True, as a participant of the testing contest, I will not be writing 'as a tester' comments since Anton sir told us not to.

How can we forget this torture to us :/

I guess the statements are so short that the author decided to call them questions.

Yes ofcourse. You have mine :)

Glad to hear that you liked the tasks,

If you are mad at anyone, you should be mad at me or the other setters. As a coordinator Anton did the best he could/was humanly possible.

Will the contest be difficult? Scoring looks fine tho

Is it a trap again? :)

I guess may be C will be interactive. As it is also divided in two significantly easy questions.

is't it unusual start time ??

I was surely going to miss this round. Thanks :)

It doesn't necessarily have to be a binary search problem, it can either be a segment tree problem or even not a certain tag.

Exactly xD

We have testers to virtual the problemset and they give feedback on whether the contest should be longer or shorter. And we kinda agreed that 2 hours 30 minutes was good.

• We hope you find the bugaboos interesting and fun! Wish you high rating!

Please read annoucement more clearly next time

It means there will be C1 and C2 bugaboos, with C1 being easier, therefore scored 750.

We do all the time, but only to the authors/coordinators. This is because we like to watch contestants suffer, so we have every motivation to avoid telling if the round is bad. And we might even lie and tell that it is good when it's not.

(Anyway, this round is good.)

I will, when I test a round (sad face) :(

bugabooset Done

I think it's a trap. They are misleading the harmless contestants to just walk into the world of Binary Search.

Not just any one of the problems. I bet it will be the interactive problem. And most importantly, it's not even surprising.

There will be 2 sub bugaboos C1 and C2 with C1 be easier as compared to C2.

cyberphobia, I guess !!

:eyes:

You mean bugaboo statements?

hahaha.. people trying hard to create memes to get some upvotes.

Your current streak is just 1.

How are these problems interesting? Just speedforces from A — C2.

Such a shitty problem!!

After watching solution you will hate yourself (for some time) xD

Weird flex but okay.

I think we really did not know about this. So sorry about that.

I am very sorry, I will try to research more next time.

Yes. C1 could be done by DP though.

yes , tbh i found it easier than B. In C you just need to remove most negative element when overall sum becomes less than 0 . Let's hope it passes pretest .

Yeah. You maintain a value h (health), ans (longest sequence), and Q (a priority queue of used potions), and at each element, if h + a[i] >= 0 you increment h += a[i], ans++, and add a[i] to Q. If h + a[i] < 0, then check whether we can improve h at that index by swapping the smallest used value in Q for the current value of a[i].

Yeah it is kinda greedy. For every i from 1 to n let's say ps = (sum of positive integers till i) and ns = -(sum of negative integers till i) and also you maintain all negative elements in a priority queue. So you will remove the least elements until ns<=ps for every i and update answer by counting i-(number of removals).

You can solve it also using a segment tree.

1430E - String Reversal

It's the inversion count

Let x be the first character of A. We can always say that the first occurrence of x in B will be the character that will go to the first slot in B(otherwise we can see that for any other occurrence of x should swap with the first occurrence of x and by doing that we are swapping two same characters which is not optimal). So now whe have the following algorithm: Take first character in A — x. Find the first occurrence of x in B. Move x to the first slot. Remove both xs from A and B and repeat until there are no characters left. You can store the list of position of each character in B and using Fenwick tree you can efficiently calculate the position of x.

If x > 11 * 111 — 11 — 111 (frobenius coin problem), answer is YES. Otherwise, answer is yes if x is obtainable by combination of 11 and 111.

I thought of it this way. If you add 111 to any multiple of 11, you are basically adding 1 mod 11. So if your given number is >= x*111 , you can shift the mod by x times. It's easy to see that if x>=10, any number is possible.

Making his DNA beautiful.

The first pretest is just the sample test from the statement. And the OK message is probably the same for all test cases (in this contest at least), but obviously the system won't show more testcases before the end of the contest.

How to solve it ? I summed the position at which each character is occurring and then put those characters whose sum is lowest in last.

Is just looking mod11, it can be done in O(1) per query with 11 if

If n >= 11*111, you are guaranteed yes, because you can subtract values of 111 until you get a multiple of 11.

Then we know we are only concerned with 1111, 111 and 11. We can brute force every combination of these that doesn't take us over 11*111, and update an answer array with YES for each found value (default is NO). Precompute this to avoid TLE.

Every number with even number of ones is a multiple of 11, and every number with odd numbers of ones is 111 + a multiple of 11. So the answer is 11x + 111y. Floor(n / 11) should be >= (n mod 11) * 10 to have an answer, i.e., trying to take every 1 in the mod with 10 elevens to form 111.

While testing, current problem B used to be A. We were made to solve it as A :notlikeduck:

Actually there are few cases at each index(from left to right). If current value is negative and reduces the total current sum to < 0, then just check minimum element from priority queue(of negative numbers which have seen and added to the sum until this step during the iterations) if this value is less than it, just ignore this step and go to the next, otherwise just pop priority queue and push this value into it, also change current sum. Note that 1) We need all the positive numbers, it's not the problem of course. 2) If current value is bad to choose this time, it's also bad idea to pop pq twice or more, it's much better to skip this value only(or swap with minimum negative to improve current sum). If current value isn't bad, then just increase total amount and change current sum too(also if this value is negative, push it into our pq of negative numbers).

// Below see code

include <bits/stdc++.h>

using namespace std;

define ll long long

int main() { ios::sync_with_stdio(false);

int n;
cin >> n;

vector<ll> vec(n);
for (int i = 0; i < n; i++) cin >> vec[i];

ll res = 0;

priority_queue<ll> pq;
ll cur = 0;
for (int i = 0; i < n; i++) {
    if (cur + vec[i] < 0) {
        if (!pq.empty() && cur - (-pq.top()) + vec[i] >= 0) {
            if (-vec[i] > pq.top()) {
                // don't need do anything, if this value doesn't improve sum
            }else {
                cur = cur - (-pq.top()) + vec[i];
                pq.pop();
                pq.push(-vec[i]);
            }
        }
    }else {
        if (vec[i] < 0) pq.push(-vec[i]);
        cur += vec[i];
        res++;
    }
}

cout << res << endl;

}

The place under the picture should be mine

Cause I still don't know how to solve A constructively, just random shuffled it

How come? You haven't even participated

A DIV1+DIV2 contest is in two days and you are eligible to participate in it. Also more DIV2 contests are likely to show up in the schedule. Many of them are announced with just a few days notice.

You can only hack those people who are in the same room with you (after locking the problems).