why contribution on profile just showing +38 even when you just got +362 on this comment only.

Agnimandur orz

One of these problems is interactive.

Oops, pressed downvote by accident

don't try to earn contribution begging for upvotes unless you're a tester, as you can see people don't like it just a useful advice)

Will the universe balance my negative contribution with positive rating deltas? ;-;

Can we testers expect for t-shirts ??

here

You did it orz

Best Of Luck

Why does this makes me feel scared?

What's wrong in having a contest in weekends like do you prefer to come back from work/college/school and participate or wake up when you want drink a cup of tea do some yoga and then participate in the contest wouldn't you perform better in that case?

we dont care about you

Be like him

https://www.youtube.com/watch?v=aEyJRW2L7Gw

yes

yes bro

Am I the only one watching Italian Grand Prix ? :)

The rounds are open and rated for everybody.

that can help in cheating

Do not make any submission. You are safe then :)

It didn't :D

still, I don't understand what it's said.

Best end points b/w each $$$a_{i}, a_{i + 1}$$$ (till where $$$a_i$$$ goes right, and till where $$$a_{i + 1}$$$ goes left) is independent of any other end points, it depends only on whether points $$$a_{i}$$$ and $$$a_{i + 1}$$$ each go left and then right or vice versa.

So we can just use an $$$n \times 2$$$ dp for this state (position, left then right / right then left), and just use two pointer to calculate the optimal split point between $$$a_{i}, a_{i + 1}$$$ in $$$O(n + m logm)$$$ time. Removing ranges that already cover some $$$a_i$$$ makes the implementation easier since every range is now strictly between some $$$a_i$$$ and $$$a_{i + 1}$$$. (consider $$$a_0 = -INF$$$ and $$$a_{n + 1} = INF$$$).

Its always worth it to either leave a element alone or merge it with an adjacent one. Elements with mex 2 will always be alone and 0 and 1 should always be paired if they are adjacent. In that case, you just need to iterate through the indices and see if the previous one is the opposite of the current one and if it wasnt paired before to merge them.

some observations are:

1. column [1, 0] and [0, 1] are the best, and they shouldn't be combined with any other columns (it won't increase the MEX sum)
2. column [0, 0] is okay itself, but combined with [1, 1] is better.
3. column [1, 1] itself yields no MEX points, but with a [0, 0] they can contribute 2

so you can greedily scan the columns from left to right, if you encounter [1, 0] or [0, 1], just add 2 to your answer, if you encounter a [1, 1] ([0, 0], respectively) column, first examine whether the next column is [0, 0] ([1, 1], respectively)

Same here

Think about the composition of your tree into buds. The optimal solution is always taking all of the buds and putting them one on top of the other. Each time you do this, the number of leaves in total decreases by one. That way you just need to count the total ammount of leaves — the total ammount of buds+1.

That sounds cool. My approach to find the two disjoint edges with smallest weight was:

Let $$$(u, v)$$$ be the edge with smallest weight. Take the next two smallest edges connected to $$$u$$$ and the next two connected to $$$v$$$. Also take the smallest edge $$$(x, y)$$$ that is disjoint from $$$(u, v)$$$. Somehow I convinced myself that the answer will be a pair of these $$$6$$$ edges. The only thing remaining is to find $$$(u, v)$$$ and $$$(x, y)$$$ for each query. For that I threw everything in a horribly messy segment tree that got TLE, but I'm guessing there is a better way.

I don't understand, why do you need dynamic connectivity?

Resubmit solution with $$$100$$$ iterations was the right decision. Probability of accept is $$$(1 - (\frac{7}{8})^{iteration})^{test\underline{ }count}$$$. Problem have more than $$$750$$$ test. With $$$50$$$ iterations probability of accept is about $$$0.39$$$, with $$$100$$$ iterations is $$$0.998$$$.

find the position of the median "pos = ceil(n/2.0)". before this position, consider all elements to be zero. So we have to distribute the sum s among the leftover places "left = n-pos+1". The answer simply will be "sum/pos". Since, if the sum can not be divided equally among the leftover positions, the remainder will be added to the positions greater than pos. As we are finding the median the array should be sorted non decreasingly at all times.

Inspired by the n=7, s=17 test case (optimal distribution: 0, 0, 1, 4, 4, 4, 4), My strategy is to put the largest possible value M over $$$\frac{n}{2}$$$ times. (to reach the median place)

more formally, let's consider 2 cases:

1. n is odd, which means we should put M $$$\frac{n+1}{2}$$$ times
2. n is even, which (by the definition of median in this problem) means we should put M $$$\frac{n}2+1$$$ times

so the largest possible median value M can be derived:

1. $$$M\leq \frac{2s}{n+1}$$$, if n is odd
2. $$$M\leq \frac{2s}{n+2}$$$, if n is even

probably big overhead of recursion + overcomplicated + O(n*8) instead of O(n) + maybe cache misses

Frankly, it looks like div3 gradient and this is not particularly good. I don't complain though

Why would n*m*log(m) not work?

but n*m*log(m) is like O(10^6), it should totally pass...

According to me O(n*m*m) will also work on an average. Don't know whether I will fst or not.

My O(n*m²) code works in just 62 ms: 128625308

Please don't spread fud, you will scare the noobs :(

But if score for D2 was 1250,the total score of D would be equal to score of E, but E is much harder than D.

Its strange that youre the second person saying that pretests were weak because a solution that should pass is passing

$$$n, m \leq 300$$$ in a given test. $$$O(n \times m ^ 2)$$$ is equivalent to $$$O((n \cdot m) \times m)$$$, we know the first term is bounded by $$$10^5$$$ and the second is bounded by $$$300$$$, the worst possible is $$$3 \times 10^7$$$.

$$$t <= 100$$$

not just slow bro, it's actually frozen on 16% :/

easy!=good...

Some time after the problems are tested. Unfortunately there were a lot of submissions this round so I think this might be slowing down the process.

Sorry if my comment was sensitive to you.

They're my opinions on the difficulty of the problem.

I'm at least happy it only went full potato mode now and not during contest :P

Does that actually work?

Meanwhile, G having 630+ (and counting) tests..

And our beloved Systests have gone somehow. Take a look at the number of tests of this submission 128675942. Maybe something went wrong due to exceptionally large datasets? isaf27

persons with lower eye sight should be given seats with lower numbers but in your sequence 3,4 is bigger than 1,2 one should consider the total sequence not for each sequence