Let's denote the sum of numbers on the upper halves of pieces as *s*_{1}, and the sum on the lower halves — *s*_{2}. If this sums are even, than the answer is obviously 0. Note, that if the numbers on both halves of piece have the same parity, than parity of *s*_{1} and *s*_{2} won't change after rotation this piece. If the numbers on halves have different parity, than parities of both *s*_{1} and *s*_{2} will change after rotation. Therefore, if *s*_{1} and *s*_{2} have different parities, than the answer is - 1. If both *s*_{1} and *s*_{2} are odd, than we should check, if there is a piece with numbers of different parities. If so, the answer is 1, otherwise, the answer is - 1.

Let's say for shortness, that we put numbers, that are painted on cubes, in piles, instead of cubes themselves.

Note, that the answer is the product of *c*_{1}... *c*_{2}, where *c*_{i} is the number of different numbers in the *i*-th pile. Let's consider that all numbers are different. In this case the answer is *n*^{2}. Now, let's suppose that we have two equal numbers and all the other are different. Then, if we put them in different piles, the answer will be *n*^{2}, but if we put them in one — *n*·(*n* - 1). Obviously, the first case give greater product.

Thinking in similar manner, you can conclude, that we should do the following. Take numbers, that appear more than once, and put one of them in the first pile, one of them in the second pile and the other put aside. After that, divide the numbers, that appears once, in two equal part and put the first part in the first pile and second part in the second pile. Finally, take the numbers, that we put aside, and separate them in two pile in any kind.

Let's see on the highest bit of *m*. If it equals to zero, than for any there is a zero on the (*n* - 1)-th position, so *a*_{n - 1} doesn't affect the answer and we can put it aside and find the answer for smaller number of elements.

If the highest bit of *m* equals to 1, than *a*_{n - 1} for some *x* will present in *f*(*x*), but for some will not. Let's consider such *x*, that *a*{*n* - 1} will present in *f*(*x*) with zero coefficient. It is obvious that . In this case *f*(*x*) will have maximum value when *x* = 2^{n - 1} - 1. Try to update the answer by this value.

Now we should analyze the case, when , find the maximum value of *f*(*x*) for all such *x* and try to update the answer by this value. Let's note, that in all such *x* there is 1 in (*n* - 1)-th position. Therefore we can find the maximum value of *f*(*y*) for all and add *a*_{n - 1} to it.

Note that if there are some girls in the begining of the line, they will never move. So let's remove them and will consider that the first schoolchildren in the line is a boy. Also note, the relative order of the girls doesn't change. Let's calculate for each girl such moment of time *t*_{i}, that after it she won't move forever. Note, that for *i*-th girl *t*_{i} ≥ *t*_{i - 1}. Let's calculate *t*_{i} in order from left to right. Let's denote *y*_{i} is the position in the line where *i*-th girl will stop, ans *x*_{i} is her current position. Therefore it is needed *x*_{i} - *y*_{i} second for girl to reach her finish position. So if *x*_{i} - *y*_{i} > *t*_{i - 1}, then *t*_{i} = *x*_{i} - *y*_{i}. Let's manage the case when *x*_{i} - *y*_{i} ≤ *t*_{i}. The girl with number (*i* - 1) will be on *y*_{i}-th position by *t*_{i - 1}-th second, so *t*_{i} ≥ *t*_{i - 1} + 1. Let's consider such moment of time *p*, when *i*-th girl stand right after (*i* - 1)-th, but not on *y*_{i}-th position. After that, in (*p* + 1)-th moment of time (*i* - 1)-th girl and the boy standing in front of her will swap their positions, but *i*-th girl will save her position. Then since *p* + 2-th second till *t*_{i - 1} both girls will change their positions. Finally, at (*t*_{i - 1} + 1) - *th* second *i*-th girl will occupy her position. Therefore, *t*_{i} = *t*_{i - 1} + 1 in this case.

Let's divide our graph on chains. Denote chain as the maximal sequence of the edges, which have the same direction. If there are more than 2 edges in each chain, then the answer is the number of such chains.

If there is a chain containing only one edge (*u*, *v*), then brute vertex, which we will take in maximal antichain (also consider the case, when we take none of them). Let's suppose we brute the vertex *v*. After that we put aside this vertex and all vertices, which is comparable with *v*. In received graph we can find the size of the maximal antichain by uning dynamic programming with *O*(*n*) time complexity. Let's show how to do this.

Write all remaining vertices in line and renumerate them from left to right by the numbers from 1 to *k*. After that we are going to calculate *d*_{i} — the size of the maximal antichain among vertices with numbers *j* ≥ *i*. So if *i* > *k*, then *d*_{i} = 0. In other case we can skip *i*-th vertex and try to update *d*_{i} by the value of *d*_{i + 1}. Also we can try to take *i*-th vertex to the answer. In this case we should skip all vertices, that are reachable from *i*-th, or the vertices, from which we can reach *i*-th, take the answer from the next vertex, add 1 to it and try to update the answer.

translation to English?

it will come soon)

In fact we have the strict inequality: for every ti > ti - 1

problem B can someone tell me why this code gives wrong answer ? the expected answer is so close !! http://mirror.codeforces.com/contest/353/submission/4736607

Read the comments here and you'll understand your mistake.

Good alanysis for problem D

I solve the problem E using Matching

That's too slow. There is no (not even closely) linear algorithm for matching, so you'd TLE. But in this case, the graph we're given is composed purely of a cycle of cliques connected in some vertices, on which a greedy idea works well.

But he did. Look at this nice solution.

It didn't TLE probably because of special structure of the graph.

During the contest, I did it using matching too, via Hopcroft-Karp algorithm, and some friend of mine did the same. Here are our submissions:

4735010

4735373

My question is, why this doesn't TLE? Why this graph particular structure allow that? In Wikipedia I could only find that

"for random graphs it runs in near-linear time.". Are the tests weak?I actually really like this round especially Problem D. It took me a before I could build up all the ideas to solve it. My track of thoughts is like:

1. The answer is the number of seconds needed to move the last girl to the finish place.

2. The number of seconds needed for

`girl i`

is at least the number of seconds needed for`girl i - 1`

+ 1.3. The number of seconds needed for

`girl i`

is at least the distance between the initial position and the finish position.4. Each

`girl i`

will move as soon as she can, so she will be right at the back of the`girl i - 1`

or take at most`dist(i)`

seconds to follow the`girl i - 1`

.so we can calculate the

`T(i) = max(T(i-1) + 1, dist(i))`

and the answer is just`T(lastGirl)`

.how you computed dist[i]?

dist[i] is just the distance between the initial position and the finish position (You already know the finish position of each girl).

E can be solved by greedy, also in

O(N). Let's count the lengths (as number of vertices) of individual chains (counted cyclically).Every chain of length 3 adds 1 to the answer. That's because we could always choose a vertex from it that doesn't belong to any other chain instead of one of the side vertices that do belong to one, or instead of not choosing any vertex at all.

Now, we're left with either just a cyclic sequence of chains of length 2 (input 01010101...01 or reverse), for which the answer is the (length of the input string)/2, or some sequences of chains of length 2, each bordered by chains of length 3 on both sides. Greedy approach works here, too: no 2 chosen vertices from any such sequence can be adjacent, and we can't choose from the border. That means we can choose at most (length of such sequence-1)/2.

Was beaten hardly by problem B!! :( A great round though, I have learnt a lot!

I use a similar approach in probem D, i try to move the boys though:

for each boy, we know his initial and final position, thats the

minimumtime for this boy to move Actually it's the# of girls initially behind this boyfor each boy, if in the moving progress he has to stand and wait next boy to move, actually he can start in a later time so that he won's stop until arrive his final position, I call this the

start timeof the boyFor step 2 we can greedily precompute, and the ans is max(start time + # of girls behind) among all boys

Here's my code: 4742830

I use same basic idea, the difference is only on implementation, I didn't store that much information. here is my solution: 4735037

http://mirror.codeforces.com/contest/353/submission/4745339 this code is wa for "011001" but it is accepted.

I get wrong answer at Div2 D — Queue with this source, but I can't figure out why because of the large input. Could somebody give me a hint?