ to maximize $$$\sum n$$$.

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      3 years ago, # ^ |
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      GOT IT

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      I used boolean vector of size n and m instead of integer vector and the code is not getting TLE on the pretests, so I would like to know exactly how fast is the initialization of boolean vector as compared to integer vector? (Submission link: https://codeforces.net/contest/1644/submission/147329366)

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        3 years ago, # ^ |
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        Maybe there is some mysterious optimization in the implementation of std::vector. In fact I don't know, either.

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        3 years ago, # ^ |
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        std::vector is something like bitmap, which is at least 8 times faster than normal version (I guess).

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How to solve D ???

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    3 years ago, # ^ |
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    process from back, keep track of used rows and columns.

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    3 years ago, # ^ |
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    Main idea is that some rows and columns get completely overlapped in operations succeding it, making that operation worthless. So just count those out, then, k ^ cnt

    For video explanation you can check out my video

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3 years ago, # |
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ConstraintForces

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3 years ago, # |
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I don't think it's reasonable that there weren't constraints on $$$\sum n$$$ and $$$\sum m$$$. Some people (including me) used O(nT) initialization in each testcase, and some of them got hacked, some didn't.

My submission ran about 1300ms so it can't be hacked due to its relatively small constant, but some submissions that ran 1900+ms got hacked because of the fluctuation of the judge.

I don't think it is constant that matters in Codeforces contest. Maybe a TL of 1000ms is more preferable for C++.

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3 years ago, # |
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Why is my O(q) approach giving tle for d?

https://codeforces.net/contest/1644/submission/147345716

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    3 years ago, # ^ |
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    Because you are Initializing row and col vectors for each testcase, which takes O(nT) and since there weren't any constraints on ∑n and ∑m, it should TLE.

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    3 years ago, # ^ |
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    There is no constraint on sum of n and m in problem

    so you can have $$$t * (n + m) = 10^4 * 4 * 10^5 = 4 * 10^9$$$ operations in your code

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3 years ago, # |
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Did anyone else do a 2-D Dp for C?

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3 years ago, # |
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solving E 3 minutes after contest finished is really big pain

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    3 years ago, # ^ |
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    it's also pain to have >40min for F but can't even get any clue

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

how would u rate b,c problems ?

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

I think it was a good round.

But it has only ~100 likes.

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

Didnt like C(too straightforward, no interesting idea), B and D was good though.

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

Any hint for E

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

Thanks for the round!I liked the tasks and they were interesting!

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

Was stuck in C for quite some time. Dropping by few hints for whosoever interested

Hint 1
Hint 2
Hint 3
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3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone tell me why my solution 147364129 for problem D is getting WA?

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

 Can someone explain why do I see this contest as unrated?

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    3 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    The rating hasn't yet been recalculated

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3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

I am waiting to see what happens if the ratings for this round are not updated before the Div1/Div2 round starts xD

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    3 years ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    it would be more funny if someone registered for div2 with rating<2100 will get delta above it for previous contest after joining next contest :)

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    3 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    Codeforces right now:

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

    Just 6 to 7 mins before the next round, the ratings got updated.

    That was close. Looks like we'll never know.

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

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

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

Is there any specific reason why ratings of Educational rounds are delayed to update?

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

147328156

Can someone help me figure out where my solution is wrong ? Please , Thank you !! (Problem C)

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    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    My god, i just took inf as -1e8 & taking it as < -1e14 , would do the trick , just because the sum can go below -1e8, feelsbadman :(

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

Huge gap between E and F. XD

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

hi, all my solutions got skipped on this round because code c is similar for other's code there must be a coincidence i didn't cheat or use other account to submit my solutions it's basic dp on best sum for len of k how can i speak to someone can help

what should i do ?

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

Why the difficulty rating of the problems havent been updated yet ?

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

The apply now link isn't working