orz

Bro u have just cancelled yourself

Please don't ban orz!

This is my ALT account. I am legendary gaand masturbater on my main account.

Make girlfriend

Sorry, it was an our mistake, fixed now. Rating also will be recalculated.

Why was this contest rated for 7-star coders, above other Starters? I felt the problems were quite easy for a "div 1" contest.

Yes

L

L

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I agree.

L

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These "science" channels are cringe-worthy. Most of their viewers are people who like feeling smart without doing shit.

This phenomenon occured due to the first comment being so close to lord Sparky_Master_WCH1226's comment. It basked in the divine glory of our overlord, and immediately got +69 delta.

L

what about telling you that it is helpful for me now after 9 years (:

nice pfp

I don't think checking for SCCs is required at all. A simple DFS should be fine.

Let $$$v_x$$$ be the node representing the value $$$x$$$, and $$$i$$$ be the node for index $$$i$$$ ($$$1 \leq i \leq N$$$).

Observe that if an edge $$$v_a \rightarrow v_b$$$ exists, then $$$b$$$ is a submask of $$$a$$$, which means $$$\neg a$$$ is a submask of $$$\neg b$$$. So the edge $$$v_{\neg b} \rightarrow v_{\neg a}$$$ also exists.

For simplicity, consider the path from node $$$i$$$ to $$$j$$$ that passes only through some value nodes:

It's easy to see that a path from $$$j$$$ to $$$i$$$ is guaranteed to exist:

Or instead: proof by AC :)

My sleep cycle literally cycles over the period of 10-12 days.

It initially starts off like a regular persons: I wake up at 7-8 AM and sleep at around 12 AM,

Then I start incrementing the time at which I go to sleep by 1-2 hours everyday until I reach the point where I sleep at 11-12 AM and wake up at 7 PM. At this point, I feel pretty retarded, so I decide its time to hit the reset button. I skip sleeping that "day", stay awake for about 30 hours, and then go to sleep the next day at about 12 PM like a regular human being.

I still hate abstract algebra

orP seems the worst to me

L

How ? I see you solved the 8 questions and global rank 4

Do 100 pushups a day until you reach Specialist.

agree +

L

Give Money to people who comment here

It is true that this contest was easier than some previous "Rated for All" Starters. But as mentioned here, we do expect it to be sometimes less challenging for the highest rated participants now. The range of "Rated for All" contests has increased, and so there will be some contests where IGMs solve all problems without too much sweat.

We just feel that there are enough users in the 7-star range as well, who find these challenging enough to have it be rated for all. For eg. in yesterday's contest, there were 25 7-star coders, and 8 of them solved all 7 problems. But 10 couldn't solve 1 problem and 7 couldn't AC 2 problems. Given that we definitely cannot have contests every month where the number of AKs is 1 or 2, we feel that having occasional easier Rated for All contests is a better compromise than having much fewer Rated for All contests.

Maybe from the next time, we'll also add something along the lines of "We expect it to challenge IGMs" or "We expect IGMs to AK the problem set" in the intro blog. Would that help?

No wonder why this braindead has negative contribution

with this trend, we're gonna make half of codeforces (finally) get a girlfriend for the first time

Use '\n' instead of endl. Your code will get accepted.

         O
stO stO /|\   Orz Orz
        / \

I didn't notice this. What problem was it that had its TL changed ?

Sorry, I made a mistake, I'll fix it soon

Probably my CP community in my university. Here's the story, when I first got into college. I had a clash with my seniors, who apparently got very outraged because I didn't show them any respect despite the harassments they caused me in my freshman year! (Yes, that's what they expected me to do, bully me into submission and hope that I will respect them for it). Well, I didn't. Thus, for next three years, not only that I received no support from the university, but now after three years have passed, they intentionally didn't let me participate in Inter-University Programming Contest and another Intra-university programming contest for which they showed me the cause that "I know what I did". About how do I overcome my obstacles; There are obstacles in life and so there are things to overcome. I have come a long way in spite of the obstacles I faced, in fact, because of them. They are my motivation and the reason that I keep pushing forward.

By doing that you wouldn't train your problem solving skill. But I think it maybe still be useful if you want to get familiar with different approaches used in editorials. As a conclusion I think if you already have decent problem solving skill and want to get grip of approaches and topics in competitive programming it is probably good strategy.

L

That shit is happening because you are not taking care of overflows in Java. You can read the editorial to see how it should be done.

If you won't reach specialist till end of July You won't quit cp untill you reach specialist and when you reach specialist you will set your goal to expert and repeat this again

Maybe before the TL changed.

I don't want to be rude, but, seeing your (very low) rating and previous blog, I don't think you have the required experience to write a useful book, even if that book is only for beginners. And no, you can't write a whole book with just a dozen comments from this CF blog.

++

++++

+

Using C++20:

#include <ranges>

vector v = {1, 2, 3, 4};
for (int x: v | views::reverse) 
  cout << x << ' ';

// or
for (int x: views::iota(1, 5) | views::reverse) 
  cout << x << ' ';

I'm not sure about who has tried that before, but I know about a lot of people that got really good at codeforces and didn't try this. (sorry for bad english =) )

They agree (and me too) that the best way to improve is to try your best solving the problem alone. Because if you get used to reading the solution you will become lazy at thinking and probably have bad results reflected in your ranking.

TIL 7:30AM happens only in Bangladesh

As far as I remember, hacking doesn't give extra points in div.3/4 and edu rounds, and in div.1/2 rounds, you can only hack in your own room, and there are hundreds of rooms so getting your alt in a room is unrealistically time-consuming and stupid.

In my opinion, they may be testing the hack system for themselves so they can hack others more smoothly next time, OR they just want to harmlessly brag about making a hack. Either way, this isn't cheating.

Cringe. Downvoted by all my accounts.

answer: no

reason: subset sums is (weakly, but still) NP-Hard ($$$O(2^{n/2})$$$ is possible tho)

or maybe yes if u prove $$$P=NP$$$ but who would even prove it so soon

orz to all 4 of them

Our IOI team is very strong this year!

to create a book, you need to be at least a master, and you are a beginner.What will you teach people if you have no experience?

he didn't insult, he spitted facts

In Thank U, Next, why is distance the number of nodes and not the number of edges in the path?

make child

This is the second time you've been caught doing this. The codes are not just similar, they are the same. Also, you are from the same location. Given all this, I refuse to believe in such an incredible coincidence.

I have another similar approach for problem F.

• Step 1: Multisource BFS can be done up to a depth of $$$d$$$ for every R.
  $$$d = time[R] - currentTime + 1$$$
• Step 2: Then make the entire row/col visited as 0 for each query R.
  $$$visited[i][j] = 0, (i/j = fixed, j/i = varies)$$$
  $$$currentTime = time[R] + 1$$$

Repeat Step 1 and Step 2 until $$$(N,M)$$$ cell is reached. $$$visited[N][M] = 1$$$
Time Complexity: $$$O(NMR)$$$
Space Complexity: $$$O(NM)$$$ (better than editorial solution)
Implementaion is easier and more intuitive idea than dp.

Code: 208794165

Auto comment: topic has been updated by SP_22 (previous revision, new revision, compare).

Got it! Thank you so much for the help.

(string)a==(string)b isn't $$$O(1)$$$, it's $$$O(\min(|a|,|b|))$$$

Spray paint "All Hail Patrick Bateman" onto the main entrance of your nearest mall.

will be a bit irrelevant but I want to ask how do you add grid/spreadsheet to your blog

make an alt account

make blog and mention you original account

infinite brain.

Productivity based on waking time varies from person to person. For some they might be more productive at late night, for others maybe at early morning. As for your question, I wake up around 7 to 8 am and sleep at around 12 or 1

Today, 07.06.2023, me , master fzx with rating 2103, sets my self a challenge to reach grandmaster rank before the end of the year.

UPD : Fixed

I think it would be worse if they weren't grown men

Good luck to you guys!

P/s: I am a little bit curious about why two of your top six guys might not be able to compete. I understand if you don't feel comfortable answering this question.

I am legendary grand masturbater too

Escaping grey means solving problem A,B (and problem C when contest is easy) of every div.2 contest. Try out as many as problems as you can, have a thorough look at Problem A,B of recent div.2 contest and also try to increase your speed by setting a time limit while solving easy maths,greedy,constructive problems in practice. Master atleast some very basic concepts like modular arithmetic,gcd properties . Solve problems in practice and learn from your mistakes, try to upsolve in every contest.

The last statement is false; regardless of whether hacks give points in this contest, this is still cheating. Using alts in contests, especially using two accounts concurrently in the same contest, is not allowed.

One can use the alt account to submit a wrong solution, making corrections until they get AC, then submit the AC solution on their main.

Hacked :)

For example managing the positions $$$x$$$ which $$$S_{1,x} \neq S_{2,x}$$$ and $$$x$$$ isn't locked, with some data structure or smarter way.

Now, you've hacked the editorialist's solution itself.

Ural FU: reshke Friska ArtyomTsepov

You can use a merge sort tree or a wavelet tree

Yes please

Here is my guess, since there is no statement in the problem which says: "the sum of all n over all test cases is less than 10^7," it is right to assume that there can be 1000 test cases where n = 10^7.

To initialize your dp array of size n + 1, it takes O(n) time. And there are t test cases. Since you are making a new array each test case, the total amount of time to make your dp array is O(n * t). n is at most 10^7 and t is at most 1000, so multiplied this is 10^10. Those many operations will certainly time out.

something like a multisource Dijkstra, idk what else to call it

BFS. Put every node x[i] in a p_queue acc to their d[i] and do bfs if d[i] become 1 for any node remove it from queue. Once your queue become empty check every node is visited or not

Sorry for the late reply. If I'm not mistaken, Vladithur had a deterministic solution while testing, I've reuploaded it here.

Spain's IOI team is:

• Innokentiy Kaurov (ikaurov) — 0 attempts left

• Alejandro Vivero (Esomer) — 1 attempt left

• Pablo Sáez (Pablo-No) — 0 attempts left

• Darío Martínez (MeGustaElArroz23) — 0 attempts left

sorry for that.. i actually mean the same.. as input < 2^15 then sum of 2^0 t0 2^14 is 2^15-1 as u said. Thanks for find my mistakes

Can you add the Country/Region?

It sometimes gives me problems that I have already solved. Pls fix this bug. Other than that, it's awesome!

Bro ,you have to reorder array A not array B ... then you can reorder B as follows : 2 4 5 2 5 4 4 2 5 5 2 4

I guess, you can use bitmasking for this approach, wait lemme write it with the code.

#include<bits/stdc++.h>
using namespace std;

int main(){
    
    int n;
    cin >> n;

    vector<pair<int, int>> pr;

    for(int i = 0; i < n; i++){
        string s;
        cin >> s;
        int len = s.size();
        int mask = 0;
        for(auto &a : s)
            mask |= 1 << (a - 'a');
        pr.emplace_back(mask, len);
    }

    int mxProd = 0;

    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            if( (pr[i].first & pr[j].first) == 0 ){
                mxProd = max(mxProd, pr[i].second * pr[j].second);
            }
        }
    }

    cout << mxProd;
    
    return 0;
}

I will do the action of asking someone; however, no promises on me succeeding (assuming this ranks at least top 3 most liked comment which I'm scared it will).

reach candidate master.

I wanted to read a blog, not to watch a video. One might get distracted by the YouTube algorithm and waste a lot of time........... say it in words so that, others like me don't suffer

"Positions"

number of palindromic numbers below 2e15 are 427

so precalculate all those number and iterate in array and see how we can form those numbers

You cheated in a contest and solved 2 Div 4 problems??? and that too in a near 90 minute gap???

You would have been better off not telling us. Fucking disgrace to all cheaters let alone people who actually try in contests.

What is orz ?? now the toxic CF community will downvote my comment...

No bro

You are a cheater !