It’s fast and smooth because we are there , monitoring every cheater, scaring them, and we built strong tests.

😁

No i just answered query separately by their color, and did some math

There also possible to solve $$$ O(n+q) $$$

My friend have similar blog https://mirror.codeforces.com/blog/entry/137774

Honestly, I really enjoy catching cheaters. There’s something satisfying about exposing people who try to outsmart the system with zero actual effort. Whether it’s someone copying from GPT or using cheap social engineering tricks, it’s like a puzzle—except the reward is cleaning up the community. Feels like digital justice, and I’m all for it.

Dude, you haven't even tested.

as a tester this round is peak

every node will have convex hull of segment[tl,tr]

when merging two convex hulls,we can save for every left child next right child that next to him in father hull, and vice versa.

now find j that $$$ t[v][j].f(x) $$$ is min ,

if(t[v*2][left[j]].f(x)<=y)
go(v*2,tl,mid,left[j]);
else
go(v*2+1,mid+1,tr,right[j])

I think this might work. segment tree of convex hulls with fractional cascading.

1.You binary search once among the root’s hull breakpoints to find the best line.

2.build the hull at a parent node, you remember, for each line on the parent hull, exactly which position in the left child hull and the right child hull you should jump to if you end up descending there.

3.overall memory $$$ O(n \log n) $$$ and time complexity $$$ O((n+q)\log n) $$$

I don't understand the problem — are you trying to find an $$$ i $$$ such that $$$ f(i,x) \le y $$$, or one that minimizes $$$ f(i,x) $$$?

problem based on non-negative Diophantine Equation 311C - Fetch the Treasure

i wrote my editorial

As Tester, this round is great.

just basic flow where edge and vertex being swapped.

are you sure you read word except?

You just need more experience. Try participating in virtual contests — it’ll help a lot.

There is JavaScript code that hides comments from selected users using Tampermonkey.

link

You can do sqrt decomposition on height, but it is worse than HLD.

I concur with this blog. AI is highly effective for problem creation, solving, and translation. However, cheating is a human tendency driven by the desire to excel, yet depending on AI for it ultimately hinders personal growth.

Wait... 6 problems and 6 dragon balls? Are we summoning Shenron or debugging our way to a wish?

D is good problem , idea graph of diameter 3.

C my English is bad, I always confuse “at most” and “at least”.

As a tester, this is my first contest as a tester, I wish luck for participants!

best contest, but in my opinion E is easier than D

D: Use array size of $$$ 10^{6}$$$ ,to clean array just do for with in $$$O(n+m)$$$

E: simple scanline

Gratz.

good contest

Can someone help with C. 238738762

I know that can be solved by $$$ O(n^2) $$$, but can be there other solutions than authors?

I used $$$ dp[i] $$$ like when $$$ a[i]==i $$$ and $$$ dp2[i] $$$ when $$$ (a[i]==i+1) $$$.

My solution is $$$ O(n*log2(k)+k*log2(k)+n^2) $$$

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you can use VS code or Sublime text

be Expert

no you use int x not string x;

int can't store so large number. 177296883

🗿

use this

#define sqrt(x) sqrtl(x)

now i will always write

#define sqrt(x) sqrtl(x)

Hide your rating button.

replace vector<vector> v(n, vector(3, 0)); to vector<vector> v(n, vector(m, 0));

What is your strategy to solve problems?

does this code work? CODE

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