The best educator of them all!

Unfortunately, I am not sure that vovuh is back :(

This group of people is always the creators of Educational Codeforces Rounds. For example, you can check the last one -> vovuh was among them, but was not mentioned in the tutorial

He also said he would return with a div 3 round in this comment

I'm done with memes. No more memes from now!

Still waiting for official Editorials. Problem setters should make editorials along with problems.

Yes, you are rigth

Yes. How many AC, then how many minutes (penalty).

Good to know, I have participated in a few educationals but did not know this one. :O What I noticed about them is that they tend to be harder than normal Div2 rounds.

which means

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Bruh Bruh Bruh Bruh

Seem like an Educational round's announcement is incomplete without someone reposting this meme.

I think entire blog is almost copy pasted, maybe so

Typical Indian households

True AF

That's why it seems peaceful to do contest in hostel room.

Damn, right in the feels.

It's way better than what I experience almost every time.

Or when they don't talk to you for the entire day, but the moment your contest starts, they decide to hold a party in your room.

So true, XD

xswl

HTML IS A PROBRAMMING LANGLUAGE!

Good idea, but they wont accept. We are doomed =(

Educational round is nice but div1 contest is rated for 2100 <= and it is a Competitive for 2100 <=

But after seeing that there code are same even comments are same my heart like "OOOH LALA";)

Having fun? :p

Spend some time on cf giving contests regularly and you will be able to answer this yourself.

He is the chosen one who sends all undeserving people like me from specialist to expert.

Yes

</almost copy pasted part>

Of all bad and misworded As, today's A was the worst.

Actually, in Educational round it doesn't matter at all)

idea was simple but problem statement could have been worded better

I would suggest don't give up and keep trying :)

i feel the same way, contestants are doing much better recently

todays div2D<<usual div2D

Off-topic:suddenly notice , after a long time you change your profile picture..

set answer, low , n to 0

iterate the string
if '+' n++, else n--
if n < low, answer += string.position, low = n

print answer + string.length

i got WA and spent so many times to realize ans can't be store in int

I think with enough practice on this types of problems, it would become intuition. I used to find these types of problems really hard, and after solving and understanding them they become easy to solve.

Yeah man, the difference between D and E was huge today.

My solution for problem E: https://mirror.codeforces.com/contest/1373/submission/85070127

The key insight is that K<=9 so you can only overflow at most once. For example if you pick the last digit to be x%10=7 with K=5, then it must look like this:

(front    ) 7
(front    ) 8
(front    ) 9
(front + 1) 0
(front + 1) 1
(front + 1) 2

So for every fixed starting digit and K, the sum of all digits is:

sumOfOnesPlace + numNoOverflow * sumOfDigits(front) + numOverflow * sumOfDigits(front + 1) = N

Where front is the variable we want to solve for and the rest are constant.

If numOverflow is 0, you can solve for:

sumOfDigits(front) = (N - sumOfOnesPlace) / numNoOverflow

If front doesn't end in a 9, you also know that sumOfDigits(front) + 1 == sumOfDigits(front + 1), which again lets you solve for

sumOfDigits(front) = (N - sumOfOnesPlace - numOverflow) / (numOverflow + numNoOverflow)

Once you know a target for the sum of digits of front, you can greedy it.

This isn't complete because I didn't cover all the cases, but I am guessing other cases are impossible via proof by AC. :)

You can check out how to come up with the solution here :D

Yeah bro I also initially wrote code for 1 based indexing and after checking for test case 1 I felt something is wrong & and read the question again and realised that it is 0 based indexing

void solve() { int n,i,j=-1; ll ans=0; string s; cin>>s;

n=s.size();

vector<int> pre(n,0);
pre[0]=s[0]=='+'?1:-1;

for(i=1;i<n;i++)
{
    pre[i]=pre[i-1]+(s[i]=='+'?1:-1);
}


for(i=0;i<n;i++)
{
    if(pre[i]==j)
    {
       ans+=i+1;
       j--;
    }
}

ans+=n;

cout<<ans<<endl;

}

Here is a detailed explanation for the problem

Nevermind, figured it out:

Sometimes it is optimal to put an 8 as the first digit after the 9s and 1s digit, rather than the remainder of the excess sum divided by 9.

The video editorial for the D problem . Please subscribe to our channel :)

How did you generate those values?

How did you map all the values, did you get any online tool to calculate that ?

How did you generate values?

Is that allowed? Or would it be considered cheating

Meanwhile setters: Am I a joke to you.

FBI wants to know your location

Codeforces be like :

Can you prove it?

I used — if (k >= 2) then apply brute to find an optimal answer, and greedily solve for k = 0,1 but got stuck at T-3

you can find a detailed explanation here in the video

Or prefix sums while maintaining min

Can you help with some intuition on, how to solve F with kadane's ? Thanks

my solution: i simply assigned all of the capacity of the i-th edge to the i-th node and greedily give the next node the cumulative excess capacity while paying attention to the capacity of the edge. i traversed through the graph twice since it is a cycle and checked for validity on the third pass. 2 passes might be enough but i was being safe and did not want to waste time on checking for correctness. in all honesty i am surprised this simple solution worked.

edit: fst :(

i also got stuck in testcase 4 of D, but it is just integer overflow, i did't realize it until the last fifteen minutes.

you can find a detailed explanation here in the video

I completely ignored flow because the limit was so large. Is it possible to pass?

Its memory limit gets exceeded anyways with Push Relabel (with gap heurestics). https://mirror.codeforces.com/contest/1373/submission/85062675 (Total 2 * n + 2 nodes are needed to model the problem into flow, with capacity taken in long long).