In Codeforces Round #485 the authors didn't include that line. We saw what happened:D

P.S. I stoled the joke from C137.

Probably 2.5 hours.

The announcement is correct, the Contests page will be updated soon.

Там участвовало 20 человек в сумме на три раунда, из которых большинство не топы, это очень маловероятно что ты случайно с кем-то обсуждал задачи.

Если хочется быть уверенным на 100%, то прочитай в начале раунда просто задачи в середине (C или D например), на предмет того знакомы они или нет, прежде чем что-то отправлять.

The successor of L :P

Probably envious because he succeeded against Kira and you didn´t, right? xD

Nobody cares about Peru-Denmark match ;)

Wanna watch this match ? Stop coding and start watching :) no one's gonna disturb you I think :)

It is a reasonable concern. However the problems were not visible to anybody, we actually tracked all the problem views on JavaBlitz (only logged in users could see the problems), and the total number of people who saw at least one problem across the two rounds is 19

We sent all of them to Mike, I assume they just won't have an option to register.

Those are the maximum scores you can get for each problem (if you solve each in 0:00 time). The score gets decreased as the time goes on, and the final score you get is the sum of the scores of problems you solved (and hacks). It's not directly relevant to your rating, as the rating is affected by your rank in the contest, not the absolute score you got.

Как написано здесь, для совмещенных Div. 1 + Div. 2 раундов распределение по дивизионам остается без изменения, то есть в Div. 1 редакцию раунда будут попадать участники от 1900 единиц рейтинга и выше, поэтому Div. 2 будут решать те, у кого рейтинг ниже 1900.

This contest is nothing but Eid bonus from codeforces for us. :)

ohh .. no .

i will die

what the joke is ?

div.1a will be easier than usual i think.

argentina vs iceland going on!!

Мы с Фетиском лучше знаем

It's same for Div.2 D and Div.2 E, which is Div.1 B and Div.1 C

same here :-(

awesome right?

then you should try E, its statement is most ambiguous

I still dont get how D works. Worked a solution based on the explanations. (Though got WA because of weak pretests even though there was 20, but after getting wa, found my mistake)

Anyway still dont know why i did what i did.

Maybe it said, there are random pair of points. Find the common number in the pairs that was given to both of them. If there are more than one common points but those 2 persons can still differentiate those then ans is 0 else -1.

Да.

There was a post a while back about what the best CP sites are, and for me this is why CodeForces is not top. I think this is regularly a problem with CF rounds, just more accentuated in this particular round (at least for Div 2).

Искать, где в решении присутствует неопределенное поведение.

Calculate C_i = how many numbers less than x from 1 to i. Now you have to calculate number of l, r such that C_r - C_l = k for each k. Do it with FFT.

kick

Just IF ELSE will DO

First I checked every vertex and the center point of the oblique square, if it is contained in the other square then the answer is YES.

Then I rotated the coordinates axes of 45 degrees ( https://en.wikipedia.org/wiki/Rotation_of_axes ) and did the opposite.

It probably isn't the best way

EDIT: Fuck yes it got accepted

Iterate through all integer point in the parallel square, and check if it is in the rotated square (can be done by calculating mahhattan distance to the center of the rotated square).

I tried this and got WA on pretest 10:

Let the square be ABCD, and we iterate over the points P in the other square. Calculate AB × BP, BC × CP, CD × DP and DA × AP; P is inside ABCD iff all of these complex cross products have the same sign, at which point print yes and exit.

Swap the squares and do that again. This was all inspired by 29.2 in the PDF in this blog post.

EDIT: Does not work because doesn't account for the case where the squares intersection region does not include any vertices of the squares. Same is true for my other post below :(

Also, calculating the areas of the four triangle formed by a point P and the sides of ABCD and comparing this with the area of ABCD should work.

Checking if a point is inside a rectangle/square is a standard problem ( check this https://www.geeksforgeeks.org/check-whether-given-point-lies-inside-rectangle-not/ ). Now for all x,y co-ordinates possible from -100 to 100, check if a point lies inside both the squares. If yes, they intersect.

My solution:

1) Mark all integer points on the edges of a second square.

2) Iterate through all points inside the first square, check if point is marked.

3) Mark points in the first square and check second square.

In Div-2C You can also use point in polygon algorithm to check whether any point from any square is inside or on the other square.

for all (x, y) iterate from -100 to 100 and see if any point is found in both square.

Complexity O( 200 * 200)

or u can find minX, maxX, minY, maxY and some if else's.

oh no T_T

Maybe karatsuba works too..?

(Although I failed to code it)

For each pair of each person, we will find out the list of pairs belonging to the other person, that two pairs only have one common number.

The answer will be:

• A digit from 1-9: only when all possible sets of 2 pairs have similar common number.

• "0": Both players can determine the number (since they know their own given pairs, while you lack such informations), but there are different digits that could be the answer (which you cannot decipher, since you lack the aforementioned info).

• "1": Other cases. Technically, at least one player cannot determine the common number.

The determination can be done by brute force. From a player's pair, iterate through all pairs of the other player. That pair won't cause ambiguity when either no pairs from the other player have exactly one common number with it, or every pair that does returns the same number.

Me. I was using cumulative sum after sorting. Thankfully pretests were good.

yes.

You can see it in "hacks" page.

It can't be TLE because you must have checked the constraints before submitting.

I had to read the statement 10 times to understand it. For me it was very confusing.

Omg, same as me.. I kept thinking for 40 minutes until I saw it.

Yeah, same. As always, reading turns out to be the hardest part of problem solving... .-.

The same mistake :(

To make coding easier?

stl map

I also didn't read that line and skipped it because C looked more promising. :(

Same here @asgar, didn't even know all pi's were different until i read your statement. Lost 10+ minutes there, plus the drain on every other question after that :'(

Yeah, coding is easier; sort from small to large powers, then iterate over the knights, each time tossing the previous knights' coins into a priority queue.

Without that condition it's much less clean to add things to the priority queue.

I did the same still got WA;(;(

You know, I made a lot of incorrect submissions (five, to be precise), and still got my rating increasing from ~1700 to ~1800. A particularly nice thing about CF is that it values correct submissions a lot, even if they are made after many wrong submissions.

My point is that you should try hard to solve as many problems as possible, even if you have already made some wrong submissions, and never ever give up!

It's okay. Given that it's "long time no see"

WTF!!!! I didn't realize that until I read this. FML...

Got 2 WAs because of this xD

I made 2 wrong submissions due to that :(

@Radewoosh xd

To make an enemy spaceship destroyed, one of our ships must stand at the middle point between them.

Technically, there are 40000 possible positions. But we're not iterating through all of them (It will guarantee a TLE verdict at large tests).

Let's iterate every ship from the left side, then with each ship iterate every ship from the right side (or you can choose the vice-versa order, doesn't matter).

Denote the left ship's y-coordinate as y_1i, the right ship's y-coordinate as y_2j, to destroy both of them one friendly ship must stand at the y-coordinate of (y_1i + y_2j) / 2 (for simplicity, you might consider only caring about the value (y_1i + y_2j)). Add this pair of enemy ships to the list of possible takedown pair if standing at the aforementioned y-coordinate.

Keep only the y-coordinates that if standing there, at least one pair of enemy ships will be destroyed. From this point, you can simply do a quadratic-time-complexity iteration to find out the answer (since the maximum number of valid coordinates cannot exceeds nm).

Total time complexity: O((nm)²).

Update: There are times when there exists exactly one y-coordinate that if standing there, a positive number of enemy ships will be destroyed (technically, every enemy ship from both sides would be destroyed). You can handle this case exclusively — the answer will always be n + m. Thanks Aeon for pointing that out (in his reply below). ;)

what is the correct answer? 2 4 4 or 2 4 6

All power levels are distinct. That means, your test is invalid.

Power of all the knights are distinct....see the input description please

All powers are distinct

That's one small thing that I think really made me believe I was supposed to print either -1, 0, or 1. Although I can't complain for not carefully reading the output section, the statement definitely could have been clearer.

This is maybe the hardest thing to do for an author. I know myself how hard it is to avoid words like that. That being said, of course they should try to watch out for that and it is a bad thing.

My advice is to read the output section anyway and not care much about wording like "you want to find this" in the middle of the statement. Authors often describe the story and only at the end they say what they really want.

I don't like,too. I use int to print ans (void write(int x)),so it failed system test 43. It make me have low rating.I think it's unfair.

Why there aren't large ans in pretests?

I do not agree that having templates make problems B and C easy