IMHO Problem B to me was like you either know it or you don't. Like, I can't get a piece of paper and start working on it and finally get an answer, but 7000+ people solving it really got me questioning my existence.

That makes sense. Thank you very much <3.

thanks

as long as it's fun for me I'm ok with it :"O

thanks ^^

thank you so much <3

thank you for the suggestion, i will read it.

Thanks :)

thanks :D, keep practicing and you will reach it

thank you

thank you, i practice both skills

Thank you, I'll definitely keep practicing

thanks, i hope so too :D

thanks, i'll keep grinding

thank you

thank you, I appreciate your words.

thank you bro, you definitely will

thank you, you too!

Thank you, i will definitely keep going!

Thank you

Thanks :D

so i see that in problem B, if we replace every A with a ( and every B with a ), it becomes a "valid bracket sequence problem", why the solution works for both, can anyone elaborate ?, thanks in advance.

the submission you mentioned looks like mine (i submitted it after the contest), and mine only passes if I use the "Ofast" optimization option.

1) don't know why the submission you mentioned is faster than mine.

2) don't know why both submissions pass.

this is my code: https://mirror.codeforces.com/contest/1661/submission/153254977

probably mine is hackable.

Can you help me with why this TLEs for problem B, I'd appreciate your help, thanks in advance.

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N;
    cin >> N;

    auto&& solve = [=](int64_t X) -> int64_t {
        std::queue<pair<int, int>> q; // pair of (number, how many operations done to reach it).
        q.push({X, 0});

        std::vector<bool> done(32768 + 1, false);

        while (!q.empty()) {
            std::pair<int, int> u = q.front();
            q.pop();
            done[u.first] = true;

            if (u.first == 0) // if the number (u.first) is congruent to 0 (mod 32768).
                return u.second; // return the number of operations done to reach it.

            for (auto nxt : {(u.first + 1) % 32768, (u.first * 2) % 32768})
                if (!done[nxt])
                    q.push({nxt, u.second + 1});
        }

        return -1; // shouldn't reach here.
    };

    while (N--) {
        int X;
        cin >> X;
        cout << solve(X) << ' ';
    }
}

[edit]: understood the idea

Thank you so much, this helps me a lot.

Thanks for the advice, definitely will start considering it while practicing.

you need to know something because i knew people would get this post wrong, i am NOT against the hardwork strategy, infact i ENJOY the process and would love to keep doing it until i am a better version of myself, but the idea is for example: person "A" reached expert in 2 yrs & person "B" reached expert in 3-4 yrs so, was person "A" Smarter in the way he approaches CP ?, we all are here to learn, you learn CP because it fits your rating, i learn how to learn CP because it fits my rating.

no brain me solving question A by shuffling the array multiple times codeforces: we wont let u cry this time, Accepted

Me After Seeing The Delay:-

is it delayed ?

I have a talent at not being able to solve questions

i learned a lot from this round, good job!