Блог пользователя MrPaul_TUser

Автор MrPaul_TUser, история, 3 года назад, По-русски

Идеи задач: MikeMirzayanov.

1551A - Поликарп и монеты

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1551B1 - Чудесная раскраска - 1

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1551B2 - Чудесная раскраска - 2

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1551C - Интересный рассказ

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1551D1 - Домино (упрощённая версия)

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1551D2 - Домино (усложнённая версия)

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1551E - Неподвижные точки

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Решение

1551F - Равноудалённые вершины

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Решение
Разбор задач Codeforces Round 734 (Div. 3)
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3 года назад, # |
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good contest for "try of pen",i like it,thanks.waiting soon for your div 2 and div 1 contests.

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3 года назад, # |
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Even though the DIV3 was harder than the last few DIV3's, I think the problem set was way more enjoying and educational for a lot of people. I hope you set more DIV3's MrPaul_TUser

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    3 года назад, # ^ |
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    But please, make it easier, as a person with <1600 rating,I can say, that at least 4-th problem was very and very hard for actual div3 participants.MrPaul_TUser Thank you anyway. ;)

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      3 года назад, # ^ |
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      actually,i dont think that this contest was harder than standart div 3

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3 года назад, # |
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Transposing the table in D2 was a nice idea! Saves a lot of casework.

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3 года назад, # |
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A slightly shorter branchless alternative implementation of A, which does the same job (adding 1 to n before division by 3 is the same as incrementing the result by 1 in the case if the remainder was 2):

Python code
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3 года назад, # |
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Please help me in this code (problem B2).
I'm not able to figure out where I'm doing wrong.
This code is failing on tescase-185 of test-2.
My submission

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3 года назад, # |
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Please help me in this code (problem B2).
I'm not able to figure out where I'm doing wrong.
This code is failing on tescase-185 of test-2.
My submission

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    3 года назад, # ^ |
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    when ur are considering the nos having frequency less than k the order also matters suppose there are 4 colors. and 1,2,3 are having count less than 4. if the arrangement is like : 1 2 2 3 1 nos in array. 1 2 3 4 1 coloring . the no 1 is getting the same color.

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3 года назад, # |
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Really enjoyed A and B! Thanks for a great contest.

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3 года назад, # |
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https://codeforces.com/contest/1551/submission/123610411 This is my submission for problem F. According to me the complexity of my solution is of degree 4 so how did this not give TLE?

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    3 года назад, # ^ |
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    you are lucky

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      3 года назад, # ^ |
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      Yeah but still the runtime is just 62ms

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        3 года назад, # ^ |
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        clear $$$n^3k$$$ time for $$$n,k<=100$$$

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        3 года назад, # ^ |
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        The O(n^3k) part of your code is setting the dp table to -1, which has a small constant and runs very quickly (10^8 operations). Your recursive solve function for every root and distance runs in O(n^2k) time since every edge can only be connect to 2 vertices. Iteration to find combination over each edge happens at most twice each edge.

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3 года назад, # |
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What if problem C was changed a bit and one letter should have more occurence than other letters taken individually, and not in total, how should we proceed then?

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    3 года назад, # ^ |
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    I misread the problem as you described during contest and was banging my head to get a solution :(.

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    3 года назад, # ^ |
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    Not sure whether this is correct, but here is what came to my mind.

    We can calculate the maximum number of words we can take considering every character as the 'dominant' character.

    Let us say we want to have 'a' as the dominant character. We iterate over all the words one by one, taking all of them in our collection (for now) and keeping track of the total frequency of every character till now. We also maintain separate sets for F(s, 'a', c2), where F(s, 'a', c2) represents this — In word s, frequency('a') — frequency(c2).

    So now, suppose we encounter a word due to which 'a' is no longer our dominant character, but instead 'b' is. We refer to our set for F(s, 'b', 'a') and remove the the word s which has the largest F(s, 'b', 'a') from our collection (and also from other sets).

    The idea is, take all words as you encounter them. But if a certain words causes 'a' to lose its dominancy to some other character (say 'b'), we will remove such a word which will reduce the total frequency of 'b' as much as possible while not hurting 'a' much, thus restoring dominancy of 'a'.

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3 года назад, # |
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The contest was great! It would be better if the editorial for C, would be more explained!

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3 года назад, # |
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It is possible to solve E with a 1D dp. Details in spoiler.

Spoiler

Submission: 123519134

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    3 года назад, # ^ |
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    $$$ (i - a_i) - (j - a_j) \le i - j - 1$$$

    seems to be just

    $$$ a_i > a_j $$$

    right?

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      3 года назад, # ^ |
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      Yeah...you can simplify the condition like that. I just thought it made more sense to present it the way I did.

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3 года назад, # |
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Tutorial for D2 seems complecated.

We can reduce to two cases: Number of rows is odd or is even.

In case odd we need to place one row of horz dominoes, max k dominoes. In case even we do not.

Then we place allways two rows of dominoes at once, until k dominos are placed. Then fill all other fields with vert dominoes.

Finding a useable letter foreach dominoe is not hard, simply check the surrounding cells. 123518743

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3 года назад, # |
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If anybody is interested and have enough time to help me, please do. I understand their are better approaches for B2 then this, but I am not able to understand why did this approach failed.

Approach :

  1. Find how the max number which we can allot colors by desired conditions , say this is x
  2. loop through k colors : 1.assign kth color to a number 2.Store that number in unordered_map, so that same color doesn't get assigned to same number 3.Maitian a count=x, which is basically the number of time to allot and decrease it everytime the color is alloted.

Here is the link for the submission — fails on tc 554 123566013

Also show no mercy if the code is bad, so that I get to learn how to code better than what I have done.

Peace out!

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    3 года назад, # ^ |
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    use this tc 1 6 3 1 2 5 3 3 3 your solution will output 1 1 2 2 3 0 but one of the correct ans will be in the form 1 2 3 1 2 3 I think your solution will also give tle for larger value of n. if you till have any doubt feel free to ask me.

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3 года назад, # |
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Is it possible to solve Equidistant Vertices in faster than O(N^2 K) (maybe with some optimization in the dp)?

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    3 года назад, # ^ |
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    Here's what I think. If you fixed the root, and then you fixed the depth of the nodes from the root, you want to find the number of ways to pick $$$k$$$ nodes from them so that their pairwise LCA is the root. Naturally, this splits up our set into some $$$m$$$ groups, from which we can fix $$$k$$$ groups and then the number of ways would be the product of the sizes of those groups. Let's say the group sizes are $$$g_1, g_2,...,g_m$$$. If we consider the polynomial $$$(x+g_1)(x+g_2)...(x+g_m)$$$, then the value we want is the $$$k^{th}$$$ coefficient starting from the largest term ($$$0$$$-indexed). Now, this product can probably be computed in $$$\mathcal{O}(n {\log}^2 n)$$$ with divide-and-conquer+FFT/NTT for an overall complexity of $$$\mathcal{O}(n^2{\log}^2 n)$$$. I haven't implemented it myself, so please point out the error if you think I'm wrong.

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      3 года назад, # ^ |
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      Oh that's a nice idea. How is Kth coefficient computed in O(N log^2(N)) though?

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        3 года назад, # ^ |
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        Well, I think the following: If we split our product into two halves and compute each half, we'll get two polynomials of roughly equal degree, which we can then multiply with FFT/NTT. The recurrence relation would look like $$$T(n)=2T(\frac{n}{2})+ \mathcal{O}(n\log n)$$$, and we can use the master theorem for divide-and-conquer to get that this works out to $$$\mathcal{O}(n{\log}^2 n)$$$. We can simply extract the required coefficient from the final answer presented in the form of a list of coefficients.

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3 года назад, # |
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I am getting TLE for test case two for prob B2. Can someone tell me is my approach wrong as I am getting TLE even though tc is NlogN. My solution https://codeforces.com/contest/1551/submission/123636719

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    3 года назад, # ^ |
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    When the number of test cases is equal to 10000 your vectors hsh[200005] and prev[200005] initializations take too much time. Try the size of vectors N like in your v[] and ans[] vectors.
    But your greedy idea to take the first vacant color is incorrect. Try test:
    1
    9 3
    1 2 3 4 5 6 6 7 7
    Your answer is:
    1 1 1 2 2 2 3 3 0
    The correct answer may be:
    1 2 3 1 2 3 1 2 3

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3 года назад, # |
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Problem B2:

let's calculate for each of k colors the number of elements painted in the color — all calculated numbers must be equal

I am not understanding how are we managing this condition in given tutorial. Please anybody explain

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    3 года назад, # ^ |
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    Take k occurrences of the numbers whose frequency is greater than or equal to k. For all other numbers, take the total sum of frequencies of these numbers (let say the sum is s), and color a*k of these occurrences s.t. (a+1)*k>s .

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3 года назад, # |
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Another approach for B2 would be doing binary search on how many elements can be painted in one color.

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2 года назад, # |
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147410960 Hii! can someone check why is it failing on test case 1, even when everything is correct according to me. Thanks!

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    2 года назад, # ^ |
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    If you look at the output, you have a situation where two vertical dominos, sharing a side have same color. This is not allowed, as mentioned in problem as well.

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20 месяцев назад, # |
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Could anyone check why this submission of mine get WA for this test case (wrong answer invalid table description) :

1
45 90 967

When I change the line $$$38$$$ from char ans[n][m] to vector<string> ans(n, string(m, 'x')) as in this submission I get AC. So it probably has something to do with iterate through an ans[i][j] that hasn't been initialized, but I manually checked the result and didn't see anything wrong.

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19 месяцев назад, # |
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Hey guys, did you realise that in problem B2 the example was referring to the irrational number (pi) in the last three test cases? Nice one :D