I hope so )

Да :) P.s. Антей — это школа такая, все совпадения с античными персонажами случайны :)

Testing was a breeze. Thanks a lot guys.

Казанский Национальный Университет?

I am really sorry for the level of my English, if you read my bad-english post attentively, you will find, that english problems are not translated by me.

problem set is level of English? do you knoving engrish?

А откуда Вы знаете, какие задачи давал Я, а какие Сережа? :) Может быть, Сереже далеко до Сережи? П.С. увы, это мой первый раунд, и я не такой сильный алгоритмист, я всего лишь пытаюсь. Спасибо за участие

"двумерная динамика" d[i,0..1] — ответ на задачу для первых i кроликов, 0 если не кормили левого соседа i-го(переход выполнен с +a[i]), 1 если кормили(переход выполнен с +b[i]).

DP problem: DP[i][j] 1<=i<=n 0<=j<=1

DP[i][0]=the answer of the first i hares when before i,feed i-1;

DP[i][1]=the answer of the first i hares when before i-1,feed i;

final answer will be max(DP[n][0],DP[n][1]); sorry for bad english.

I'm still not sure if this solution will be accepted, but here's a simple one:

State is the current hare and a flag whether the last hare that was processed was fed before the current one: 1 if it was fed before, 0 if not. So if the last hare was fed before the current one the value for this state is max(b[current]+state(current+1, 1), c[current]+state(current+1, 0)). Otherwise it's max(a[current]+state(current+1, 1), b[current]+state(current+1, 0)). The reason why this works (I hope) is because the only thing you care is whether the current hare was fed before the current-1 and whether it was fed before the current+1.

Also I found a bit tricky that the sequence of operations doesn't necessarily end in 0, so you can do whatever you want with the last numbers.

I am one of the victim :( .

oh my god...i missed that... so sad......

I missed the exact same point!

Но решается по-другому)

It was Dima's problem)

http://codeforces.net/blog/entry/9326#comment-149365
сверху уже отвечали на такой вопрос

Важно лишь для каждой пары соседей кто первее поел, будем рисовать стрелочку из А в Б, если они соседи и А поел первее Б (любое расположение стрелочек коректно). а[i]-=b[i] c[i]-=b[i] (b[i] будем считать нулем, в конце прибавляем сумму b[i]) . Дальше ДП ans[k] — ответ для зайцев 0, 1...k.

The input of pretest 3 is clearly seen:

2 0 1

So, I advice you to check it again more carefully.

yeah i figured it out, nothings wrong

Not really, it would have to be at least 3*10^5 if the answer was yes, but it's no.

no, the problem only guaranteed total length of all words doesn't exceed 10^5 which is valid for that test, and that doesn't mean it also guaranteed all words + '<3' also will not exceed 10^5

Not really. It does not say that the text message will contain all the read words.

the only way there is a self-intersection is, when there are 4 points of the form i j i j, meaning that a semicircle goes from point 1 to point 3, and also another goes from point 2 to point 4.

There are four cases, handle them and you get AC:

let A,B be the last two points entered, and C,D be each pair entered before. These patterns are the only one producing entanglement:

ACBD BCAD CADB CBDA

(Note that if you have XYZ entered before, you handle the pair (x,y), then (y,z), etc)

it is easy to check when they do not intersect, so you have [a1,b1] and [a2,b2].

they do not intesect if b1<a2 or b2<a1- the case when EXAMPLE: E1:([1,4] and [5,10]) and
E2:([5,10] and [1,4]) and when one interval is subinterval of the other so: (a1>=a2 && b1<=b2) || (a2>=a1 && b2<=b1) example:
E1:([4,7] AND [1, 8] ) E2: [1,8] and [4,7]) so you check every connected pair of points.

Thank you, everyone! :) I got it! Consider x1, x2, x3, ... , xn. Make pairs (A,B) for all x(i),x(i+1) such that:

A=min[x(i) and x(i+1)] and B=max[x(i),x(i+1)]

Now, consider 2 segments (P,Q) and (R,S).

P ---------------------- Q

'R' can lie either within PQ or outside it.

1. If R lies outside,

R---P--------------Q

S can either lie outside or inside. If S lies inside PQ, then there is an intersection, else there is not.

R---P------S-------Q (intersection) R-S-P--------------Q or R---P--------------Q---S (no intersection)

1. If R lies inside PQ,

P -------------R-----Q

If S lies Outside PQ, then there is an intersection, else there is not.

P -------------R-----Q---------S (intersection) P -------------R--S--Q ( no intersection)

So, it is just 2 cases we have to check, if we make a min/max pair.

What's the problem, then? The sample clearly explains that question. That's what the samples are for — clarifying the problem statement (and testing the program, of course).

And it kind of makes sense, if you imagine the happiness values given per number of full adjacent hares, which is at most 1 for the border hares.

Yes, it involves Euler tours on an undirected graph.

First, vertices: for each possible K you can extract a graph of which points were directly visited: Assuming the upper-leftmost point is at (sx,sy) then exactly all points of the form (sx+i*k, sy+j*k) are intermediate positions for the victim.

The remaining '1's indicate moves from one intermediate point to another, so if there is an unbroken line of points between (x,y) and (x+k,y) or (k,y+k) then an undirected edge should be added between them.

A valid sequence of kicks corresponds to an Eulerian path through this graph, so you just need to apply the standard algorithm for verifying that one exists.

Then Dima inserts a random number of small English characters, digits, signs "more" and "less" into any places of the message.

costed me 3 WAs, but finally ACed

Так именно что пересечений нет. "В единственной строке выведите «yes» (без кавычек), если линия имеет самопересечения. В противном случае выведите «no» (без кавычек)."

Спасибо :) Очень приятно!

For: 3 i love you <<<<3333iloveyou

Your code outputs yes while it should output no. The problem with you code is that it does not check if the characters are in the right order i.e (<3word1<3word2...), it just checks if they are there.

It's something similar to this: If your program would check if two words are the same, the words "love" and "elov" would be the same.

In Java a + b is slow for strings (copies both a and b), you have to use StringBuilder.append() instead.