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Rayan Programming Contest 2024 - Selection (Codeforces Round 989, Div. 1 + Div. 2)
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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 30.11.2024 01:29:54 (f1).
Десктопная версия, переключиться на мобильную.
При поддержке
div1C не хватает
Теперь хватает
Каноничной подписи на аватарке тоже не хватает
"В хорошей матрице либо в какой-то строке не больше двух различных символов, либо в каком-то столбце не больше двух различных символов."
Не в какой-то (каком-то), а в каждой (каждом)
Ваше утверждение, конечно, равносильно, но это не сразу очевидно)
"A fat fish can't dangerously eat a fish with smaller weight. Indeed, even if all the smaller fishes eat each other, the resulting fish would be too small. We can conclude that the danger is not greater k−t." How can you conclude that? All fat fish may as well take part only in fights with bigger eels, case in which the argument you have doesn't imply a "lost fight". They as well might fight with each other (you have around t/2 non-dangerous fights from here, and from then on, there's no argument as to why those newly created eels will not take part only in dangerous fights). Could you give a more formal proof for that observation?
Let's take fat fish, say it is k-th in sorted order. Let's give each fish its mark: it will be 1 for (k - 1)-th fish, 2 for each fish smaller than (k - 1)-th, and 0 for all bigger (including our fat fish). The result of the fight will be fish with mark min(x, y) where x and y are marks of fishes in the fight. In the end we will get fish with mark 0, so it is easy to see that there will be a fight between fishes with marks 0 and 1 at some moment. I claim that this fight is non-dangerous (any fish with mark 0 is more than twice bigger than any possible fish with marks 1 or 2). Also it is not hard to check that these fights are different for all fat fishes.
Укажите пожалуйста на мой косяк в Div2 B 48038355
посмотрите 19-й тест. исходные квадраты точно укладываются в большой общий квадрат (поэтому вам нужно длину квадрата умножить на два — чтобы получить, к примеру, длину левой и верхней стороны, на которые потом будет ориентироваться София), а у вас лишнее прибавление единицы где-то произошло.
Can someone explain their easy approach for div2 D with example ?
graph : https://imgur.com/a/Iac4Lm2
Here we see,that s(1) = 1 ==> a(1) = 1.We can say,that a(3) = s(3) — a(1) and a(4) = s(4) — a(1). But can we reduce the answer?Yes.We can put on vertex 2 min(s(3) — a(1), s(4) — a(1)).And this value will embrace all children of vertex(2).So a(1) = 1, a(2) = 2, a(3) = 0, a(4) = 1. Ans = 4.
how a(4)=s(4)-a(3) ???? cant get this part
and also plz explain when will be the answer not possible
1)Fixed.2) If s(u) < s(v), where u is a child of the v
please provide proof of problem E- Nice Table.
If a row contains three characters, two neighbor rows of each row will be same. If a column contains three characters, two neighbor columns of each column will be same. Thus, in a good matrix, either each row contain at most two different characters, or each column contain at most two different characters.
How to find amount of integer solutions of inequality with Euclidean algorithm div1 E ?
https://codeforces.me/blog/entry/53007
My solution for div 2 F: Cookies gives WA verdict on test 6. I followed the same approach as given in editorial. Can someone kindly point out my mistake? Thanks in advance. 49932059
IN problem Sum in the tree 10 1 1 2 2 2 3 7 7 7 1 2 -1 -1 3 2 -1 2 3 4 As per question for this test case solution should provide ans as 7 but most of solution is giving -1 as answer. Correct me if I am wrong.
In problem 1098D Eels, if we use
setorpriority_queueto maintain each half-interval, isn't the complexity O(log^2 n) pre query(instead of O(log n)) ?