### awoo's blog

By awoo, history, 11 days ago, translation, 1792A - GamingForces

Idea: BledDest

Tutorial
Solution (Neon)

1792B - Stand-up Comedian

Idea: BledDest

Tutorial
Solution (awoo)

1792C - Min Max Sort

Idea: BledDest

Tutorial
Solution (Neon)

1792D - Fixed Prefix Permutations

Idea: BledDest

Tutorial
Solution (awoo)

1792E - Divisors and Table

Tutorial

1792F1 - Graph Coloring (easy version)

Idea: BledDest

Tutorial
Solution (BledDest)

1792F2 - Graph Coloring (hard version)

Idea: BledDest

Tutorial
Solution (BledDest)  Comments (54)
 » slow tutorial :)
 » Isn't it possible to have an O(n) solution for problem C (Min Max Sort)?
•  » » yeah, there's an easy $O(n)$ solution for problem C, involving using the array $pos[i]$, to get the position of index $i$ in the permutation, then start at the middle value and finding the $LIS$ to both side. My submission: 190340250
•  » » maybe you read only the last line.read the third last line.
•  » » » Indeed I did. Thanks :)
 » Thanks Man Searching for long time
 » In problem A, why the answer is n — cnt1 / 2?
•  » » Basically, we are assuming that it will take n spell casts. for n=10 and cnt1=2 so initially it was taking 10 spell casts but because of cnt1=2 it will take 10-1=9 spell casts. 8 of second type and 1 of first type.
•  » » Here, the first operation only makes sense when we have 1 as the health, in all other cases we will try to use the second operation . Try to do rough work, you will get a fair idea.Now by using the first operation, we will try to eliminate all pairs of 1. hence, number of operations = cnt1 / 2.
 » can somebody explain how to come up with the equation in problem B.
•  » » 11 days ago, # ^ | ← Rev. 3 →   Here, the Optimal strategy to solve the problem is by first saying all the jokes that both Alice and bob likes. (type 1 jokes) then by saying jokes alternately that (Alice likes,bob doesn't) and vice versa. Equation: a1 + min(a2,a3) x 2 Final strategy is to say remaining jokes : first by saying remaining type 2, type 3 jokes and then by saying the jokes that both doesn't like (Type 4). this should be compared with the points that he already acquired hence a1 is taken. We take a1 + 1 as the judges goes when points become negative. Hence we add 1 to make it -1. Finally, We take the minimum of both these values to form the final equation. Equation: a1 + min(a2,a3).2 + min(a1 + 1,abs(a2 - a3) + a4) 
•  » » » great explain
 » How to solve q3 with binary search approach ? And what is the logic behind it.
•  » » 5 days ago, # ^ | ← Rev. 3 →   I used binary search. Here was my logic:I noticed that the last operation has to be x = 1, and y = n. You have to swap those two, because otherwise another two numbers will be on the outside. We can extend that logic further to say that the last operation has to be (1, n), the second to last has to be (2, n — 1) and so on. Therefore, the maximum number of operations is n / 2, and we can see what the lowest value k is such that if you do the operation on pairs (1, n), (2, n — 1), ... (k, n — k + 1) is sorted.Here is the submission: https://codeforces.com/contest/1792/submission/191351639
 » Does anyone have an idea on proving the efficiency of bisearch-on-divisors approach for prob. E? Never expected it will be this fast = =
 » Can somebody explain what is the meaning of run binary search on k in problem C tutorial?
•  » » 11 days ago, # ^ | ← Rev. 2 →   Remove value out of [k, n + k — 1] from array. If the remaining numbers are increasing （from [k, n + k — 1]), then the number k is appropriate. You can use binary search to find k from [1, n]. You can see this submission, for more details: https://codeforces.com/contest/1792/submission/190334713
 » 10 days ago, # | ← Rev. 2 →   For whatever it's worth, here's my $O(n \log n)$ solution for problem F.First, we notice that the property for the set $S$ is the same as having a one-colored cut in $S$ (in other words, the set of vertices can be split into $S_1$ and $S_2$ so that all edges between $S_1$ and $S_2$ have the same color). The only issue is that a given set $S$ can possibly be split into $S_1$ and $S_2$ in several ways. So I introduce $f(n)$ to be equal to the number of graphs on $n$ vertices where the global cut is blue. Then the answer is $2 f(n) - 2$, since the global cut may be red, but $2$ extra cases arise when all edges have the same color.To find $f(n)$ we need to partition the set of $n$ vertices into $k \geq 2$ subsets so that vertices from different subsets are connected by blue edges, but the subsets themselves obey the rules recursively (they have global cuts of red edges). Since $f_{\text{red}} = f_{\text{blue}}$, we have $f(n) = \sum\limits_{k \geq 2}~ \sum\limits_{a_1 + \ldots + a_k = n} C(a_1, \ldots, a_k) \cdot f(a_1) \cdot \ldots \cdot f(a_k),$where $C(a_1, \ldots, a_k)$ roughly means the number of ways to choose subsets of sizes $a_1, \ldots, a_k$ from a set of size $n = a_1 + \ldots + a_k$. If $b_1$ occurs $c_1$ times, $\ldots$, $b_m$ occurs $c_m$ times within the multiset ${ a_1, \ldots, a_k}$, then $C(a_1, \ldots, a_k) = \frac{n!}{a_1! \cdot \ldots \cdot a_k! \cdot c_1! \cdot \ldots \cdot c_m!}.$Define $h(n) = \frac{f(n)}{n!}.$ The base values are $h(0) = 0, h(1) = 1$. Via $H(x)$ we denote the generating function of $h$: $H(x) = h(0) + h(1) \cdot x + h(2) \cdot x^2 + \ldots$. After a careful examination (I don't know how to prove rigorously it though) we obtain $e^{H(x)} = 2H(x) + 1 - x.$Everything else is the standard approach of how to solve these recurrences: if we know $H(x) \bmod{x^{m}}$, then from the equality above we can obtain $H(x) \bmod{x^{2m}}$. Underneath we need FFT (NTT) and an exponential generating function. Each step from $m$ to $2m$ takes $O(m \log m)$ time, so the overall complexity if $O(n \log n)$.
•  » » I also found this solution, thanks for sharing it!
 » I don't get why we have to make sure that no set of vertices is connected by both colors in F1. Doesn't the lemma proved it impossible to connect a set of vertices with both colors?
•  » » 9 days ago, # ^ | ← Rev. 2 →   The lemma proves the other thing: it is impossible to have a set of vertices which is neither blue-connected nor red-connected.
•  » » » Thanks. I made an oversight while reading the lemma.
 » A-F1 video editorial for Chinese:BiliBili
 » 10 days ago, # | ← Rev. 3 →   In fact, we can calculate the convolution-like sequences (such as those in problem F) in $O(n \log^2 n)$ or $O(n \log^2 n / \log \log n)$ or even faster. One can find the approach from Elegia's report. The implementations usually has lower constant factor in time than those of the Newton iteration (if exist).Here is my implementation, which imitates Elegia's implementations for other problems.(but it's just as fast as a brute force lol)
•  » » Do you have this report as a PDF?
•  » » »
•  » » » » Do you know how to find ref.  (罗煜翔。(2020)。浅谈 Nimber 和多项式算法。IOI2020 中国国家集训队论⽂集。) ? I wanna see how to do semi-relaxed multiplication in $\frac{n\log ^2 n}{\log \log n}$(if that's what it says). And what nimbers have to do with it xd.
•  » » » » » Let's discuss about it here, actually in that report, most contents are just putting the general algorithm into the nimber framework. Nimber does not play an important role in the relaxed convolution.
 » "The array may large" in E made me realise how I don't trust my gut feeling at all.
 » Where does the log(divs(m)) of the complexity in question E come from Isn't it only need O(divs(m)⋅z(m)) to calculate the dp array?
•  » » oh i get it,because you can't find the element within O(1)
•  » » then why not use hash
 » 9 days ago, # | ← Rev. 2 →   can somebody explain why in D if we use lower_bound ans EITHER result or the previous one? why not just result? thanks in advance
•  » » It depends on the value immediately after the longest common prefix. While searching for $p$, you find some inverse $q_1, q_2, \dots, q_m$ that starts with $p_1, p_2, \dots, p_k$ for some $k$. The next value is different. If $q_{k+1} > p_{k+1}$, then lower_bound will point at $q$ (or one of inverses with such prefix if there are multiple). However, if $q_{k+1} < p_{k+1}$, then $q$ will be smaller than $p$, and lower_bound will jump over it. The next inverse has to be greater than $p$, so you only have to look one step behind.
 » For D, I thought while $n=5\times 10^4$, the number of subsets of $\{1,...,m\}$ is $10^3$.So I constructed a map $M$ mapping each subset $L\subset \{1,...,m\}$ to $\{(a_i[p_1], ..., a_i[p_t])\ |\ i\in \{1,...,n\},\ \{p_1<... •  » » 8 days ago, # ^ | ← Rev. 2 → I thought of the same thing during the contest but midway realized that its gonna give TLE :( •  » » You can implement it with Trie in$O(2^m\times n \times m)$. I'm not sure whether it is fast enough.  » I have some questions in problem F1 Why iterate k — the number of vertices whick are in the same 'red' component as 1 but not iterate the blue component, is this because you are counting the blue component? What about the edges between vertices in the same componet as 1 and the rest vertices, are they must be the same color? how to proof? Hope someone can help me, thx a lot •  » » We are considering the case when the whole graph is blue-connected, so there's no need to iterate on the blue component. The case when the graph is red-connected is symmetric to it. It's easy to see that all these edges are blue, since any red edge between any vertex from the "red" component of vertex 1 and any vertex outside this component means that we haven't picked the whole component •  » » » I understood!Thanks a lot  » Can someone give a test case on which my code is failing? I couldn't view the 156th item of 2nd test case. Is there any way to view it? •  » » Click the "Click to see detail" button in the bottom of the page to see the detailed test cases •  » » » I did that but the 156th item is not visible.  » has anyone solved problem C with binary search ?  » NEED PROVEMENT FOR PROBLEM F •  » » Nevermind  » F1's solution is flawed I guess. Consider case with n=3, when A1=B1=1, A2=2, B2=1, you get B3=4 which is incorrect. •  » » 4 days ago, # ^ | ← Rev. 2 → So what's wrong with the$B_3=4$? •  » » » A3 which is the answer for n=3 is 6 in the pretest case, but surely 6 is not 4 * 2 :P •  » » » » Well, maybe the tutorial forget something or i just didn't see.But don't forget the limitation 1 and 2.The$A_3=8$contains the case that all blue and all red. •  » » » » » I noticed that too. I'm just saying that define "An" as "the answer for n" is not very strict and could lead to someone's misunderstanding(like me) •  » » » » » 另外看兄弟id八成也是国人，直接说汉语就好了吧:) •  » » » » » » 哈哈 •  » » Sorry, I initially stated the problem without the constraints "at least one edge should be red" and "at least one edge should be blue". I wrote the editorial for that version, but then decided to introduce these constraints. So, the actual answer is$A_n-2\$ since we need to discard the case "all edges are red" and the case "all edges are blue".