Auto comment: topic has been updated by chillingjellyfish (previous revision, new revision, compare).

Автокомментарий: текст был обновлен пользователем chillingjellyfish (предыдущая версия, новая версия, сравнить).

Hope I can become Master in this round :) (I failed last time)

UPD: Success!!

Not that 6 — 7 problems joke again...

GLHF!

i shall get stomped like the goombah i am

Let's see how this goes!

Hope to don't lose my Specialist, please......

Good luck guys!

What is the difference between educational round and regular round other than scoring system? Is the problem "easier" and more "training" like than regular round?

Hope to reach 1700

i thought it was on russian till i saw that was translated

I absolutely HATE people who reply with "what". Seriously, are you a runtime error? A segfault in the human language compiler? Imagine spending your precious CPU cycles crafting a beautiful explanation, laying out your thoughts with the clarity of a competitive programmer optimizing their bitsets, and the only response you get is "what".

Like, excuse me, is your brain currently experiencing buffer overflow? Did you accidentally AND your listening skills with zero? Maybe if I rephrased this explanation as a Codeforces editorial starring your beloved bitset waifu, you'd suddenly comprehend everything. Honestly, next time someone hits me with a "what," I'll just reply, "Sorry, buddy, didn't know your attention span had a time complexity of O(1)."

Do better.

Why is the writer list on the contest page still empty so far

UPD: its ok now

what

What happen? A judge pause for at least 10 min.

InQforces ._.

Can any1 check the server pls??

Now! A judge pause for at least 30 min.

My submission was queued and got processed after 17 minutes and after 17 minutes I got to know I have submitted code for D in B problem. Queued again LOL

Is it just me, or does statement D kind of suck? (It’s written really unclearly)

problem tells you: take a number, write its digits, add its digits, repeat until it becomes a single number… now arrange all the digits so the result of the magic game becomes some number..meh does that mean math or magic?nobody understands... but actually, nice contest, it's the first time problem D didn’t risk your rating.

Speedforces&Queueforces

the queue was way too long mid contest

time to train Lagrange multipliers, I forgor how they work

In problem D, for each component is it not enough to simply check if there's an odd length cycle? Iff not, add by x/2, where x is the number of nodes in the component.

How did so many people solved D?? was it so easy? Even after solving 3 problems(acd) my ranking is like 9k

There are sooo many cheaters. For instance, I am quite sure that sonsimpmiku cheated, who got every problems accepted in just 35 minutes. That's incredible.

How to solve C..I tried but WA 367013755 here is my submission after contest... What i am doing wrong!

i don't understand what even question 4 is saying .

I am depressed , it was my birthday today and I solved three in 25 mins saw 1k rank thought inflation would yield 2k rank but getting 5k at end has made my day bad... is D that easy , i was unable to even understand test cases

Why is D's statement so unclear? I don't think problem statement language is supposed to be the bigger hurdle in a div2D.

Problem E edge cases cooked me bad.

plz help me finding flaw in my logic in problem C...

#include <bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

#define endl '\n'

using namespace std;

void solve() {
  i64 p, q, r, n;
  cin >> p >> q >> r >> n;

  i64 onlyp = n / p;
  i64 onlyq = n / q;
  i64 onlyr = n / r;

  i64 pnq = onlyp / (p * q);
  i64 qnr = onlyq / (q * r);
  i64 rnp = onlyr / (r * p);

  i64 pnqnr = n / (p * q * r);

  i64 ansp = (onlyp - rnp - pnq + 2 * pnqnr) * 6 + (rnp + pnq - 2 * pnqnr) * 3 +
             (pnqnr) * 2;

  cout << ansp << endl;
}

int main() {

  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  int t;
  cin >> t;
  while (t--) {
    solve();
  }

  return 0;
}

Screencast with commentary (once YouTube is done with processing it)

Preventing $$$O(n log$$$^$$$2$$$$$$n)$$$ solutions in F feels unnecessary.

You cannot simply turn a forest graph into a general simple graph in the middle of the round :(

That's too critical to do so and I wonder how this had not been noticed before the round started.

how do u do e?

For problem E: Today I learned that using sscanf ("str" -> "int") and sprintf ("int" -> "str") to a short temporary buffer_string and do string manipulation is slow (TLE:367013757) on codeforces machine and better do manipulation on the integer directly: (AC:367016186)! Though that's not the case on my PC:

• my TLE solution 367013757 ran 76ms on my PC
• my AC solution 367016186 ran 106ms on my PC

weird x_x"

https://mirror.codeforces.com/profile/Ser_arthur Ser_arthur it looks is cheating , a newbie unrated who just joined and is in top guys in standing

felt like a div3

It looks like https://mirror.codeforces.com/profile/Ser_arthur Ser_arthur cheated, unrated guy who just joined some days ago is already in top standings !! and did all!! *

Pretty nice E and F, but imo E should be D and F should be E

for problem D is this sequence valid? 1 --> 2 <-- 3 --> 1 for test case 1 3 3 1 2 2 3 1 3

I find it interesting that 366982885 gets rte, but 367037092 passes. The only difference is allocating the large arrays on stack / heap. (Had to wait 20 min during the contest to find out)

Won't we talk about cheaters ? It is really frustrating, they affect the contest rating changes and this is not a fair play. I hope you do something about it

This time the tasks were easier than I thought.

felt like a div2

How to solve F? If we had only one k, then the greedy approach is to choose always one element in a subarray, more precisely element with smallest value of a[i], then decrease it until it becomes 1/1, then increase numerator. We can calculate for each element it's contribution. But how to solve this problem for many k's?

Hello guyz!

man e was such an easy solution idk why i got stuck on attempting brute force as last digit assuming last digit can be anything from 1-9 and then proceeded backward propagation man fml