qwexd's blog

Codeforces Round 1121 (Div. 2)
By qwexd, 3 weeks ago, In English
1121. Another Round
time limit per participant
2 hours
memory limit per participant
1 brain

This is an interactive problem a contest with an unusual start time.

You are given a Codeforces account and $$$n = 6$$$ problems, authored and prepared by qwexd and jeroenodb.

In one operation, you may choose a problem $$$i$$$ ($$$1 \le i \le n$$$) and submit a program intended to solve it.

The maximum scores are given below. Problem E is divided into two subtasks.

Problem A B C D E F
Maximum score 500 1250 1500 2500 2000 + 1500 3000

The points awarded for a correct submission may be lower, as determined by the Codeforces scoring rules.

Your task is to maximize your total score before the time limit expires.

Input

The input consists of the problemset of Codeforces Round 1121 (Div. 2). It became available on Sep/13/2026 20:05 (Moscow time).

The round is rated for participants with a rating below 2100. Participants with a rating of 2100 or higher are welcome to participate out of competition.

Output

For each problem you choose to solve, submit a correct program. You may solve the problems in any order.

Printing YES is not sufficient.

Note

It has been shown that every problem has a solution. The proofs are now available in the editorial.

The following participants were the first to construct correct solutions:

We would like to thank:

Thank you for participating!

  • Vote: I like it
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  • Vote: I do not like it

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3 weeks ago, hide # |
 
Vote: I like it +124 Vote: I do not like it

As a tester, I may have started a trend with announcement posts :)

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3 weeks ago, hide # |
 
Vote: I like it +15 Vote: I do not like it

Thank you for making my bedtime 2 a.m.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Before you add the score distribution put it in a Scoring section instead of where it is right now

Also I think you should note the unusual start time

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3 weeks ago, hide # |
 
Vote: I like it +47 Vote: I do not like it 
/* 
 * Author: temporary1
 * Problem: 1121. Another Round
 * Time Created: 01:24:40 Sep/12/2026 (UTC+0)
 */
#include <bits/stdc++.h>
using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n = 6;
	vector<string> contest;
	for (int i = 1; i <= 6; ++i) {
		string problem;
		cin >> problem;
		contest.push_back(problem);
	}
	mt19937 rng(1121);
	shuffle(contest.begin(),contest.end(),rng);
	for (string problem : contest) {
		cout << "#include <bits/stdc++.h>" << '\n';
		cout << "using namespace std;" << '\n';
		cout << '\n';
		cout << "int main() {" << '\n';
		cout << "	ios_base::sync_with_stdio(false);" << '\n';
		cout << "	cin.tie(NULL);" << '\n';
		cout << "	cout << \"as a tester, i tested this problem\" << \'\\n\';" << '\n';
		cout << "	return 0;" << '\n';
		cout << "}" << '\n';
		cout << '\n';
	}
	return 0;
}

as a tester, i tested the round announcement

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3 weeks ago, hide # |
 
Vote: I like it +15 Vote: I do not like it

didn't have to call me out like that smh

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

thank you for making me not have to get up early

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

as a tester, im glad i didnt end up on the shame board.

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3 weeks ago, hide # |
 
Vote: I like it +4 Vote: I do not like it

please highlight the unusual start time, someone might not notice!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

As a tester, you shouldn't be saying "We would like to thank you, for participating in the round" because I will not participate.

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3 weeks ago, hide # |
 
Vote: I like it +23 Vote: I do not like it

its bed time for me

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3 weeks ago, hide # |
← Rev. 2  
Vote: I like it -11 Vote: I do not like it

heuristic problem in codeforces!

also, i want examples, i don't read statements, only examples and notes

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3 weeks ago, hide # |
 
Vote: I like it +9 Vote: I do not like it

As a tester, the best thing I did wasn’t solving the problems it was bringing in a better tester than me, shoutout Rayo my goat

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3 weeks ago, hide # |
 
Vote: I like it +23 Vote: I do not like it

Good time for Chinese students

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3 weeks ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

one more outside the box announcement!

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3 weeks ago, hide # |
 
Vote: I like it +16 Vote: I do not like it

Hope to not get a Brain Limit Exceeded verdict :)

By the way, creative announcement though. Initially, I thought I had entered into a problem statement haha...

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

As a tester I would like to say that this is my very first comment.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Guess I'll be doing Codeforces from midnight Monday, what a good way to start a week.

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3 weeks ago, hide # |
 
Vote: I like it +9 Vote: I do not like it

As a tester who didn't test, my memory limit was eaten by the zombies.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

No not midnight again!!!

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3 weeks ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

tested

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

creativity

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Nice way of announcement!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Very pioneering!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

The pressure for creativity is going up for the next announcement.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Thank you for making my bedtime 3 a.m.

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3 weeks ago, hide # |
 
Vote: I like it +9 Vote: I do not like it

I guess the next announcement will take the form of a chat message.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Love to see an announcement getting more creative

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Bro thinks he's nik_exists

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3 weeks ago, hide # |
← Rev. 2  
Vote: I like it 0 Vote: I do not like it

Really Amazed by the Announcement Style!!! What an Idea! WOW!!!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

memory limit per participant should be 1 brain and 0 ai

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Well i was not gonna participate but now i will after this creative announcement.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

As a nontester :(

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3 weeks ago, hide # |
 
Vote: I like it +16 Vote: I do not like it

Good news: I have no brain.

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3 weeks ago, hide # |
← Rev. 2  
Vote: I like it 0 Vote: I do not like it

Hope to get back to CM!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

why E and F are ordered that way if total score for E is bigger than for F? even in 1120 div2 we have B > C1 scores so its fine, feels like problems should be swapped

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3 weeks ago, hide # |
 
Vote: I like it +2 Vote: I do not like it

Is Shakespeare really dead?

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

the most creative contest announcement ever

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

It can be shown that this problem is orz

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Amazon SDE II Interview Experience (Bangalore, Selected) ~ Aug 2026

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

As a participant, my goal is to solve two problems. Good luck to everyone!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

all the best guys

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3 weeks ago, hide # |
 
Vote: I like it -10 Vote: I do not like it

Thanks everyone for participating in Codeforces Round 1120 (Div. 1, Div. 2). Although it was initially difficult to understand from the editorial when I first participating in Codeforces contests. However I found that once I got used to it, the explanations felt very easy to understand, accurate, and highly academic. I'm just saying little advice, I mean do it when there is the editoral. We should read the hint and solution before reading code (Sorry for my bad English) I came from the past, not the future

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Interesting annoucement! Thank you for making my bedtime 3AM :(

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Bad timing for me

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Does this have any interactive problems?

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

speedforces?

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3 weeks ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Thank you for making my bedtime 2 a.m.

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3 weeks ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

Ahh man!! Had to unregister because of goofy timing

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

I hope I reach rating 1500+ after this round good luck to everyone

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Awesome contest anouncement this is so creative bravo authors

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Why is the competition held at midnight?

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

i have to become specialist by the end of november

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Oh no!!!I have to go to school!!I can't get up at 01:05!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

allow hacking

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

btw why unusual time?Any specific reason? qwexd

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

RIP my sleep

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3 weeks ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

Users in the announcement are displayed with their max rating color instead of their current rating... orz

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Bad time for Chinese programmer!!!

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3 weeks ago, hide # |
 
Vote: I like it +6 Vote: I do not like it

wtf was that D

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Very difficult round. C was itself very difficult as compared to other D2 C's (atleast for me). Disappointed!

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3 weeks ago, hide # |
← Rev. 2  
Vote: I like it +8 Vote: I do not like it

How to solve F? I tried Lagrange interpolation on the first 50 terms of the deranged sequence and then represented it in binary(by using (n/n+n/n)), but it ended up taking 323612 characters, which is way over the 10000 character limit

upd: the answer is n! / e

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    3 weeks ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    The problem is writing $$$e$$$ and $$$n!$$$ in few enough symbols and have enough precision on $$$e$$$

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    3 weeks ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    You can also use D_n = n * D_{n - 1} + (-1)^{n} and tweak it a little bit to work for all k <= n.

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    3 weeks ago, hide # ^ |
     
    Vote: I like it +16 Vote: I do not like it

    The number of derangements is given by $$$n!/0! - n!/1! + n!/2! - \dots$$$

    which can be expanded to be $$$1 - n(1 + (n-1)(\dots))$$$, and you can adjust the sign afterwards using $$$\mathrm{round}$$$, which avoids computing $$$e$$$ entirely

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3 weeks ago, hide # |
 
Vote: I like it +1 Vote: I do not like it

damn, why did I even bother staying up

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

For $$$D$$$, when we put three 1's, I just guessed something about the sizes of the groups of 0's and it passed. Could anybody prove it?

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    3 weeks ago, hide # ^ |
    ← Rev. 3  
    Vote: I like it 0 Vote: I do not like it

    lets say we have all 0's

    this will give us order of len2 segments (every L and every R is a valid — len*(len+1)/2).

    so breaking it into half (with only one 1 in the middle) gives us (len2)/4

    similarly breaking into more parts will further reduces the segments count

    now look at the powers of 2

    1 2 4 8 16 32 64 and so on

    mod3 = 1 2 1 2  1  2  1 . . .

    U know the divisibility property of 3 is the sum of digits should be divisible by 3 in decimal, doing same here in binary its sum of modulo3 for powers of 2

    Now we need to choose the indices where we don't try to avoid having three 1's in the mod3=1's position or three 2's in mod3=2's position or one in 1's and one 2's, if its possible

    and given it is atmost 3, placing two 1's is always better than placing one 1; depending on the length required we try to place 1's they creates least number of numbers whose mod3==0

    now based on length of the binary string there is a pattern that repeats that u need to catch

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Modulo with negative numbers...

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Is D too non-trivial or am I dumb? First three looked way more intuitive

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3 weeks ago, hide # |
← Rev. 2  
Vote: I like it +3 Vote: I do not like it

solved D after long time, but I don't know how it works

  • so I just brute forced till n=20 and observed the pattern, n=10 and n=16 was a bit different but that also worked for n=22 so I just guessed it will continue to work.

this time I might get good delta due to D but then next time it will cause me big negative delta .. lol !!!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to solve E1?

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    3 weeks ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    The only thing that matters in a subsequence is the maximum and the minimum value which can be found using dp, then you count.

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    3 weeks ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    We can precalculate the X value for any two number using a dp-form thing. like g(a,b) for the max x we can get after operations so that a->x and b->x. Secondly, f(b)=g(min b_i,max b_i). So we can enumerate over all min b_i and max b_i and calculate the corresponding contribution.

    (I'm not sure if this is correct)

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

I hate those problems like C, in which I spend half of the time finding the integer overflow somewhere. I found the recurence in like 10 minutes and then spent 20 minutes debugging. I think I would be able to find solution to D with that aditional time.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Solution sketches for A-D (I didn't get cooked):

A. We will only have to operate on indicies that are out of place ($$$i \neq a_i$$$), so simulate the operation on all out of place

B. We can compress the formula to $$$b_m \cdot m$$$ minus the rest of the elements. Use a std::multiset to track the $$$m - 1$$$ least elements behind each potential final element.

C. It's a somewhat complex formula, see my submission, but basically we're sorting the array and taking each element's contribution.

D. Write a brute force, and notice for sufficiently large $$$n$$$, there is always a unique solution with three ones that ends with 1. We can find constructions that end with 1 for $$$n = 6 \dots 12$$$ and notice that due to the fact that $$$2^k$$$ is cyclic modulo $$$3$$$, we can construct a similar solution for any $$$n$$$ and $$$n + 6$$$. Store the solutions for $$$n = 1 \dots 12$$$ and construct similar solutions for $$$n \gt 6$$$ from $$$n = 6 \dots 11$$$. See my submission for more details: 390645815

Carrot broke for me, can someone check my performance?

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Math contest, good problems!

Solved D fast:)

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Great Contest

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

i'm so happy i managed to solve C for the first time!!!!!!!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

I am curious about how checker for D works

The best I can think is that first we can count all the substring whose all element is 0 after that for every index i we will check for the nearest 1s in string and check whether they are multiple of 3 or not and add them in this way i think checker could work in o(n) , but there has to be good and efficient way

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    3 weeks ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    some kind of dp

    like 'how many substrings end at index i with r mod 3' .. I feel this will be linear... didn't think very much though

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    3 weeks ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it 0 Vote: I do not like it

    since there are at most 3 1s, the f value can be calculated within O(1) complexity.

    (p.s. specifically, this is because only the parity of the index of 1s matter when checking if a string is illegal.)

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

madly difficult C; had it not been for the unusual start time i may have solved it. good round tho, really challenging for once!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

How to solve Problem C?

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    3 weeks ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    C Madamant's Skating Dynasty O(N log N) Approach

    First, sort the array so we can process the elements in order.

    Then, we calculate suffix sums. This lets us quickly find the sum of all the elements to the right of the current position instead of calculating it again and again.

    We also precompute factorials up to N, since the formula uses factorials such as (N-1)!. For the division part, we use modular inverses because everything is calculated modulo 998244353.

    After that, we go through the array once, calculate the contribution of each element using the formula, and add everything to the answer.

    The sorting takes O(N log N), while the rest takes O(N)

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      3 weeks ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      What is the formula

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        3 weeks ago, hide # ^ |
         
        Vote: I like it +1 Vote: I do not like it
        $$$\frac{(n-1)!}{n-i} \times \left( \sum_{j=i+1}^{n} a_j - (n-i)a_i \right)$$$

        Breakdown:-

        • $$$n$$$: The total number of skaters.
        • $$$i$$$: The current skater's index in the sorted array.
        • $$$a_i$$$: The rating of the current skater.
        • $$$\sum_{j=i+1}^{n} a_j$$$: The sum of the ratings of all valid parents (every skater strictly to the right of $$$i$$$).
        • $$$(n-i)$$$: The exact number of valid parent choices for skater $$$i$$$.
        • $$$(n-1)!$$$: The total number of valid dynasty trees that can be formed.
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        3 weeks ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        (n — 1)! / (n — i) Call this Bi

        Bi x ∑(a[j] — a[i]) for i < j <= n Call this Ci

        Final answer is ∑Ci for 1 <= i < n.

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        3 weeks ago, hide # ^ |
        ← Rev. 2  
        Vote: I like it 0 Vote: I do not like it

        after sorting in desc cost until ith elemet = (i-1)*cost unti i-1 + (prev no of trees) * fixed cost per tree fixed cost per tree= sum till i — i*ai prev trees= prev tree * (i-1) update cost

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      3 weeks ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      Also it can be solved with dp too.

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3 weeks ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

Nice contest excellent problems I like it thanks to authors for contest!

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

E2 question: I solved E1 with an O(n^2) approach and tried submitting the same solution to E2, but it TLE'd on pretest 3. I understand that E2 needs a better complexity, but I'm having trouble seeing the optimization. What was the key observation for making the E1 approach fast enough for E2? I tried submitting it during the contest because I wanted to see whether my E1 solution would surprisingly pass E2 as-is, since I already had the E1 solution working. I didn't want to wait until after the contest because I was curious about the actual result. Also, the solution was completely my own I was just experimenting with whether the E1 complexity would survive the E2 constraints.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

C: I tried to % 998244353 before sorting it XP

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

i like you so much this round is so good <3<3<3<3<3

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3 weeks ago, hide # |
 
Vote: I like it +2 Vote: I do not like it

It was a really exiting contest.

I enjoyed it.

Thank you, Codeforces.

I love you.

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3 weeks ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Hi!

What was the accepted time complexity for B?

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    3 weeks ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    $$$O(n \log n)$$$

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      3 weeks ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      I have an O(N log N) solution using 2 priority_queues, but still get Runtime error, and I don't get it. Could anyone check it out, please?

      // this is code

#include <bits/stdc++.h>

using namespace std;
int v[200005];
int main()
{
    int T,N,M;
    cin>>T;
    while(T--)
    {
        cin >> N >> M;

        for(int i = 0; i < N; i++)
        {
            cin >> v[i];
        }
        priority_queue<int> pq1;
        priority_queue<pair<int,int>> pq2;

        long long sum=0;
        for(int i = 0; i < M - 1; i++)
        {
            sum += v[i];
            pq1.push(v[i]);
        }
        for(int i = M - 1;i < N; i++)
        {
            pq2.push({v[i],i});
        }
        long long sol = M * pq2.top().first - sum;

        for(int i = M-1; i < N-1; i++)
        {
            while(pq2.top().second <= i)
            {
                pq2.pop();
            }
            if(pq1.top() > v[i])
            {
                sum -= pq1.top();
                pq1.pop();
                sum += v[i];
                pq1.push(v[i]);
            }
            sol = max((1ll * M * pq2.top().first - sum), sol);
        }
        cout<<sol<<'\n';
    }
    return 0;
}
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        3 weeks ago, hide # ^ |
         
        Vote: I like it +1 Vote: I do not like it

        maybe pq is empty (m=1)?

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        3 weeks ago, hide # ^ |
         
        Vote: I like it 0 Vote: I do not like it

        wait so do u have to use a priority queue? i lowkey havent learnt that yet so i was wondering if its possible without.

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          3 weeks ago, hide # ^ |
           
          Vote: I like it 0 Vote: I do not like it

          yes, with the formula I reached:

          sol = max(sol, M * Bm — S)

          where S = B1 + B2 + ... + Bm-1

          I basically had to determine the best pair o minimum S sum and maxim Bm

          So you do need 2 priority_queues or a multisets (but the priority_queue is a bit better in time complexity)

          I used the first pq for the B1, ... , Bm-1 so I can always have the smaller elements <= i, and if something smaller appears I can quickly take out the bigest element I had and replace it (because the priority queue keeps the elements in decresing order)

          The second pq is for Bm, to always know the biggest element I have remaining in the interval from i+1 to N. And I only remove elements if they are not in the range any more.

          I hope that I answered your question clearly :)

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            3 weeks ago, hide # ^ |
             
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            yeah I think i get the solution now. What I did wrong during comp was that I missed the simplification at the start where the sum is just the max element * m — (b1 + b2...bm-1).

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              3 weeks ago, hide # ^ |
               
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              yeah, me too, I only realized near the end

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            3 weeks ago, hide # ^ |
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            also, pq does push operation in O(logN), and top(), pop() in O(1), so it is way more convenientthan sorting every time

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3 weeks ago, hide # |
 
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I really hate E1 with 5th test

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    3 weeks ago, hide # ^ |
     
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    I liked the problem and have no issues with it; I just noticed that many solutions failed on test case 5 (including mine), so I decided to write about it.

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      3 weeks ago, hide # ^ |
       
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      Its not a TLE error, logical error. Mine failed there, but I was checking if the primes dividing x and y are the same, and not if the primes dividing y is a subset of those dividing x

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        3 weeks ago, hide # ^ |
         
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        I realize the fault is entirely mine—I had a brain fart—but it sort of turned into a local meme that Test 5 fends off all attacks from incorrect solutions.

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3 weeks ago, hide # |
 
Vote: I like it -54 Vote: I do not like it

I was used to the 17:35 time and had it in my head that it started at 20:35; then I logged into the system and was shocked to see that the contest had actually started at 20:05.

Please don't do this again—only hold contests at 17:35.

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    3 weeks ago, hide # ^ |
     
    Vote: I like it -15 Vote: I do not like it

    Why are you downvoting? You could have just written that you disagree with my opinion and that this contest time works for you.

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3 weeks ago, hide # |
 
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Good contest, really bad time for the round :(

Its bedtime here for me lmao

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Deleted

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3 weeks ago, hide # |
 
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I didn't read 'exactly m' in problem b and then lost 40 min implementing a solution with two segment tree for frequencies and sums and coordinate compression. Anyway really cool contest, i wished i had more time for D.

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3 weeks ago, hide # |
 
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good contest

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3 weeks ago, hide # |
 
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we need more div 4 and div 3 contests please for newer

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3 weeks ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

I'd like to thank nik_exists for such a lifechanging contest!

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3 weeks ago, hide # |
 
Vote: I like it +11 Vote: I do not like it

Thank you nik_exists for this beautiful contest! Before this contest I couldn't get past outskirts in rain world and now I've beaten the game on hunter with my limbs removed. I enjoyed how every problem was about obscure number theory algorithms. Thank you.

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3 weeks ago, hide # |
 
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loved the round ❤️

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3 weeks ago, hide # |
 
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strange scores for problems and very hard D (at least for me and my friends ) but cool A-B-C thanks for the contest but I hope to have a more balanced D next time :)

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3 weeks ago, hide # |
 
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Yesterday I give this contest and Now I got a message regarding getting same answer copying from other user. I really do not understand why I get this message because I obviously solved Problem B by my own. It might be coincidence that user loly.talal might get same thinking process for this process, it doesn't say that we cheated on this problem. This is a div2 B and I do not think there might be different solutions for this problem, so it is common to get same solution for this kind of easy problems.

Please remove the “skipped” tag from my contest participation; otherwise, it may negatively affect my rating and could result in my account being banned.

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    2 weeks ago, hide # ^ |
     
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    I completely agree with you! This was a straightforward Div. 2 B problem, and it's totally normal for multiple people to have the exact same thought process and standard approach independently. There was no copying at all, and it's just a coincidence. Hopefully, the status gets resolved and our ratings/submissions are restored properly.

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how I can get Accepted with this problem?

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3 weeks ago, hide # |
 
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Hello,i'm not sure but System warned me for my D problem and i will try to clarify my state so i will explain in this blog(System said "If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details.").First of all thanks for not blocking my account and giving me a chance to clarify the state.Firstly i accept that some cheaters used same logic with me and their code is AI written(Which they got banned).I want to state that this code was written by me and in my style(You can check my old submissions).Also i tought that n<=2e5 for all test cases which is why i used memorization(also no other people used memorization for this problem).Normally i was not gonna solve this problem but i saw that "Dr. Agos wants a pattern with at most three lit pixels that minimizes f(s) among all binary strings of length n, including strings with more than three ones.". After i saw this i can divide the string in 3 which would reduce answer because a string which it's length is x and only have 0 contributes very very much to answer(x*(x+1)/2) so i need to minimize x.Then i tried all N values from N=1 to N=14(Which is why i spent 50 minutes on this problem)(and i want to upload my notebook page but i don't know how any help will be appreciated.) .After trying i had a general idea and implemented my idea.Also i want to state that Editorials solution uses my idea but it tries more than 1 possibilities.Well i can't prove proofs rn but if i manage to upload fotos this will definitely be helpful for me.Also i want to say that i really tried to get AC because i was saying a lot "Tomorrow i will be Specialist" to my friends from Olympiad and i spent a lot of time on B and C and according to Carrot i couldn't be Specialist and next contest is in 5 weeks and today schools have opened and i want to enter to new school year as a Specialist so i had a really big motivation in solving this problem (Even my friend mhdkr2012 said me that he was gonna send me all of his money to me if i manage to solve problem D.)So in the end i manage to solve this problem because of a big motivation.Also a lot of my friend knows me and they would help me with the case.TIA.

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3 weeks ago, hide # |
 
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I found some people thanks nik_exists for the great contest, but realizing that he is not even the author...

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@MikeMirzayanov

I received a plagiarism/coincidence warning for my submission 390640446 for problem 2264C, which was reported as coinciding with BlackAugust's submission 390638218.

I want to clarify that I do not know this user and did not copy their solution or share my solution with them. I independently solved the problem.

I understand that there are significant structural similarities between the two implementations, particularly the suffix sum, factorial calculation, and modular inverse calculation. However, these came from my own implementation of the solution.

My submission also contains an additional duplicate-value check and differs in several implementation details.

I would appreciate it if the submissions could be reviewed manually before any penalty is applied.

My submission: 390640446 Flagged submission: BlackAugust/390638218

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3 weeks ago, hide # |
 
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wait how am i yellow in the blog

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2 weeks ago, hide # |
 
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Subject: Explanation for Code Coincidence (Problem 2264C — ArefinMahir)

Hello Codeforces Team,

I am writing regarding the plagiarism warning for Problem 2264C (coinciding with user 'SuperIntelligence'). I did not intentionally share code or collaborate with anyone during the round.

The code coincidence occurred due to two unintentional public exposures during the contest:

Public Online Compiler: I experienced local setup issues running C++ directly in VS Code, so I executed my code using an online compiler without changing the default privacy setting from public.

Public GitHub Repository: I configured automatic code pushes to a public GitHub repository as part of learning Git and GitHub.

Someone likely scraped the public compiler output or public GitHub repository during the round and submitted the code.

I understand that public code exposure is my responsibility. I have changed my GitHub repository to private and configured an offline C++ compiler in VS Code to ensure this does not happen again.

Submission ID: 390651278

Matching Submission ID: 390651632

Thank you for your understanding.

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[Deleted]

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2 weeks ago, hide # |
 
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Hello Codeforces Team,

I sincerely apologize for the violation related to my submissions in the contest.

My Codeforces account 23h51a05w2_codeforces has been disabled, and I am currently unable to access it. Therefore, I am posting this comment from another account only because I cannot access my original account.

During the contest, I was experiencing problems with my PC. Because of this, I used Ideone to run and cross-check my code. I now understand that using Ideone with publicly accessible code could have resulted in my code being exposed and could explain the significant similarity detected in my submissions.

I sincerely regret this mistake. I understand that even if the code leakage was unintentional, I was responsible for following the Codeforces contest rules and for ensuring that my code was not publicly accessible.

I did not intend to gain an unfair advantage or share my solutions with other participants. Nevertheless, I understand the seriousness of the situation and accept responsibility for my mistake.

During the contest, I was experiencing problems with my PC. Because of this, I used Ideone to run and cross-check my code. I now understand that using Ideone with publicly accessible code could have resulted in my code being exposed and could explain the significant similarity detected in my submissions.

I respectfully request the Codeforces administrators to please review my case and consider restoring my account 23h51a05w2_codeforces.

I have learned from this incident and will be much more careful in the future. If my account is restored, I promise to strictly follow the Codeforces rules and will never use a public external service for contest code again.

I understand that the final decision is completely up to the Codeforces administration, and I will respect whatever decision is made.

I sincerely apologize to the Codeforces Team and the community and kindly request one opportunity to correct my mistake.

Thank you for taking the time to review my request.

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2 weeks ago, hide # |
 
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Again massive cheating, just block everyone from India.

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11 days ago, hide # |
 
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Hey someone kindly give this post an upvote it's currently at 1299 make it 1300.

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