ok

Let me try my luck as well

You deserve it. <3

why not

As a tester wasn't it your responsibility to check the bug in B one correctly.. plaese correct me if I am wrong.. I think that is real job of tester is to check there are no bugs in the preparation of the round (intended solution, validator, checker, interactor, etc) rather than asking for contributions..

This is an OP way to get contribution.

No.. Just like other global rounds.

The only ones that possibly lose something more in combined rounds are div1 people that get beat by div2.

Let's not think about rating but exprience and learning something

Yes

well same to you xD

:ghosthug:

I don't understand on which basis you decide which one is trusted and which one is untrusted, which one is interesting and which one is boring?

When you use 100% brain.

Wrong answer on test case 300*

we missed "that's stupid!" in your comment! that's rare! :v

Actually not sure about that. Considering that A and B are just one minute tasks, we have a round with 7 problems. With decent problem difficulties it looks like ordinary Div 1 with 5 problems, except that for the hardest task you can choose between 3.

I'm curious to know how do you approach the situation when you are stuck on a hard problem. Thanks!

So now after seeing the actual problem set, where do you draw the line for "easy tasks" for today? A-G? Or do you see it differently?

And what do you think about the problem set in general?

I guess this contest favoured you a lot. Lol

It varies a lot but I think authors try to keep the consistency of first 6 problems the same as div2, ie combined B is around the level of div2B

i hope distribution look like div.3(2)+div2(4)+div1(3) kind of distribution, otherwise it seems very unlikely to have contest of 9 problems.

come on growup man!

Done Bro!! Enjoy. Wish us good luck!

*negative

Yeah,It is rated for every Division.

500 and 1000 respectively, the post is updated with the score distribution now.

Ideally, we are looking for fulltime candidates, but internships are also possible. Regarding the number of candidates: we don't have a fixed number, if there are 2 very strong candidates who both fit, I would imagine we will take both, but hard to say upper limits and it depends on the candidates.

And What is your requirement for these positions?

We test cognitive abilities, problem-solving skillz, we gonna ask about previous experience, education (are you from STEM or not?, bachelor / master / phd?) but we don't have "must know language A, library B" requirements. Although for a more senior position we gonna expect you to know some specific stuff

why not?

Its rated.

I feel like this round is around Div 0.5

Don't worry 4000 is gonna get a huge drop after the system tests. Lot of masters and GM are also failing system tests for B, this indicates that pretests were very weak.

Same here couldn't solve B, I was trying very hard but nothing happened, very ashamed.

WA11 for me :)

Lol you atleast got B, I gave up on that and started doing D. And didn't get it. Now Imma go cry in a corner. Thought I'll become CM today. Predictor says I'm gonna lose 100. F.

#MeToo

Very briefly,

For components with size 4,2,1, first merge 1 with one of element in 4

[4,2,1] ==> [3,2,2]

then merge the two 2s

Basically you separate into powers of 2 and greedily merge them.

So 7 decomposes into [1,2,3,4], [5,6], [7]. It's pretty easy to get all of those to be equal to each other. Now, merge [4,7] to get [1,2,3], [5,6] and [4,7] as your buckets. Then, merge [5,4] adn [6,7], so you have [1,2,3], [4,5,6,7] as your buckets, which is at most 2 distinct elements, as desired.

Notice that if n = 2^k, there will always be a way to make n elements be identical. In general case, we can make elements in [1..2^k] identical and do it one more time with [n-2^k+1, n], k is the largest number that 2^k <= n

As for me, I couldn't understand E within 20 minutes) So I just skipped it.
And F is just seems pretty easy to spend more time on it (and to solve it)

You can make $$$2^k$$$ elements equal in $$$k 2^k$$$ operations, by first making the first $$$2^{k-1}$$$ elements and last $$$2^{k-1}$$$ elements equal, then applying the operation to the first element of both sides, then the second element of both sides, and so on.

To have at most two values, just take the largest k such that $$$2^k \leq n$$$, and apply the above process to the first $$$2^k$$$ and last $$$2^k$$$ elements.

Let's define $$$p$$$ as the largest power of $$$2$$$ not greater than $$$n$$$. Now merge $$$[1,p]$$$, and merge $$$[n-p+1,n]$$$.

When I tested the round there was problem C in place of B.

It was added after the testers feedback,so testers are not aware of it.

greedy solution worked for me

For each (robber, projector) pair calculate pair (upMoves, rightMoves). Now sort this list n * m pairs and answer is some prefix of robbers moving up and suffix of robbers moving right. This can be calculated with suffix maximums in O(n * m).

You can do 5 moves right and 1 move up.

I did binary Search

Binary search for each time T and if the position of x > position of y increase T else decrease it..!! ~~~~~

long double lo=0,hi=l,mid;
while(hi-lo > eps){
    mid= lo + (hi-lo)/2;
    long double x=diff(mid,l,a);
    if(x>=0){
       lo=mid;
    }
    else{
       hi=mid;
    }
}

Here diff() function tells the difference of positions (x-y)

Hint: calculate a vector {c[i]-a[j],d[i]-b[j]} then sort it.

In problem D sample input 4, you have to make 5 right moves to save (7,0) robber from (11,5) light and you have to make 2 up moves to hide (3,5) from (6,6), so the minimum answer should be 7, not 6. I have read problem statement multiple times. There will 5 moves when we increase ai of all robbers and 2 moves where we increase bi to same all of them. I can't understand where I am wrong.

wanna see a hard D...

Binary search across time and calculate position for each car, stop when the difference in position is within 1e-6

Let's say initial positions are 0 and L. Now, assume that the first car will reach the flag closest to it before car 2 reaches the flag closest to it (car 2). So, calculate that time and add it to answer. Also, update the positions of both the cars accordingly, by p1 += time * v1 and p2 -= time * v2 (note time will be same).

Do this till you see that there are no flags between them. Then say x is the position where they meet. Solve (x — p1) / v1 = (p2 — x) / v2 (this says that they travel for same time), which gives x = (p2 — p1) / (v1 + v2). Now calculate time and add it to answer.

More like arrayforces in A and B

For C, you can do a binary search on time required for both to meet.

Same. Didn't initialize variables in G submitted at 2.48, got wa on pretest 1, fixed and contest over in selecting file to upload. ;_;
Edit: AC in practice but after fixing a bug. ;_;

Actually E and F (and maybe above) are insanely good. Just poor preparation for cakewalk problems.

Why nobody hacks my B?

Something like this

1
5 2
1 1 1 1 1

Hint: calculate a vector {c[i]-a[j],d[i]-b[j]} then sort it.

Think of a graph where you have one node per each set and one node per each number occuring.

Can you explain your solution?

I solved with just physics

By just physics you mean relative velocity?

I got the equation, but decided writing BS is easier than solving the equation :P