Yeah, we get it. Kinda common as an East Slavic name tho.

Forget about +ve delta dude. CF is brutal.

me too

Also you will be negative

Maybe they haven't decided between two tasks and only one of them will be in the contest?

If an interactive task is coming, I hope it's not E or F again, it's always sad that I never get to try the interactive ones because they are obviously beyond my competetence...

It kinda partially happened in my contest 4 years ago, I stated that interactive problem(s) might have appeared, and in fact the only interactive found was Div1E (and no interactive in Div2 at all).

I hope you get +204

i hope. to become green ,,, quite noob right now

Get a cloth first.

Or interactiveforces?

Not just you.

No because contest is after 3 minutes

ага.. пора банить индусов

how c in 20 minutes?what was the observation?(sample cases?)

Same, I didn't know why my solution to C of LCM is right until I read the tutorial

I have proof for the binary search solution, but i dont have proof for the lcm solution

LCM. (I just somehow guessed it, couldn't prove it in-contest either)

it involves basic math

Let $$$S=\sum x_i$$$

$$$x_i*k_i > S \implies x_i > S/k_i $$$

Sum over all $$$i$$$: $$$S > \sum{S/k_i}$$$

Divide both sides by S: $$$1 > \sum{1/k_i}$$$. If this condition is true $$$S$$$ can be anything, otherwise no solution. But since we need integers simplest is to take LCM

Sad I was one hour-late, problems look interesting

Just check if sum of money = $$$10^9$$$ is possible or not. I don't have any proof either.

can you please prove your solution for problem c

how did you come up with the idea of B?

I work in mathematics related field, trust me the number of times I have read incorrect "number of balls in a bag" in a probability example is probably infinite.

I just became Newbie, solved A and B on my own thought this going good tried my hand on C thought it was tough and then the RESULTS. Only way to over come this cheaters is actually equipping ourselves to solve that particular problem.

Hint: Use hashing

My crumbs of thoughts:

• If the string is already valid, apply $$$p = n$$$.
• Due to the nature of the operation, the suffix of it (after $$$p$$$) should already be valid, or at least only invalid at the end (not yet having k similar digits). Iterate backwards, find the first occurence after the last chain of similar digit that causes incorrect pattern, choose a certain $$$p$$$ to keep the suffix correct, apply the operation, and check if the result is valid — if yes then yes, if not then immediately -1.

I think this was a really nice problem.

Ai = i xor x

Bj = j xor y

If Ai == Bj=> (i xor x) == (j xor y)

=> i xor x xor y == j

We can replace x xor y = k

So we want

i xor k == j

(i + 1) xor k == (j + 1) and so on.

And now one nice thing is that if i xor k == j

And (i + x) xor k == (j + x)

Then (i xor (i + x)) & k == 0

The above statement basically means that all the bits changed from i to i+x are unset in k.

Using this we can find the answer.

I think the line of reasoning can be like this:

1. Look at examples, and see that all answers are powers of 2. Googling "log2(33554432)" gives an integer.
2. Looking at the notes on the bottom (or by writing a test script), notice that xoring all integers will not cause any duplicates or missing numbers, but numbers are instead shuffled and gradually increase. Choosing $$$x=1$$$ causes adjacent elements to be swapped. If $$$x$$$ is even and $$$y$$$ is odd, then they cannot have more than 1 consecutive equal element. Testing powers of 2, choosing x=8 will swap adjacent blocks of size $$$8$$$, giving $$$[8,9,10,11,12,13,14,15,0,1,2,3,4,5,6,7,...]$$$. If y does not have the same bit set to 1, then any equal subarray between the two can have at most size 8.
3. This leads to the strategy: pick the smallest bit $$$i$$$ that differs between $$$x$$$ and $$$y$$$, and take $$$2^i$$$. Trying this for the examples, it yields the same results.

Lmao how? This proves guessforces is real.

how did u solve E.

It is a standard LCM problem.

I also didn't find C obvious.

It was leaked on internet. I wasn't able to do it, but I guess I was close

Yeah, saw that and gave up on implementing it.

You can try to rotate the grid 4 times and each time looking for one direction only.

My implementation was also terrible. However, it is my own fault. My casework started with two cases and didn't look that bad. Then two turned into four, and four turned into eight. At this point, I was too invested to turn back.

Looking back, there were multiple ways to eliminate a big chunk of the casework. You can take this opportunity to learn useful techniques and avoid extensive casework in the future. Or you can shift the blame to the setters and fall for the same mistake next time. The choice is yours.

great passion...

Shit, that is an essential hint. I still know how it will help me to reduce the time complexity but alleast now I have something to think lol

Just use hashing for checking if the string after doing a given operation is equal to 000111000111... or 111000111000111... You have to do like rolling hashing or something like that in order to get the hash of the transformed string at each step

Just brute-froced with hashes, I hope they won't FST :p

You notice final string should be K size blocks like 111000111(for k = 3). Then you notice transformation is taking prefix, reversing it and addind to the end of the string. Then it is kinda obvious prefix must end on the first symbol where something gets wrong(because else this "wrong" part will get between good blocks and we can't do anything about it). Also if you see a block with size greater than k you should leave k last symbols of it and take the rest.

Trick is to release that only one possible location can be flipped to create a k-complete string.

find the first group whose size is $$$!= \ k$$$
if group $$$size$$$ is $$$< \ k$$$ consider the group in operation
if group $$$size$$$ is $$$< \ k$$$ consider first $$$k \ - \ size$$$ elements of group in operation
then check if the resulting string is good

we can notice that if triangle is correct and $$$(x, y)$$$ is one of point of this triangle, than $$$(x + \frac{d}{2}, y + \frac{d}{2})$$$ also belong to this triangle. now all we need is to check that third point exist.

Transform cords by [x-y][x+y]. Now distance function is max of dX and dY. Aka the square. Then for point A with cords [X, Y] point B could be [X, Y + D] and C [X + D, (anything between Y and Y + D)]. Same with minus and Y cord.