Каждому члену команды

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Тренировка сдвинута, спасибо тренерам МИСиС.

No hacks. ACM ICPC rules

Недавно был переход какого-то региона РФ в другой часовой пояс вроде, но я как клятый москаль не следил за деталями конечно)

Исправил, спасибо.

Fixed now. Thanks.

results of Buuble cup onsite will be revealed after the competition.

No matter

Well, I promised a simple rage post but instead I tried to put my Bubble Cup experience in a bigger picture.

Difficulty is that both divisions can participate. If you competed in last year's mirror, than i would say that difficulty is the same

Click Here

да

Yeah :D but as it is online mirror, so there is no way of being rated :D

you have to reverse the given sequence at first. then multiply every term with non reverse one and at lust add all the product. thats the answer.

You just need to use magic numbers!

20534967

You can transform the problem in to :

Given some intervals, each interval cover some points and has a weight. Select a subset of the intervals such that the sum of weight is maximized and there isn't exist any point being covered more than x times.

And this is a classic min cost flow problem.

Node i represent position i in the crossword. Let the interval as from a to b with w weight.

So you should calculate each interval. If crossword[i .. j] equals the word, add interval (i, j + 1, points of the word).

First add edge with x capacity and 0 cost from source to node 0 and from node n to sink. Then add edge with infinite capacity and 0 cost from i to i + 1 for every i < n. Then for each interval add edge with 1 capacity and -w cost from a to b.

The answer is the negative value of min cost flow from source to sink.

This is correct because there is only x flow present. So it restricted each position will not be covered more than x times. And the flow tends to go to the negative edge, so it will give out the minimum value.

To win the game, the xor of the pile sizes must not be zero. So we find the probability that xor of pile sizes is zero and subtract it from 1.

Consider dp[i][j] which denotes the probability to get xor j for first i piles. The straightforward recurrence is:

dp[i][j]=sum dp[i-1][j^k]*p[k] from k=0 to 100

where ^ is the bitwise xor operator.

But this is not good enough for the problem. We can do better. To calculate dp[i][j], divide i into two piles of size i/2. Then,

dp[i][j^k]+=dp[i/2][j]*dp[i/2][k]

Above is true when i is even. You need to modify it a little when i is odd.

As there is always a solution and the size of it doesn't matter,

Do a normal DFS traversal , print each vertex you visit and change the color of it . Once you finish from a child print the current vertex again and change it's color because you will back to it , and if this child's color is pink then visit it again (without it's sub tree)

you have to stop at node 1 or one of it's children , after visiting the last child of node 1 if color[1] is pink then back to it otherwise stop at this child .

this code may help : 20538605

Your p[100] array is too short, iterating k up to 128 causes undefined behavior.

The compiler does warn you about this. Use and read the warnings.

wait for editorial, all explanations will be there

Because pretty much anything in this problem works so you can either spend hours to solve it properly, which in this case is a solution with guaranteed probability, ir just try to submit anything and get AC.

Because the expected number of "crossing" edges in a random team and conference distribution is equal to .