"6 задач и 2,5 часа" тоже))

OK, yesterday in the announcement it was mentioned that Delinur translated the problems, today it is written that Tatiana_S did the translation, I want my up-votes back, it's not my mistake :(

Solved A,B,C,D :3 Told you it's a beautiful round :D :D

although i was late a bit :3

I wonder if anyone is going to get raided by treat or trick while competing.

Anyways, Sunday round should be much more better, and I am pretty sure that many people are going to hang out on Hallo... Wait...

I agree, Sunday is the safest option.

I believe it's much better than the usual time for the Chinese:)

Only treat you will get is negative change

Canada cup was for DIV.1 and DIV.2.

I think it is just a coincidence that recently the problemsetters have sets of 6 problems :)

updated

I am not sure if this incorrect translation can be remedied at this point, but please consider any possible steps you may take in order to prevent quite a few people like myself being victims to errors in translation. (there are 4 people in my room who considered the phrase "English vowels" to only mean 'A', 'E', 'I', 'O', 'U')

As it's explicitly written in the statement which characters are vowels as you said, there's no point of arguing about it, everyone should follow the statement in order to solve the problem. Also https://en.oxforddictionaries.com/explore/is-the-letter-y-a-vowel-or-a-consonant.

Why so butthurt dude?

Dang, I knew I'd seen a similar problem before! I guess that's why you should always upsolve...

I use greedy and prefix sums!

You should choose m segments such that i-th segments sum is equal to b[i].

Consider first element of ending array. It must be formed by first elements of original array (else where does first monster go?). So it suffices to solve the problem in the case of ending state a single monster (then apply this to each range by precalculating partial sums). This can be done by identifying a monster of maximum size that is bigger than at least one of its neighbors (if none exists, then solution does not exist), and having it eat everything.

Notice that you can only convert the initial array into arrays which contain elements which are contiguous sums of elements of the first array. Now, in order to generate the procedure, all you have to do is in every subsequence, find the largest element and go left and right to consume all the others (as it will only become larger, it will not create problems). You also have to take care of cases when the largest cannot go either left or right.

Тест не подходил под условия (вероятнее всего)

It will be min(a,b,c)/2

D is enough to maintain hash of (pairs -> list of third dimension) for all 3 possible pairs (first sort the dimensions for each brick).

D

Imagine the shape described by the problem. It becomes obvious that the sphere is limited by the edge that has minimum length. You have block A that you are considering. To use A and B (another block) the value of some area must be the same (both in size of the area and shape). But there's something more: if you put A and B together, you are basically increasing the length of the edge that you didn't use to get the area. So this length that you are increasing needs to be the minimum length of block A and B, otherwise some length you used on the common intersection is already limiting the volume of the sphere and you already have the maximum volume you could get using this union. The max volume in this case will be limited by the minimum between the sum of the edges that A and B aren't using and the minimum edge of the intersecting area (because if the sum is greater than this, now this is the limiting factor of the volume of the sphere).

Now, to get the answer from this you have a vector of (max_area, minimum side of area, side not used in area) and sort it. The blocks that you can glue together are next to each other, so you can easily consider all the unions.

edit: now that I think about it, you just need the dimensions sorted and check if the 2 maximum sides match, but the thinking process is the same.

For problem D: Basic observation is that if you have parallelepiped (lol I copyed it from statement :D) of size a * b * c, then the radius of the inner sphere is min(a, b, c) / 2. Also you can glue two parallelepipeds together only and only if they have two common edges of the same length. After that you should use a data structure like std::map<pair<int,int>, pair<int,int>>. In this DS you keep the maximum possible 3rd edge length for given a and b edge lengths, after filling this DS you can use a simple loop to get the answer. Complexity is

For a rectangular parallelepiped of dimensions a×b×c, the radius of the inscribed sphere is min(a, b, c) / 2.

The jumping root is regular, consider current step is right, so we jump right until reaching the first left step(now all the steps we have visited is left), then we jump left until reaching the first right(now all the steps we have visited is right), and so on until jump out, the answer has two parts, here consider the left part(distance we go at left side of current step), (curIndex — leftIndex1) + (curIndex — leftIndex2) + ...= totLeftNum*curIndex — (leftIndex1 + leftIndex2 + ...)

To calculate the answer in O(nlog(n)), precalculate several arrays, lcntUp(number of up steps from left), lsumUp(the sum of every up index from left) and rcntDown, rSumDown. for every step there are 4 cases.

Sorry for my poor English.

Consider position i. Let char be 'U' there. The movement will be oscillatory. You will move to first 'D' above i, then to first 'U' below i, then to second 'D' above i, then to second 'U' below i and so on. I advise just running through an example on paper to see this. So, if you know the count of Us below and Ds above, you get to know whether you will exit from top or bottom. Then just calculate the total distance travelled. The case for when str[i] = 'D' is analogous. So each i can be dealt with in O(1). :)

Code

The variables are global, so initialized to 0

WOW , how could that pass the pretests ?
it should completely change the answer

However my O(n^2) solution submitted and the result is TLE in pretest 7.

yup,i agree with u.

I agree. It is more of a problem of one's familiarity with English language.

I think that on the one hand, the current hacking system is not fair. According to who is in your room, you may hack between 0 and n >> 1 solutions. Getting 0 hacking point or 1000 hacking points may be the result of your luck, more than of your hacking skills.

On the other hand, I'm not sure letting everyone hack anyone would be a great idea. Indeed, we may see one day someone winning the contest with one problem solved and 70 hacks.