I think copy pasted part has been skipped.

I asked the question when post didn't mention it. Now it's updated.

More like 4 except for a one-day gap.

Did you account for up coming round "Dasha Code Championship"? :)

I think combine round is different from separating them. People usually earn more and lose less rating. :)

Combined rounds should be used only for events (usually with prizes). Normal rounds are much better. Problems fit participants more and hacking is more balanced.

It's literally midnight (i.e. 12 AM) in China when this starts

Maybe around 8, I guess.

There are known knowns, there are things we know we know. We also know there are known unknowns, that is to say we know there are some things we do not know. But there are also unknown unknowns, the ones we don't know we don't know.

Вот

Поорал в голос с формата вывода в A:

"Выведите максимально возможное число в двоичной системе счисления. Выводите двоичные цифры, разделяя их пробелами." — это единственное, что надо знать про авторов.

Графов много не бывает

you have the results of: a1*a2,a1*a3,a2*a3

a1*a2 = r1

a1*a3 = r2

a2*a3 = r3

a1*2 = r1*r2/r3 --> a1 = sqrt(r1*r2/r3)

Hope it works after test

why im so stupid to use gcd ?

I'm not sure but... If we have A and B, where A is odd and B is even we can create a cycle with odd length. So we will use recursion There are two possible ways: delete all even numbers in an array and leave recursion or delete all odd numbers, divide all numbers left by 2 and do recursion again. We continue recursion and divide by 2 because there can be something like this: 2, 4. We can create cycle 0 — 2 — 4 — 0. So we can solve this problem with O(n log n)

you have to realize that:

A and B can be in S if and only if the maximum power of 2 that divides A is equal to maximum power of 2 that divides B.

It asks you to delete the minimum number of vertex, so you could delete either 1, or 2.

It's not a traingle. 3 elements in set B doesn't mean there are 3 vertices. There are infinte vertices actually

Can you please tell O(n^2) approach?

One fact is you can only pick two numbers that are divisible by same power of two (for example you cant pick 6 and 40 but you can pick 10 and 22).

in the set there can only be numbers of the form

2 ^ s * i , where i — any odd number, and we can iterate s

AAAAAAAAARGH

Maybe it's finally time to write a geometry library using bigints only. Or something. -.-

(At least it's gonna be useful on GCJ geometries.)

I tried something like: consider several small primes, get possible values of X and Y modulo them, combine it using CRT. I didn't have enough time to think it through but maybe it is the intended solution.

Checking if distance from center of mass is exactly as needed makes solution faster. After this check you can try about 100 points around the intersection and not get TLE.

Having finally solved this, I want to strangle the next person who mentions geometry to me in person. I tried so hard to make calculating cos() from cosine theorem precise enough, but in the end, I gave up and did all subtraction in bigints.

I'm scared for system tests now.

No.

What? Really? Noticed it right now :)00)

I think the statement was very poor in problem D :/

[Deleted]

I spent 1 hour thinking a different problem :c I didn't notice that until now xD

Ow maan, but, $$$i, j \in \mathbb{Z}$$$...

any node in a cycle or in the path from $$$s$$$ to the cycle, can be always taken by $$$s$$$->cycle back-> $$$s$$$. all the nodes can be treated as $$$s$$$. so remaining graph is a tree. find longest path to some leaf.

Notice that element $$$a$$$ is an ancestor of an element $$$b$$$ when it's minimum on a subsegment from $$$a$$$ to $$$b$$$. Count amount of ancestors for every element in initial permutation. Now, when you remove element $$$l$$$ from the left, all elements between it and the leftmost element smaller than $$$l$$$ now have one ancestor less. When you move it to the right, all elements between it and the rightmost element smaller than it have one ancestor more. You can store amounts of ancestors in a segment tree with lazy propagation. Solved.

Just consider the smallest char already seen. The game will not last longer than 1 move :)

Only thing that matters is the first character on the left that is smaller than s[k]

The game either has 0 moves or 1. If a smaller alphabet occurs before the start pos, ans is Ann, else Mike.

Not sure though.

Thanks everybody. It was seriously difficult problem...

I first used suffix array but failed on test as "abas" as for k = 3 the answer is Mike not Ann but using suffix array would print Ann as suffix "as" is bigger than "abas" but after that I realized the simple solution considering the minimum char already seen!