？？？

很蔡盛liang...

Durant ＆ lrving. will. ak. this. game. .

No.

I can promise no anime pictures in this round!

The contest will start at the time when I arrive home after getting off work :(

Does that sound harsh??. It might but that's the truth. 300 upvotes to 700 downvotes. That's what happened with arsijo's round.

It happened a week back

That's because Hello 2019's contribution has been halved.

cn dota best dota

Thank you very much!

Oh, yes. You are right. I am sorry, I didn't notice it:(

or for Russian(

You are experienced, so the gap will not fact i guess :)

qianqian ddw!

(〃'▽'〃)

Chtholly is the best!

(It's an old picture, the Chinese words are: Normal people's CF, aces' (or maestros') CF, my CF)

You registered 16 months ago.

Here

Instead of saying "I'm new here", just ask "Is it rated?"

Your wish is my command!

Seriously how does this keep happening? How is it possible that every other round turns into a speed contest like this one? None of the easy problems were even interesting this time, and E/F don't seem fun either :/

Yup, I missed that and realised it quite late.

Such as?

• For each y remember all x's
• Iterate through all y from biggest to lowest
• Now you are at y'. Add all current x's to your segment tree. (If ST[x] is already 1 then skip it).
• You know count of all points (which y >= y') at any segment. It is easy to calculate the number of rectangles that include your point(curX,y')
• num = (cntPoints(prevX + 1, curX - 1) + 1) * (cntPoints(curX+1,1e9)+1), where cntPoints(l,r) — number of different x's at segment [l,r]

store all special items in an array, and keep one variable li: the last index of the page you are viewing. initially it is k. now run a while loop and in each iteration remove all elements from back of array that are less than li, if you removed t elements, li+=t; now flip pages till arr.back() (last element in array) is less greater than li (flipping page is li+=k) .

edit (thanks @spookywooky) : code https://codeforces.me/contest/1191/submission/56918688 note: idk if it will get hacked, but I dont think there are any corder cases

use the following function: page(x,deleted) = (x-deleted)/k if ((x-deleted)%k != 0) and (x-deleted)/k-1 otherwise.

Sort all special items.

After this just use this function to delete as many as have the same page as the first item, then actualize deleted. And do it again untill all are deleted.

a map in C++ should do the job, plus the count of removed elements till now :D

It is not flipping but setting an interval to some value. So, this is a valid move for A 0011 -> 0011.

players don't flip the numbers, but they assign all 0s or all 1s to them, so the first player can just assign 0 to the first number, so it remains 0011.

You can flip a card $$$0$$$ to $$$0$$$. Not necessarily changed.

Yeah, after wasting a few minutes I realized that it's painting rather than flipping. I asked a "question" saying the statement is confusing and flipping means changing 0 to 1 and 1 to 0. Meh.

The first player can play to not change the state at all. Change the first card (which is already 0) to 0.

Hint: No one can get the opponent into a situation that his opponent can't make a move. (Except when n=1)

I think that the final sequence is 0 1 2 3 .... for sorted numbers before the very final move of the game, so with some boundry cases, the major part is reducing the current sorted array to that state

Solution (Div2D / Div1B)

• If first player cannot make any of possible first move, first player loses. To check whether it is yes or no, just brute force what pile will the first player chooses. To reduce time to $$$O(N log N)$$$, you can use a data structure: for example, map.
• Otherwise, the final state of the number of stones for each pile will be {$$$0, 1, 2$$$}, {$$$1, 2, 0$$$}, {$$$5, 4, 2, 3, 9, 6, 1, 8, 0, 7$$$} or something, that appears number from $$$0$$$ to $$$N-1$$$ exactly once. So, you can determine the number of steps. If odd, the first player wins. If even, the second player wins.

• $$$a = $$${$$$0, 0, 1$$$} : Second player wins because first player cannot make first move.
• $$$a = $$${$$$1, 2, 2$$$} : Second player wins because first player cannot make first move.
• $$$a = $$${$$$1, 3, 3, 3$$$} : Second player wins because first player cannot make first move.
• $$$a = $$${$$$2, 5, 5$$$} : First player wins because the number of steps is $$$(2 + 5 + 5) - (0 + 1 + 2) = 9$$$, and it is odd.
• $$$a = $$${$$$1, 2, 4$$$} : Second player wins because the number of steps is $$$(1 + 2 + 4) - (0 + 1 + 2) = 4$$$, and it is even.
• $$$a = $$${$$$8, 6, 9, 1, 2, 0$$$} : First player wins because the number of steps is $$$(8 + 6 + 9 + 1 + 2 + 0) - (0 + 1 + 2 + 3 + 4 + 5) = 11$$$, and it is odd.

In div2,this situation also appears.

The solved of porblem D was 1229 but the solved of E and F were both less than 100.I think that we should prepare problems that can differentiate the coders whose ranks are between 100 and 1500.

Despite this,the contest was really nice!

Overflow shouldn't change the remainder mod 2. How can you hack it?

How do you solve DIV2 D?

Bonus 2: two consecutive game problems with different things to print. In case you already remembered what means what.

Bonus 3: if something then print quailty...

B Print "sjfnb" (without quotes) if Tokitsukaze will win

C: print "tokitsukaze" (without quotes) if Tokitsukaze will win

Just code without reading about what to print if the first player wins, then test the examples to get what to print.

#define first sjfnb

#define second cslnb

Well, geometry problems on integers and with thinking are fine. This one was pleasant in terms of precision issues but still quite standard. What I wish for sure is that they got rid of details like (0, 0) point or two points in the same position. It's so easy to make a problem slightly nicer.

Masochists

If having to use single acos makes your life miserable then it is a bad sign. I hastily copypasted my thousand lines geometry library but ended up using just a definition of a point, function norm and wrapper for calling atan2 (both are oneliners). I can ask the same question about neverending problems with data structures. I actually enjoy geometry problem and that one was super pleasant even for people that hate geo.

Hope you won't fst :>

Good luck to you

Same here! Barely clutched it out.

In this case, wich one is the winner?

Like this :

4

1 1 4 4 or 1 1 1 4

the answer is cslnb because it's impossible to remove any stone.

i hacked my own solution after locked but no one else). it Must have to fail system tests(just waiting for failation)

I agree with you for Div-2 ABC but I find D to be interesting, but then again there is difference between your ability and mine.

I agree with the second and fourth ability, but it is hard for me to agree with first and third ability(especially the first one). Those are also a part of solving problems.

Last two problems seem fine, but probably most contestants spent too much time on dealing with corner cases in other problems to get to E and F :(

I think the problems are good, the bad part is output format, pretests, the order of problems and so on.

The main point I disagree with this round is that when I cannot solve some problems, I cannot solve because of bugs or corner cases. Usually, for most of the other rounds, I cannot solve because I cannot think of a better solution, or I cannot think of a way to implement something.

This is why I dislike this contest :(