kostia244's blog

By kostia244, 4 years ago,

Hello Codeforces! I got some interesting tricks to share with you.

1. a_i := (a_i + x)%mod Range Updates, Max Range Queries

Apply usual range addition, but after each update run the following code:

void normalize(int l, int r) {
while(getmax(l, r).max >= mod) {
set(getmax(l, r).pos, getmax(l, r).max%mod);
}
}


What is the complexity of this code? We all know that each time we %%=mod\$ number which is $\geqslant mod$ it halves, that means each number will be updated $O(log)$ times, we will perform $O(n\cdot\log{n}\cdot\log{a})$ operations in total. So each update will be $O(\log{n}\cdot\log{a})$ amortized!

2. $O(\log^2{\log{n}})$ Segment Tree

Segment tree queries are queries on path. Instead of doing them in the usual brute force-ish way we can use HLD! Since maximum path length is $2\cdot\log{n}$, thus operations are now operations are $O(\log^2{\log{n}})$, which is better than $O(\log{n})$.

3. Faster Dinic's

We know that after $i$ operations Dinic's algo finds flow which is at least $\frac{i}{i+1}\cdot{maxflow}$. In some problems we just want to check whether flow is at least $x$. Using the fact above we can run one iteration of the algorigthm which will get us $\frac{1}{2}$ of maxflow and multiply that by two. Now just compare $x$ and maxflow.

4. Linear FFT

Usually, we use roots of unity ($e^{\frac{\tau\cdot t}{n} \cdot i}$) or primitive roots (for ntt). But why would we limit ourselves to those?

Instead of roots of unity, we can choose roots of something else, that looks complex enough. This, for example:

${e^{\sqrt[x]{6 \cdot \pi ^ 2 \cdot \prod_{1 \leqslant i \leqslant 10} {(x - i \cdot x^{\frac{1}{i}})}} + i\cdot\frac{\pi}{2}}} = i$

Let's choose a few roots of this thing which we'll use for FFT. We can notice that $(-1)^{\frac{i}{2\cdot i - 2}}, 1 \leqslant i \leqslant 10$ work. So using them we have $O(10\cdot n) = O(n)$ FFT!

5. Linear Interpolation(Actually Point Evaluation Without Finding The Polynomial)

We can interpolate in linear time if given x coordinates of given points are n consequential integers (check out 622F editorial for details). We can use this approach to perform general interpolation in $O(n)$. Let $f$ be the polynomial interpolated, n be the number of points given, $v$ be the x value in which we want to evaluate $f$, $x$ be x coordinates, and $y$ be y coordinates of given points. Let's introduce a new polynomial $g$, such that $g(i) = x_i, 0 \leqslant i \lt n$. We can solve $g(x) = v$ easily using HS math in $O(n)$ time. Now, let's interpolate $h(x) = f(g(x))$ instead of $f(x)$. We just evaluate $h(g^{-1}(x))$ since input of $h$ is n consequential integers we can do it in linear time. Easy as that!

Thanks for reading, I hope your rating will skyrocket after applying those in contest! If you have any questions feel free to leave commments or DM me(kostia244) or AryaPawn

• +429

| Write comment?
 » 4 years ago, # |   +76 Me: Scrolled down the comments section to see if this was a joke :').
 » 4 years ago, # |   +25 Thank you sir
 » 4 years ago, # |   +33 Showing yourselves as grandmaster and your friend as specialist will surely get you more DMs than him/her.
 » 4 years ago, # |   +233 Thanks! Very helpful techniques!I would like to point out that the complexity of the segment tree can be improved to $log^2\left(log^2\left(log \left( n\right)\right)\right)$ if you apply Centroid Decomposition on the segment tree before applying HLD.It is actually a well-known technique in China, I'll write a detailed blog about it soon.
•  » » 4 years ago, # ^ |   +11 Now that China is into the picture, I am no longer sure that you are fooling around!!
 » 4 years ago, # |   +32 It's also way faster to write sum += a[i] * c[i] than if(c[i]) sum += a[i].
 » 4 years ago, # |   +5 whats HLD??
•  » » 4 years ago, # ^ |   +53 Learn how to use the urban dictionary.
•  » » » 4 years ago, # ^ |   +4 anyways thanks :)
•  » » » 4 years ago, # ^ |   +6 I actually looked up HLD in urban dictionary. :facepalm:
•  » » 4 years ago, # ^ |   +6
 » 4 years ago, # | ← Rev. 2 →   +33 Me who haven't use any of them
 » 4 years ago, # |   +18 Excuse me... how can the first trick works if I update [1, n] +p each times... the time complexity will be $O(nlogn \times Q)$...
•  » » 4 years ago, # ^ |   +21 SpoilerApril fool( As far as I know , not even one of those tricks is real)
•  » » » 4 years ago, # ^ |   +16 Damn, you are a real demoralizer! SpoilerThe fact about flow being at least $maxflow\cdot\frac{i}{i+1}$ after $i$ iterations is true for matchings. The "trick" was inspired by "Um_nik's Algorithm" problem from ptz.(38th ptz day8 pA).
•  » » » 4 years ago, # ^ |   +11 Sure, I'm quite stupid yesterday. :)
 » 4 years ago, # |   +19 I believed the first trick, but after I read the rest of them, I got confused, really really confused.feels like how did I not know them before, wtf are these?And I scrolled down to the comment section...;)
 » 4 years ago, # |   +29 If you use bitset they all speed up by factor of log n
 » 4 years ago, # |   +24 I wanna point out, kostia244 was orange when he wrote this blog and used some trick to make his name appear red. AND NOW THAT BECAME LEGIT ORZ
 » 3 years ago, # | ← Rev. 2 →   +28 Do you have more papers on that Linar FFT?I don't understand really.
•  » » 3 years ago, # ^ |   +39 Here is a more detailed explanation
•  » » » 3 years ago, # ^ |   +18 Thx
•  » » » 3 years ago, # ^ |   +52 I was fooled QAQ
 » 3 years ago, # |   +5 Is this is how red coders joke arround ?
 » 3 years ago, # |   +8 even the red coder's jokes are above my level!
•  » » 3 years ago, # ^ |   0 I second this.
 » 3 years ago, # | ← Rev. 2 →   -35 .
 » 3 years ago, # | ← Rev. 2 →   +57 Another thing that will supercharge your programming prowess is to replace all for-loops with a #define REP ... macro.Also when you're in the kind of contest where you run your code locally, turning off the heating in your room will cool down your CPU.
•  » » 3 years ago, # ^ |   +7 Whenever I try to see some submissions and I see #define REP my brain just stops working :( .