What F*ckhead created problem C ?

You showed us how to get a negative contribution in the fastest time.

o..orz

It looks like a kneeling man
Similarly, there is "sto"

Thank you <3 And the video editorials will be uploaded on our YouTube Channel. The links will also be in the editorial.

$$$\color{red}{\text{orange}}$$$ !

oh god no not pupil again

The prizes are only for Indian participants to avoid logistic issues faced during international shipping.

[DELETED]

Well, well, well.. how the turn tables

One of the best? Seriously bro? Its undoubtedly the best one

Hey!

I am really sorry. This year our logistics allow us only to deliver T-shirts within India. For future years, we will try our best to bring international prizes too.

Hey!

Thank you for asking this, I realize that the prize distribution which is written in the post is indeed a little unclear.

The purpose of giving T-shirts to Indian participants only is to avoid the logistic issues that we have to face during international shipment. Hence, by the Indian participants, we mean the people who have a home in India, so that we can deliver the T-shirt there. People who have their Country set to India in their CF profile will qualify as Indian participants.

Apologies for this confusion. I will update the blog soon with this detail.

That's what she said

That's what she said

Becoming specialist today means speed-forcing A,B,C. Let's see if you can?? Hope for the best . Prepare for the worst

High Sir!!

No god please no, I don't want to be specialist again

hint?

Just move a 1 to the right if you can, and then to the left if you can.

ans is 10 but you code gives 11

I started codeforces recently and got super mad at cheaters but now I stopped caring. It happens every contest man. That’s why no one responds anymore. There are too many cheaters to do manual check, and they just find ways to trick the plagiarism detector anyway. It stops matter once you solve D or above

did without segment tree code

How is D giving AC for some solutions with O(n^2) time complexity for other with completely similar logic

Do a classic sweepline. The main observation is that if at some moment during the sweepline there are open intervals of both colors, then all those intervals will be in the same connected component. Therefore, if there's a moment with both colors present, it's enough to unite all active intervals, and keep the interval with largest right endpoint of each color. Notice that during each moment in the sweepline, the set of active intervals will be either unicolor, with each interval belonging to a different connected component, or both colors will have one active interval. Therefore, when encountering a new interval it's enough to unite it with all intervals of the opposite color, and the complexity will be $$$O(nlogn)$$$.

I like overkilling tasks, so here is my approach, but it's basically the same as previous answer. Let's create a segment tree, for every red segment we will add it to vectors of nodes covering this segment. So it's standard adding on range operation, instead of increasing some variable, we're pushing to vector. Now, for every blue segment iterate down the tree. When you're in some node, add edge to every red segment in this node, now all of them are connected, so it's enough to leave only one of them in vector.

Repeat this approach for blue segments stored in tree and queries for red vertices.

Now you'll get something like n * log(w) edges, where w is maximum coordinate of segments. So you can do standard DFS now.

Just tedious casework...

        if(L && R && L + R <= k && cnt > 1) {
            cout << ans - 11 << "\n";
        } else if(R && R <= k) {
            cout << ans - 10 + !(L || cnt > 1) << "\n";
        } else if(L && L <= k && (R || cnt > 1)) {
            cout << ans - 1 << "\n";
        } else {
            cout << ans << "\n";
        }

Here $$$cnt$$$ is the number of $$$1$$$ bits, $$$L$$$ is the distance from the beginning to the first $$$1$$$ bit, $$$R$$$ is the distance from the last $$$1$$$ to the end. And if $$$cnt=0$$$, output $$$0$$$.

sum is minimum only if there are 1's on both the end points, for example: 100001 is better than 010100. Just check for closest 1's to index 0 and index n-1 Also priority should be first 1 at n-1 index than at the 0th index

Every element except 1st and last element contributes at ones place and tens place. Last element contributes only to ones place. First element contributes only to tens place

B can be proven by noticing that if there's at least an element that's unique in the array, then you can't swap it with anything. Why? Swap with a smaller is disallowed, and swapping with a larger element is also not allowed, bc the larger element now has the smaller element. Then just swap indices with the same integers, a way to do it is in the example.

For C, any 1 that is not at 0th or n-1 index will add exactly 11 to the sum. So you just have to try to move a 1 to the back first, and then to the front.

Cuz if u flip the idx of a section with an odd amount of people, the middle person in the section has the same shoe as before. In ur example, person 4 gets his own shoe back.

Bro read community guidelines. Don't post full code directly. Post in either spoiler or post link

Failing testcase: Ticket 8946

Failing testcase: Ticket 8941

I think you have done mistake in case when total 1 count is 1. Try this 7 2 0000100

Case 3 of B is n=1

ok i missed

I got it... I actually am retarded.... I have print instead of println and I misprint.

wrong output format Expected integer, but "-1-1-12" found (test case 3)

Your pretest 2 fails not because of the judge but seems like it's the compiler. You are probably running your code with no optimizations. Try adding -O1 (or 2,3). It segfaults. Exploring why this happens further.

plz never use unordered_map.

Except problem C , it was not good

You need to check all intervals [i .. j] where either a[i] or a[j] is the largest number in the interval

Someone hacked you before me :( Here is my testcase https://pastebin.com/6aBxXZuP

I can't believe I got beaten to it by tourist because I somehow copied my testcase incorrectly. I'm so mad right now.

Don't use unordered_map.

https://codeforces.me/blog/entry/62393

Don't use unordered map

https://codeforces.me/blog/entry/62393

Unordered_map can be made very slow by carefully selecting certain input values.

https://codeforces.me/blog/entry/62393

You would still, probably!

Not sure about the simple part but people are discussing it above

Nope, D has 4 times more solves than E. It's your personal opinion.