Round rotavirus skips -> good round

Anime is shit, change my opinion!

this was a wise decision....but XD

So what? Its a piece of art.Change my mind.

So? It's art, bro

Even better

Neither did I so I made it myself.

completely shocked when I entered the main page, haha

This post shows ,it already have

The problems will be very good so no bad memories will be associated :sunglasses:

nice contest~

What about jujutsu kaisen ?

NO_WORRIES

Didn't know Ari was Vietnamese :O

More like northern American authors

I believe the registration page is not updated yet. I will contact to fix the issue :)

up

Me too~

Normie

hope you don't feel the same way with problems

If you have no clue, then think of us:( May be an editorial over comment history be a new update:)

You should be judged by KAMI SAMA for addressing an Anime as a cartoon :D

ưhat's up :)

I prefer a positive delta change of > +50 in an Anime theme round XD

you definitely mean one match of Dota 2 with Techies

.

No I don't

Monogon literally just held his contest...

I think so

Don't judge a book by it's cover.

maybe

why not

According to the score distribution seems that Div2 D = Div1 A

Div2 D = Div1 A

press on this star

You will man :)

you will~

you can , i believe you~

Naruto Uzumaki~

me too

chapter 139 had me in tears ngl,eremika forever!!

i got

hi

I can feel this comment.

how to solve div2C?

It is the place where green's rest.

try 8 1 2 4 4 5 9 9 10

Same! I simply traversed a sorted circular array n times for each i and a reverse sorted circular array. Thought it was right, but got stuck on pretest 5

Choose two strings which have at least $$$n$$$ zeros or ones in common (clearly at least one such pair must exist). Take $$$n$$$ of these common, then just place the remaining $$$2 \times n$$$ elements in the same relative order.

Out of the 3 strings, you can find a pair of strings such that both have >= n zeroes or both have >= n ones. Treat the 0s (or 1s in other case) as common subsequence. Now you can combine these 2 strings (Leaving this as an exercise).

DP

Can be solved using DP. At the end of the altered array you have two options either max(1--n) or min(1--n) for optimal ans. So at each step you have two options. Lets make a vector of all speeds in a sorted manner and solve the dp,

dp(l,r) = min(dp(l+1,r), dp(l, r-1)) + v[r]-v[l]; ans = dp(0, n-1);

Submission Link

Since it is about the differences we can need to sort a[].

dp[i][j] is min cost to have runner i to j of the sorted list started in any order.

dp[i][i]=0;

for all i+1<=j:

dp[i][j]=min(dp[i+1][j]+a[j]-a[i], dp[i][j-1]+a[j]-a[i])

So, we start with any runner, and take as next the next faster or slower one.

At each point you've to either place group of smallest elements of group of largest elements in the end (try to prove yourself).

So you do a dp[l][r] on the sorted list of distinct elements and at each point decide to place either all a[l]s or all a[r]s in the end of the final list and then do l++ or r-- and continue

Sort the speeds then compute the minimum cost to include racers with speeds si...sj.

• Initialize dp[i][i] = 0 for all i (cost of including single racer is 0)
• At each step k, dp[i][i+k] is the minimum cost of including all racers between i and i+k, which can be computed from the previous dp[i][i+(k-1)] and dp[i+1][(i+1)+(k-1)]
• You can actually use 1D DP because at any given k you're only using the results from k-1 before overwriting them.

I solve it using dp,i sort the distinct value of all input,then we consider the last operation we can make , neither we put the max or min value , then dp[i][j] = min(dp[i+1][j] + count[value[i]] * diff,dp[i][j-1] + count[value[i]] * diff),here is my code: dp solution

Suppose there are two strings $$$a$$$,$$$b$$$ such that number of zeros in $$$a$$$ and $$$b$$$ are greater than $$$50$$$%. Then we can choose string $$$a$$$ and insert number of $$$1$$$'s in string $$$b$$$ in string $$$a$$$ such that $$$b$$$ is substring of whole string.

Same case when number of $$$1$$$'s in $$$a$$$ and $$$b$$$ are greater than $$$50$$$%.

When both $$$1$$$ and $$$0$$$ are equal to $$$50$$$% in both $$$a$$$ and $$$b$$$ then also it can be solved with same logic.

Let say $$$a=11111100,b=11110000$$$ then final string will be $$$111100001100$$$

Since there are 3 strings, we can always pick 2 out of them, such that either both of them have at least n times 0 or both of them have ar least n times 1. Let's call this the mode. It is 0 in the first case and 1 in the second case. Then we can create a string with n times the mode. Now we matched at least n characters from the first string, and at least n characters from the second string. We are left with n characters from the first and n characters from the second string. We can simply add them by traversing both strings to get a string of length at max 3n.

we can prove the at least there two strings which is have at least n characters are same , '0' or '1',for any case ,we assume it is '1',we just can replace n '1' in any string with another string ' s '1',don't change the order of other characters ,we can make it;here is my code: my code

Couldn't participate in the contest today. But upon reading , you just traverse through the string and check that at no point , the count of M should be greater than the count of T. then repeat the same thing , while going backwards , that is from n-1 to 0. And then just see if the count of T is 2 times the count of M. If the string satisfies all the above conditions , print yes

we can check two pass from left to right and from right to left,every pass ,we must make every m can get a t in it's front ,otherwise we can't make it ,by the way ,we always have the number of 'T' is two times of the number of 'M'

I feel like implementation is the biggest pain in Div2D / Div1A, took 5-10 mins to come up with the solution, 20-30 mins to code it correctly. Maybe I'm just bad at implementation.

Read every submission from jiangly and your implementation will become CLEAN and FAST.

Hints: There must have 2 string where no of 0/1 in both two string >=n.

approach:

when any of two string no of 0/1 >=n. Then take one string fully , so other string n element already taken, so taken the rest <=n elements of the other string with some tricks which was not common. its always <=3n operation.

implementation

A simpler solution (at least in my mind) with complexity $$$O(m\log m)$$$:

• If $$$\frac{n(n-1)}{2}-m\ge n$$$, answer is just MST.

• Else, $$$n=O(\sqrt{m})$$$. Naively process will give $$$O(m\sqrt{m}\operatorname{dsu}(n))$$$, but note that only $$$n$$$ out of these $$$m$$$ edges are useful (run MST with only fixed edges, only used edges are useful). Now the complexity is $$$O(n^2\operatorname{dsu}(n))=O(m\operatorname{dsu}(n))$$$.