### awoo's blog

By awoo, history, 14 months ago, translation, 1613A - Long Comparison

Idea: BledDest

Tutorial
Solution (awoo)

1613B - Absent Remainder

Idea: BledDest

Tutorial
Solution (Neon)

1613C - Poisoned Dagger

Idea: BledDest

Tutorial
Solution (Neon)

1613D - MEX Sequences

Idea: BledDest

Tutorial
Solution (Neon)

1613E - Crazy Robot

Idea: BledDest

Tutorial
Solution (awoo)

1613F - Tree Coloring

Idea: BledDest

Tutorial
Solution (BledDest)  Comments (53)
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 » In my opinion , E might be a little easier than D , so I recommend swapping D and E .
 » 14 months ago, # | ← Rev. 2 →   Neat and Clean explanations especially on F. Thank you!
 » This was a great contest, learned about FFT technique from problem F which seems quite useful
 » great problems! great editorial! I learnt a lot.
 » 14 months ago, # | ← Rev. 3 →   i wrote this dp solution for problem D Spoilerconst int mod = 998244353; void solve() { int n; cin >> n; vector arr(n); for(int i = 0; i < n; i++) { cin >> arr[i]; } map, ll> dp; for(int x: arr) { // mex == x - 1 dp[mp(x - 1, x + 1)] = dp[mp(x - 1, x + 1)] * 2 % mod; dp[mp(x - 1, x + 1)] = (dp[mp(x - 1, x + 1)] + dp[mp(x - 1, x)]) % mod; // mex == x + 1 dp[mp(x + 1, x + 2)] = dp[mp(x + 1, x + 2)] * 2 % mod; dp[mp(x + 1, x + 3)] = dp[mp(x + 1, x + 3)] * 2 % mod; // mex == x dp[mp(x + 1, x + 2)] = (dp[mp(x + 1, x + 2)] + dp[mp(x, x + 1)]) % mod; if (x == 0) { dp[mp(1, 2)] = (dp[mp(1, 2)] + 1) % mod; } else if (x == 1) { dp[mp(0, 2)] = (dp[mp(0, 2)] + 1) % mod; } } ll ans {}; for(auto p: dp) { ans = (ans + p.ss) % mod; } cout << ans << '\n'; } 
 » Problem F can be solved in $O(n\log n)$ time: SpoilerConsider maintaining the product for those vertex such that $d_i \ge k$ and loop $k$ from $n$ to $1$. Each time we only need to multiply $(1+kx)^{\sharp_k}$, where $\sharp_i$ denotes the number of vertices such that $d_i = k$. The complexity can be bounded by $\sum_{k} \left( \sum_{i} [d_i \ge k] \right) \log n = \left( \sum_i d_i \right) \log n = O(n\log n)$There is also an much more complex algorithm that solve F in $O(n)$: SpoilerWe define linear functional $L: x^n \mapsto n!$, our goal is to evaluate $\displaystyle L\left(\prod_{i=1}^n (x-d_i)\right)$. Let $\sharp_k = \sharp i \mid d_i = k$, let $\displaystyle P(x)= \prod_{k < B} (x-k)^{\sharp_k}$ and $\displaystyle Q(x) = \prod_{k\ge B} (x-k)^{\sharp_k}$.Since $\displaystyle L(PQ) = \sum_{i\le n/B} L(x^iP) \cdot [x^i]Q$. Noticed that $\displaystyle \sum_{i=1}^n d_i = n-1$, evaluation of $Q$ has an upper bound $O\left(\frac{n\log^2 n}{B}\right)$. (This estimation might be improved)Since $P$ has an P-recurrence with order $B$, we can evaluate $L(P)$ in $\tilde O(\sqrt n \operatorname{poly} B)$ by the fast algorithm for P-recursive sequence. Then we show that $L(x^i P)$ has an recurrence respect to $i$.Noticed that $L(P) = P(0) + L(P')$ for all $P$, we have the recurrence of $L(x^iP)$ \begin{aligned} L (x^iP) &= L(ix^{i-1}P) + L(x^{i}P')\\ & = i L(x^{i-1}P) + L \left(x^i P \cdot \sum_{k •  » » Umnik: "Idk how Elegia's mind works" •  » » » Those interested in how um_nik's mind works can read how to solve it inO(N log N)$in simpler notations here •  » » Captain EI txdy txdy txdy •  » » by the fast algorithm for P-recursive sequence I've never heard of this before, could you link to a tutorial / paper or something? •  » » » The brief idea is to evaluate$\mathbf M(x+1)\mathbf M(x+2)\cdots \mathbf M(x+\sqrt n)$for$x=\sqrt n, 2\sqrt n, \dots, \sqrt n \cdot \sqrt n$where$\mathbf M(x)$is a matrix with polynomial entries. Min-25 gives an algorithm that achieves the smallest log factor currently. Though his original blog was not available for some reason, there is a detailed description of this algorithm in the editorial of CF Round #700 div1. E. •  » » » » I read that editorial, but unfortunately don't see how it is connected to this problem.Oh, also, what's the P-recursive equation for$P$? •  » » » » » Since$\displaystyle \log P = \sum_{k
•  » » it would be very interesting to see the code for the $O(n)$ solution
 » damn those are some very elegant solutions
 » I don't understand why the runtime is $O(n \log^2 n)$ in problem F.We can write the divide and conquer runtime as $T(n) = 2 * T(n / 2) + O(n \log n)$. This is case 3 in the master theorem of divide and conquer which yields $T(n) = \theta(n \log n)$, no?
•  » » Do you have an implementation? This sounds cool.
•  » » » 14 months ago, # ^ | ← Rev. 3 →   I didn't solve this problem, I'm simply reading the solution from the editorial and wondering why their implementation is $O(n \log^2 n)$ instead of $O(n \log n)$. I think my mistake is probably that the 3rd case of the master theorem doesn't apply here, and the master theorem cannot be used. To use the master theorem case 3 there must exist a constant $\epsilon > 0$ such that $n \log n = \theta(n^{1 + \epsilon})$, and it might not exist.
•  » » » » 14 months ago, # ^ | ← Rev. 2 →   Because the $f(n) = \Theta(n \log(n)) = \Theta(n^{c_{crit}} log(n)^1)$, this recurrence actually falls in case 2 of the master theorem (if you hold on the order of cases in wikipedia). This solves to a running time of $T(n) = \Theta(n \log(n)^{1+1})$.
•  » » » » » I was referring to the cases from CLRS (page 94) and the case 2 is more specific there, and it doesn't fall into it.Indeed if I look at Wikipedia the case 2 is more generic and it does apply in this case, thanks!
 »
 » In problem D,Why the ans should become ans — 1?
•  » » 14 months ago, # ^ | ← Rev. 2 →   We initialize $dp1_{0,0}$ with a $1$, which represents an empty subsequence (which has MEX equal to $0$). At the end we accumulate the answers and count it in. Since we are asked to count only non-empty subsequences, we should subtract it.
 » 14 months ago, # | ← Rev. 2 →   I cant understand why the number of colorings that meet $k$ choosen violations is $(n-k)!$ in problem F, can someone explain it in more details please?
•  » » 14 months ago, # ^ | ← Rev. 3 →   I can give an intuitive idea behind that. For every violated constraint, a node of the tree is already defined by its parent (it's just their parent's number minus one). So basically out of the $n$ nodes, we can focus on the $n - k$ nodes which aren't tied to their parents and think about the relative order between them. Let's build a sequence of these untied nodes in which the first node has the highest value $n$, the second the next highest value and so on.Of course you can choose any of them to have the highest value $n$, then choose any other node to have the next highest value (which depends on the previous node as it might have induced the value $n - 1$ to one of its children and maybe more values to its descendants) and so on. The thing is you can choose any of the untied nodes to go next in this sequence. Therefore you can have any possible relative order between them, so the number of colorings is $(n - k)!$
•  » » » Thank you, it was very helpful !
 » 14 months ago, # | ← Rev. 2 →   I feel like F should have a "divide and conquer" tag based on what's written in the editorial, is it really so? UPD: Thanks for adding the tag, hope it will help people in their practice later ^_^
 » Can someone help me in E, 137783013 it keeps getting TLE but it's just a BFS.
•  » » Try replacing your endl by "\n"endl is slower. It also flushes the output
•  » » » Thanks a lot! In a million years I wouldn't have thought of that
 » One thing I can't understand in the tutorial for D: The tutorial says that [0,0,1,1,1,2,3,3] is a MEX-correct sequence. However, this violates the definition of MEX-correctness. For example, let i = 7. Now MEX(0,0,1,1,1,2,3) = 4. |0 — 4| = 4, |1 — 4| = 3, and |2 — 4| = 2. These are all violations of MEX-correctness as described in the question. May you elaborate on how the above sequence is MEX-correct?
•  » » Please read the second announcement. You got a wrong understanding of MEX-correct sequence.
 » 14 months ago, # | ← Rev. 2 →   https://atcoder.jp/contests/abc213/tasks/abc213_h The above problem from atcoder have the solution where one has to apply DNC FFT stuff. While The Problem F seems like a polynomial multiplication where one prefers to multiply shorter polynomial first. We can do it along the similar line of merge-sort or simply use priority queue and multiply shorter polynomials each time. 137796885. How is it DNC FFT? (Or what is DNC FFT exactly?)
 » I couldn't understand the editorial this much but can I solve E like this?? we implement a bfs or dfs from lab and just change to + the cells of grid which we can control and we can reach them with a path from the lab that we can control?? we can control means that the cell has two paths or less.
•  » » 14 months ago, # ^ | ← Rev. 2 →   You can start a BFS from the lab. Mark lab as + for generalization sake. During bfs, do the following: 1. If a cell you are looking at has at most one neighbour cell which is not blocked and not with plus, then mark that cell as +. 2. If a cell is marked + now, push all of its neighbour cells to the queue unless the neighbour cell was visited from this particular cell. In other words, keep visited not just for some cell, but for a parent-child relationship. If cell a already pushed cell b to a queue, then don't let a to push b again, but some other cell c can still push b. 3. Go to the next queue element. Pretty sure there is no way to solve this problem using DFS.
•  » » » There is a dfs solution, check my submission: 137662230. Consider the lab is the root, we follow the four "hallways". At any square we find it connects to more than 1 '.', stop there.
•  » » » thanks I understood:)
 » Thanks to div2, I became a pupil
 » Can anyone please tell me what's wrong in my submission for E?I am getting TLE on : 137850453 Also this is the exact same solution which got passed in about 200 ms : 137675491
•  » » 14 months ago, # ^ | ← Rev. 3 →   Dont use std::endl, use "\n" instead.Flushing std::cout is very slow, especially when you need to do it 1000000 times.see this: 137852168 and 137853185
 » In problem E (Crazy Robot) I am getting wrong answer over test case 8If anyone can help me outThanks in advance
•  » » Your mistake is using stack qx; stack qy;. it should be stack to not overflow as a char overflows quickly.As a tip, this is how I debugged your code. Use assert() and insert it at different points of your code to see where things went wrong. I assert()ed that your values were valid just before pushing it onto the stack, as well as assert()ing the values at the start of the function. It was okay before pushing it onto the stack, but not when you used it in the function. So I can deduce that something must have went wrong when you passed it into the function, and that's how I found out.
•  » » » Thanks bro
 » 14 months ago, # | ← Rev. 7 →   [deleted]
•  » » No, $1 \leq x_2 \leq 10^6$. Your input $1000000000000$ is too large.
 » 14 months ago, # | ← Rev. 2 →   An alternative solution for C without using binary search.Let $d_1, d_2, ... d_{n-1}$ be the differences $a_{i+1}-a_{i}$ sorted in non-descending order. Now it is easy to calculate the damage to the dragon as a function of $k$ if we observe that for $k = d_{i}$ it is equal to $\sum_1^{i} d_i + (n - i) \cdot k$, and for other values it is linear with the slope 1. https://codeforces.com/contest/1613/submission/137715371This solution would have an advantage if there were multiple queries for different values of $k$, since it would be $O(1)$ per query if we precalculate prefix sums.
•  » » How did you come up with the above function ?
 » Thanks for the great editorial for D!
 » Can anyone please help me , Why am I getting memory limit exceeded on 8th test case in problem E ? https://codeforces.com/contest/1613/submission/138290812
•  » » Never mind It got resolved after passing multidimensional vectors by reference which helped me removing some unnecessary global containers.
 » In 1613D — MEX Sequences, for the 1st test case, why [0,2,1] is not in the answer? MEX([0,2,1]) = 3 0 — 3 = -3 <= 1 2 — 3 = -1 <= 1 1 — 3 = -2 <= 1So the correct answer should include: , , [0,1], [0,2], [0,2,1]Thanks for any hint. If I understand it correctly, in the last test case:0 1 2 3The answer should include:, , [0,1], [0,2], [0,1,2], [0,1,3], [0,1,2,3]Please correct me if I am wrong.Thanks heaps :D
•  » » [0 2 1] is not correctin ith step we get mex from (x1, x2.... xi)i = 0; we have (0) now mex is 1. so |0-1| = 1 <= 1.. right. i = 1; we have (0, 2). now mex is 1. so |2 — 1| = 1 <= 1.. right. i = 2; we have (0, 2, 1). now mex is 3. so |1 — 3| = 2 > 1... wrong;
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