> stalked random one
> grandmaster
> another one
> grandmaster

is it rated?

my plank PB is 4:30 but i can barely do 5 push ups so sad

I used to be 166cm, but because of Covid, my sitting pose got worsen and worsen everyday. I was diagnosed with Scoliosis and now im 164cm

Remind me to take the handle 164cm first this new year 😠

please add some pictures on the site or at least some emoji, it's too much for me to read all text

will start in 24 hours 😳

https://tlx.toki.id/contests/tsoc-april-fools-2024/problems/F

problem G is the same as this :>

can I trade Indonesian Indomie for Indonesian Indomie

Fixed (ෆ˙ᵕ˙ෆ)♡

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[Sorry for necroposting]

lol Overkilled the Div. 1 C with small to large merging + $$$O(n)$$$ memory binary lifting from https://mirror.codeforces.com/blog/entry/74847.

I've got the observation that the new diameter can be obtained by connecting two middle points of two trees, also got the observation that the new diameter is whether to connect the two trees' diameter pair, or the old diameter pair of both trees.

Didn't know that the diameter will always be sum of ceil of both. So I coded the online merging lca-able tree sets that maintains the diameter pair.

231305622

I've graduated for almost a year now, resigned from 1 job just so I can practice full time, turned down 3 job offers. Not a whine but yes, I do feel a bit disappointed

More time to practice

https://github.com/spaghetti-source/algorithm/blob/master/geometry/_geom.cc#L729

my bad. I think it's $$$N^2$$$

Wow I got chills, literally about 10 hours ago I was looking for a valid Subquadratic Polygon Triangulation after seeing the lecture 2 days ago.

But I found this code by Takanori Maehara which does the same thing in $$$O(N)$$$

real triangulate(vector<point> ps) {
  int n = ps.size();
  vector<int> next(n);
  for (int i = 0; i < n-1; ++i) next[i] = i+1;
  auto is_ear = [&](int i, int j, int k) {
    if (sign(cross(ps[j]-ps[i], ps[k]-ps[i])) <= 0) return false;
    for (int l = next[k]; l != i; l = next[l])
      if (sign(cross(ps[i]-ps[l], ps[j]-ps[l])) >= 0
       && sign(cross(ps[j]-ps[l], ps[k]-ps[l])) >= 0
       && sign(cross(ps[k]-ps[l], ps[i]-ps[l])) >= 0) return false;
    return true;
  };
  real area = 0;
  for (int i = 0; next[next[i]] != i; ) {
    if (is_ear(i, next[i], next[next[i]])) {
      area  += abs(cross(ps[next[i]]-ps[i], ps[next[next[i]]] - ps[i])) / 2;
      next[i] = next[next[i]];
    } else i = next[i];
  }
  return area;
}

..

// ==UserScript==
// @name         Fix CF ICPC Challenge Bar
// @version      0.1
// @author       hocky
// @match        https://mirror.codeforces.com/*
// @icon         https://www.google.com/s2/favicons?sz=64&domain=codeforces.com
// @grant        none
// ==/UserScript==

(function() {
    'use strict';
    $("a[href='/icpc2023'] img").css('height', '15px');
})();

      ╱|、
  (˚ˎ 。7
    |、˜〵
    じしˍ,)ノ

I have 3 moωe ωluogss in my dωwaftt UωU
You can tωy to tωansωate the next one ૮ • ﻌ — ა ごめんね~ :p

⸜(｡˃ ᵕ ˂ )⸝♡ https://github.com/hockyy/cf/blob/main/DynamicDiameter.md

As a tester, I think this contest is ██████ ████ ████ ██████ ██ ███ ██ ████ █████████ ████████, ██ ███ ██ ███████████ ███ ████ ██ ██

If you have solved the basic DP tasks from atcoder, the D is a no-brainer problem. 😶 https://atcoder.jp/contests/dp/tasks/dp_t

uwu

lol I hopcroft-karp'ed the E

Upd: mine got tle lol. guess it's not my lucky day

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Uwp Uwp Like a Bawoon 🎈

I definitely can't be the next lead coordinator as I feel I'm not capable of writing legit analyses on complicated problems, even maintaining these current qualities of contests. 🙇‍️ 🙇‍️

I think we can expect Pyqe or rama_pang to be the (next) coordinator of TOKI contests.

I wunna say twank you too ヾ(＠^∇^＠)ノ

I've been working with prabowo since TROC #9, ~28 months ago. I leawned so much from him as I consider him my mentor, and I successfully led COMPFEST 13 — Finals Online Mirror months ago because of these experiences.. Been a long jouwney with lots and lots of fun authows. (′ꈍωꈍ‵) 💖💖💖

I also wanna thank fushar for TLX, steven.novaryo for the one with who I can spill my sad and happy thoughts.

There have been times I also felt so lazy, but I really had fun preparing problems, illustrations, artworks, editorials, with all of the past authors.

I don't know why I post this comment as I don't know what to do next. lol

Credit: ICPC Live Stream, steven.novaryo for capturing the moment

My intended solution was like this.

I've added the fifth sample to solve this ambiguity in the future.(ಥ﹏ಥ)

I see.. Thank you for the feedback!

We were trying to fit in such vary skill range in our official contest. Hope you liked the problems ^^🐾~!

The editorial will be posted soon

How was the contwest UωU? I thought pweople would find K easiew than othews easy pwoblems (ಥ﹏ಥ)

Thank you for noticing.... I've fixed it ☆*: .｡. o(≧▽≦)o .｡.:*☆

あははははは~🐾! Thank you UωU

I'm just twying to be fwiendlyy 😔😔, maybe pwepwle luvs fwiendly bwoggg ૮ • ﻌ • ა

I.. I'm not a weeb.. Baka ﾟ+.(*ﾉｪﾉ)ﾟ+

my random string generator can't reach it smh my head

Every asian rounds ever

I accidentally submitted for A at 00:54. ╥﹏╥ I swear my hand autopilotted me to submit A.

Dark mode user can't enjoy this

It will start in less than 2 hours! NGL I enjoy all the problems :D

CSES' testcases are weak. I stresstested my code and it got wrong answer on simple testcase.

https://cses.fi/problemset/task/1685 The exact same problem.. This is a joke :/
The quora one got me wrong answer on TC 6.

I can't seem to imagine such a slim font on codeforces :/. It just looks off to me. UPD:

I think if the logo is going to be changed, the whole web's font must change as well. It's like a face with a one off feature.

What if we put some old spices to the logo..

Sure! it's unrated, anyway. glhf

Yes, you can!

Outstanding move.

tmw orz

this comment has the same energy with comments after global round 7

Problem C is one of the sign we humans have reached our advanced civilization.

Never again.

:)

Tbh the contest is really good. Regardless the hard C.

15 more minutes!!

UPD : The contest has started! Yet you can still join! Rating is nothing but a number!

..

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Bump! The contest will start soon!

Thanks for reminding me. Here's TSOC — April Fools 2018! This contest will be available in English only, the same as 2018's.

lol i definitely don't deserve to be CM

Hate to say this, but 185217's a real solid one. Here we go again.

Dude.. page 6 of 264525 actually made a pretty good wallpaper

Example Solution Author: prabowo

Editorial will be posted soon!

antontrygubO_o Congratulations!

Benq on Div 2:

When youre late, but still able to register and compete in the contest

Me realizing the contest will start in 5 minutes but havent typed my template yet

Bumpy bump! The contest will start in 15 minutes!

This is the exact solution first I thought of after seeing the One-Head Equation. The approach is dynamic programming like.

Let $$$R_{0}$$$ be the EV when your current sequence doesn't have any leading H, for example "...HTT_"
Let $$$R_{1}$$$ be the EV when your current sequence has 1 leading H, for example "...TTH_"

$$$R_{0}$$$ and $$$R_{1}$$$ can be written as:
$$$R_{0} = \frac{1}{2}(R_{0}+1)+\frac{1}{2}(R_{1}+1)$$$ (1st Equation)
$$$R_{1} = \frac{1}{2}\cdot1+\frac{1}{2}(R_{0}+1)$$$ (2nd Equation)

You can substitute the 2nd Equation to the 1st
$$$R_{0} = \frac{1}{2}(R_{0}+1)+\frac{1}{2}(\frac{1}{2}\cdot1+\frac{1}{2}(R_{0}+1)+1)$$$
$$$R_{0} = \frac{1}{2}(R_{0}+1)+\frac{1}{2}(\frac{1}{2}\cdot1+\frac{1}{2}R_{0}+\frac{1}{2}+1)$$$
$$$R_{0} = \frac{1}{2}(R_{0}+1)+\frac{1}{2}(2+\frac{1}{2}R_{0})$$$
$$$R_{0} = \frac{1}{2}R_{0}+\frac{1}{2}+\frac{1}{2}(2+\frac{1}{2}R_{0})$$$
$$$R_{0} = \frac{1}{2}R_{0}+\frac{1}{2}+1+\frac{1}{4}R_{0}$$$
$$$R_{0} = \frac{3}{4}R_{0}+\frac{3}{2}$$$
$$$4R_{0} = 3R_{0}+6$$$
$$$R_{0} = 6$$$

Yet, you can do not only two-heads, but a whole pattern match.

62833307 Can't forget this

Can relate

My solution has O(n) complexity :O Capped by the input tho. only took sqrt(n)*log^2(n) to binary search the grid. Upd : nvm. i got sort.

Oh wait, my wip solution actually works when the Pi can be updated too. I was overthinking on how to merge 2 intervals, where you have to store 2 value for each end. Turns out you just need a function to compute each interval value and can lazily update the checkpoint. Guess I'm still far away from being a master :<

Can Div. 1C/2E be solved by segment tree and merging its left and right equation?

Mathforces in a nutshell

I'll just finish my homework then

Div1A reminds me of this CF995A — Tesla when people go derp mode real hard.

Contestants tried heuristic solutions, dijkstra, weird heaps, while you only need loops to solve it.

This is beyond my knowledge. Never knew LIS would be this powerful. This not only works for 3 arrays. The intuition is you want to maximize the amount of element to stay in its segment. What an elegant solution. to restore the steps of the answer, youll simply make the LIS related elements stay and move the rest greedily.

Probably the scientific judges don't know about the technical issues or site environment-related things (stack size, speed, OS). So you have to figure it out by yourselves. Last regional (not Bangkok) I attended, showed compile flag and if I'm not mistaken they said the stack size was set to default, but not quite sure how big.

Showing which problems to solve is a good thing because there are many teams and weakest ones who can only hope to solve 1-2 easiest problems. It becomes much harder if you have 12 problems and have no idea about the difficulty. If you want a contest with a completely new format and rules, just hide the leaderboard or show only the number of solved problems, but it will change the contest significantly and shouldn't be used as regionals for ICPC.

I agree.

I think somehow they shuffled the order of the problems so that the participants can't decide the easy problems through the scoreboard. Based on here

About the stack size and computing speed, I think it's your thing to test it during the practice session.

Well I guess the rest is decided by the hosting university.

Good luck!

Alhamdulillah

For those of you who are looking for a good and clear randomization implementation 49733774 Correct me if I'm wrong.

Thank you AkiLotus GreenGrape cdkrot neal