First: We apologize for any inconvenience caused. On the second day, we'll try to correct all today's mistakes.
We still hope that you enjoyed the contest.
You can upsolve here
Let's discuss the problems.
Upd: Day 1 unofficial results here
Upd: Day 1+2 unofficial results here
Upd: Day 2 now available to upsolving.
Upd: Editorials.
Can you say how to solve A?
Let's assume that we are at point $$$p$$$ now. We have two options.
1) Go to $$$t$$$ directly, paying extra $$$|p - t| * a_p$$$ money;
2) Go to another "optimal" point. There are two choices for that. Either it is minimum $$$r$$$ such that $$$a_r < a_p$$$ and $$$r > p$$$, or maximum $$$l$$$ such that $$$a_l < a_p$$$ and $$$l < p$$$.
So run Dijkstra, maintaining minimum distance. Update answer by the first option. Or go to the next optimal point paying $$$a_v * dist$$$ ($$$r - p$$$ or $$$p - l$$$) money. We are multiplying $$$a_v$$$ because each time we are going to the index with a lower value.
I also did djikstra, but there is no need for it. The graph you get is a DAG, so simple dp is enough.
If I'm not wrong, it can also be solved by the Convex hull trick. The naive solution will be using DP. Let's say $$$dp_i$$$ is the minimum time needed to reach the point $$$i$$$. Recurrent relation will look like $$$dp_i = dp_j + |i - j| \cdot a_j$$$. It is easy to notice that the next point's value will be always less than the current's (i.e $$$a_{cur} > a_{next}$$$). So the $$$a's$$$ will be decreasing.
I also solved it with a convex hull trick. I used two convex hulls. One for queries for the left (i < j) decreasing "ladder" from the start, the other for the right (i > j).
Could you share your code, please
Can you say how to solve B?
To solve in n^2logn, we can notice that actually if we reassing our element from L to R, their id to from 1 to R — L + 1, it is easy to notice that actually after each separation their indexes goes left shift. So having M different groups, we can answer the sum with the formula. Then we can notice that 1 to R — L + 1, does not matter and we can just add their initial ID. So with that we can do MO in Nsqrt(N)log(N), but it will also be TL. Then we can notice that it is enough to keep the last Position where such division starts, so just by fixing R we can answer L easily with that information. My realization was with Trie and Fenwich. Nlog^2
I Don't understand what you mean by "such division starts". I had trie and Mo's algo but I still got wa-s(and also TLE :P). Here's what I did, it's probably the same thing. You are splitting the array into multiple subarrays based on prefixes in base 2. For 2 equal values you are trying to get to the point where the prefixes of the indexes differ. Another thing is that if A and B have a common prefix of some length, A — X and B — X will also have a common prefix of the same length. In my VALMAX trie I added indexes from the original array(thanks to the observation it still works) based on the value of respective index. I counted all the nodes that have at least 2 paths going through them, because that means you have to split up to that point(maybe even further down). I also had an unordered map which included all the points in which I have to split. From multiple values would've collided. It still got WA, so it's probably wrong.
Can you say how to solve C?
Can you say how to solve D?
not today
Is the second problem reference to this post?
Yes. To meme and author
Wowowowow, it's extremely cool, thanks :D :D :D
Is Aizhan a real person?
Is that meme a real life situation?
DavitMarg foresaw that he wasn't going to win a medal, so he decided to bring potatoes instead. He didn't bring potatoes either, though :D
Will be official results published today?
Unofficial results already published.
Are all the participants official in the standings?
yes
Lol who was responsible for the 6th subtask of 3rd problem? It genuinely costed me 20 minutes :P
somehow, i had "ai = i; bi = i + 1" from the start of the contest, or i accidentally reloaded statements. im not sure
The broken statement is only at English statement, if im not mistaken.
How to solve day2 A for
n <= 30
andn <= 2000
?Do you know the cutoff for the medalists?
Or, at least, what it was (roughly) in the past years?
18 is gold
43 is silver
80 is bronze
Okay but how many participants were there?
Something around 290
There were 295 this year, yes. But how many were there in 2020, I heard somewhere that the no. of participants became quite bigger this year, right?
161 participants
Dies from cringe(
Why its not working in C on n <= 1e3?
Put your code in a spoiler, please.
Как?
Иконка кфа -> spoiler
вот так?
Thx
Я не смог найти вкладочку
Explain your idea
So. Let's for every vertex i and L border find max R, where in F(S[l, r]) no contains i. So clear, that with increasing L, the R monotonically increasing. So let's do this with 2 itterators, and check this. Condition is met, If there is a vertex in the subtree and depth LCA set <= d[i]. And i do this
Или более человечески по-русски. Для каждой вершины найдем все моменты времени, когда она находится в этом "обтянутом дереве". Для этого для каждой левой границы найдем минимальную правую границу, для которой это выполняется (понятно, что тогда выполняется на всем суффиксе). И что правая граница не убывает. Тогда будем это делать двумя указателями, а проверкой будет следующее условие: в поддереве этой вершины кто-то есть, и LCA этого множества выше, то бишь его глубина меньше, что, как мне кажется, у меня написано
ты писал стресс? мб поможет
Мне было лень( И я ручками 1.5 часа дебажил...
Имхо дебагать полтора часа +25, когда у тебя 0 и при этом по задаче с прочтения набирается 30 баллов — не очень хорошая идея :/
Ахаххахаха, я условие неправильно понял, тут было без шансов)
I don't speak russian, but it with google translate it seems to me like you found the mistake yourself. Good job?
Yes, im find mistake, wrong read statement)
Nice code, everyone in Novosibirsk writes so beautifully?
Yes
putin style bruh
Will there be editorial or author's solutions?
It should be. Editorial is likely to be published on IZhO like previous years.
I have never seen any official editorials for past years :(
There is no editorial even at page that u send....
Editorial is published.
Thank you!
Day 2 upsolve?
Added
Isn't there an easier solution for day 2 B (rooms)?