### MrPaul_TUser's blog

By MrPaul_TUser, history, 3 years ago, translation,

Ideas: MikeMirzayanov.

1551A - Поликарп и монеты

Tutorial
Solution

1551B1 - Чудесная раскраска - 1

Tutorial
Solution

1551B2 - Чудесная раскраска - 2

Tutorial
Solution

1551C - Интересный рассказ

Tutorial
Solution

1551D1 - Домино (упрощённая версия)

Tutorial
Solution

1551D2 - Домино (усложнённая версия)

Tutorial
Solution

1551E - Неподвижные точки

Tutorial
Solution

1551F - Равноудалённые вершины

Tutorial
Solution
• +73

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 » 3 years ago, # |   +11 good contest for "try of pen",i like it,thanks.waiting soon for your div 2 and div 1 contests.
•  » » 8 months ago, # ^ |   0 Nice Pen. Is it magic pen ? what ever you draw becomes true ?
 » 3 years ago, # |   +66 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
•  » » 3 years ago, # ^ |   +3 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. ;)
•  » » » 3 years ago, # ^ |   +3 actually,i dont think that this contest was harder than standart div 3
 » 3 years ago, # |   0 Transposing the table in D2 was a nice idea! Saves a lot of casework.
 » 3 years ago, # |   +11 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 codefor i in range(0, int(input())): n = int(input()) c2 = (n + 1) // 3 c1 = n - c2 * 2 print(c1, c2) 
 » 3 years ago, # |   0 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
•  » » 3 years ago, # ^ |   +3 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.
•  » » » 3 years ago, # ^ |   0 got it. Thanks!
•  » » » 3 weeks ago, # ^ |   0 why this solution for b2 isgiving tle pls help ~~~~~ void _paint_the_town_red() { ll n, k; cin >> n >> k; vl v(n); fa(v) cin >> val; vl temp2 = v; vl cnt(1e6); for (int i = 0; i < n; i++) cnt[v[i]]++; vector temp(k); ll t = 0; sort(all(v)); ll i = 0; while (i < n) { ll j = min(k, cnt[v[i]]); while (j--)//this is main problem { temp[t].pb(v[i]); t = (t + 1) % k; } i += cnt[v[i]]; } ll maxi = 1e18; for (int i = 0; i < temp.size(); i++) maxi = min(maxi, (ll)temp[i].size());//o(n) for (int i = 0; i < sz(temp); i++) { while (sz(temp[i]) > maxi) temp[i].pp; } map mp; for (int i = 0; i < sz(temp); i++) { for (int j = 0; j < sz(temp[i]); j++) { mp[temp[i][j]].pb(i + 1); } } for (int i = 0; i < n; i++) { if (mp[temp2[i]].size() > 0) { cout << mp[temp2[i]].back() << " "; mp[temp2[i]].pp; } else cout << 0 << " "; } cout << _endl;} ~~~~~
 » 3 years ago, # |   +1 Really enjoyed A and B! Thanks for a great contest.
 » 3 years ago, # |   +8 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?
•  » » 3 years ago, # ^ |   0 you are lucky
•  » » » 3 years ago, # ^ |   +3 Yeah but still the runtime is just 62ms
•  » » » » 3 years ago, # ^ |   +1 clear $n^3k$ time for $n,k<=100$
•  » » » » 3 years ago, # ^ |   +1 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.
 » 3 years ago, # |   +5 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?
•  » » 3 years ago, # ^ |   0 I misread the problem as you described during contest and was banging my head to get a solution :(.
•  » » » 3 years ago, # ^ |   0 Same story :(
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 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'.
 » 3 years ago, # |   +3 The contest was great! It would be better if the editorial for C, would be more explained!
 » 3 years ago, # |   +4 It is possible to solve E with a 1D dp. Details in spoiler. Spoiler(Everything below is considered to be 1-indexed)Note that in your solution, if $a_i$ is the last fixed point you have, the number of elements you must have deleted is $i-a_i$.Let us say ${dp}_i$ is the maximum number of fixed points you can have if you keep the $i^{th}$ element. Initially, set all ${dp}_i=-\infty$. Now we need to update it by traversing ${dp}_{1...i-1}$. We need to check some conditions.If $a_1==1$, ${dp}_1:=1$.If $a_i>i$, we can skip computing it.Now, ${dp}_i:=1$ at least for sure. Suppose we are considering ${dp}_j$ right now, so $a_j$ is going to be our previous fixed point. Obviously, ${dp}_j>0$. Next, the number of elements you are required to delete for $a_j$ to be a fixed point mustn't be more than the number required for $a_i$, so $j-a_j \leq i-a_i$. And finally, there must be enough elements in between to make $a_i$ a fixed point. So, $(i-a_i)-(j-a_j) \leq i-j-1$. If all the conditions are satisfied, ${dp}_i:=\max({dp}_i, {dp}_j+1)$.Now all we need to find is the minimum of $i-a_i$ over all $i$ such that ${dp}_i \geq k$.Submission: 123519134
•  » » 3 years ago, # ^ |   0 $(i - a_i) - (j - a_j) \le i - j - 1$seems to be just $a_i > a_j$right?
•  » » » 3 years ago, # ^ |   0 Yeah...you can simplify the condition like that. I just thought it made more sense to present it the way I did.
 » 3 years ago, # | ← Rev. 2 →   +3 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
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 In case odd we need to place one row of vert dominoesI think there should be horizontal dominoes instead of vert..as in one row only horizontal dominoes can come.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +1 Yes, sorry, that is a typo. I changed "vert" to "horz".
 » 3 years ago, # | ← Rev. 2 →   +1 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 : Find how the max number which we can allot colors by desired conditions , say this is x 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 123566013Also 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!
•  » » 3 years ago, # ^ |   +4 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.
•  » » » 3 years ago, # ^ |   +2 Thank you Astro-naut! Understood due to the tc! Now will go for the better approach thanks a lot !
 » 3 years ago, # |   0 Is it possible to solve Equidistant Vertices in faster than O(N^2 K) (maybe with some optimization in the dp)?
•  » » 3 years ago, # ^ |   0 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.
•  » » » 3 years ago, # ^ |   0 Oh that's a nice idea. How is Kth coefficient computed in O(N log^2(N)) though?
•  » » » » 3 years ago, # ^ |   0 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.
•  » » » » » 3 years ago, # ^ |   0 Ah I see. Nice solution! Could this be done faster with generating functions?
•  » » » » » » 3 years ago, # ^ |   0 Sorry, no idea.
 » 3 years ago, # |   0 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
•  » » 3 years ago, # ^ | ← Rev. 4 →   0 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
•  » » » 3 years ago, # ^ |   0 Thanks a lot ! I got it where I was going wrong.
•  » » » » 3 years ago, # ^ |   0 You Should upvote him for helping.
 » 3 years ago, # |   +1 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
•  » » 3 years ago, # ^ |   0 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 .
 » 3 years ago, # |   0 Another approach for B2 would be doing binary search on how many elements can be painted in one color.
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 Can you please elaborate your idea of binary search because I was trying to do it through binary search but got tle on tc5.Although B1 passes due to low constraints . Here's my submission https://codeforces.com/contest/1551/submission/124039451 Can I improve my code or do I need to think of solving it through another approach ?
•  » » » 3 years ago, # ^ |   0 You need to change your good() function. You should use something like set which will keep the pairs sorted. You can seelink my check() function
 » 2 years ago, # |   0 147410960 Hii! can someone check why is it failing on test case 1, even when everything is correct according to me. Thanks!
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 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.
 » 23 months ago, # |   0 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 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.
 » 22 months ago, # |   0 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