According to the score distribution, it seems a bit easier than last div2 Round ?!

So now the LGM tag isn't even good enough for the "And You for participating!" line. They've trampled all over the meaning of the word legendary.

Believe it or not, I tested.

As a participant, i will try to solve A within 5 minutes.

Hope we can have easier E and F to solve. E and F for last Div.2 are not fit for rated participant(but it has nothing to do with me thinking they are good problems, especially E is interesting), only two can solve E and no one can solve F is abnormal.

what is the contest duration

I think this is the new shortest official contest blog, beating the previous contest's blog.

Hoping for the contest to have >2 solves on E, F! (oops)

B seems harder than usual

If I don't become specialist in this contest then I'll

I think it will be interesting round !

This has probably been answered before, but why not have all the rounds on weekends? Americans can not compete in standard-time weekday contests.

Like the last contest, I don't want to mess up this one either. :)

hope i will become specilist i this contest

hard contest IG... 500-1250

"You will be presented with at least 6 and at most 6 problems" interesting description.

КЛЯНУСЬ МАМОЙ ВОЗЬМЦ ЦИАНА

score distribution looks so puzzled, especially second and last ones. Will test!

As a participant, I will try to beat rainboy ;)

believe it or not, i wish good luck to everyone(also to me, i hope ill finally reach pupil~)

Im very sure that this will be one of the contest of all time

Guys I think there are 6 problems

What will be a 1250 score div2b like be?

I just want to break the curse of problem c. I'll definitely solve it this time ^.^

You will be presented with at least 6 and at most 6 problems.

will be ending myself if i cant solve D.

My First contest. Hopping to solve at least one!!

how many question i have to perform for becoming specialist?

hopefully pupil

Can't wait to participate.

I will solve at least 0 problems and at most 6 problems.

Wish everyone have a good time!

hoping to solve C this time and D if possible

It's nice

Hoping I could get into 1600++ :D

good luck

Serious disappointment after waiting for a week.

insane gap from c to d

wtf diff gap A *900 B *1400 C *1500 D *2200

B > C

rip

Why continue to make problems like C,it serves no purpose, extremely disappointed by the quality of problems.

first 56 minutes: Solve Problems

last 1 hour 4 minutes: stare at standings 💀

B >>>>> C, the order of B and C should be swapped. Who started with C will very likely to win a lot of rating in this contest

Also I have absolutely no idea how to solve D sadly. (last div2D was way better tbh)

Am I the only idiot that can't even figure out what B is asking. I swear the diagram they've given is wrong but maybe I'm just dumb.

problem C is such a piece of shit

    int n; cin >> n;
    vector<int> ans(n);
    iota(ans.begin(), ans.end(), 1);
    ans[n - 1] = ans[n - 2] = 1;
    for (int i = 0; i < n; i++) cout << ans[i] << " \n"[i == n-1];

why this allowed?

Problems were pretty difficult but also really nice.

Thanks for the great contest!

This is (in my opinion) one of the best div2 rounds I have ever participated in. Thank you for the brilliant problems. I enjoyed every problem (except B). Ran out of time in F2 after getting the main observation in last 10minutes :(

FastForces intensifies and C << A << B

I couldn't solve D, but it was an amazing problem imo.

For D, is it all odd lengths are automatically the answer calculate in O(n) for even length you sort the array of (value,index) in increasing order and then solve the equation

max(l.second, r.second) - min(l.second, r.second) = r - l

Use some Data structure to calculate this for all pairs having array[i].first == array[(i+1)].first

thank you for contest!! D was good problem. However, I am not too sure about the point of question C

I think graph was introduced in problem B just to scare the participants.

6000 solved C, 500 solved D nice complexity management!

E seems way easier without the equality in the 2nd condition. How to solve with the condition having $$$\leq$$$?

What a big gap bitween C & D!

Can someone tell me which test case my code failed in B. https://codeforces.net/contest/2056/submission/301455594

I thought I would solve at least 3 problems. I was really sure that C will give me hard time but well life can be very surprising sometimes.

tf is problem B?? problem D: wow small question(length wise) lets try it, tried it and was able to solve it in first attempt with time compl of O(n^3logn),optimized it and submitted still time limit exceeded.

Can anyone tell me what's wrong in my code?

// this is code
void solve() 
{
	ll n; cin >> n;
	vector<string>arr(n);
	vector<int>pos(n, 0), ans(n, 0);
	for(int i = 0; i < n; i++)
	{
		ans[i] = i;
		pos[i] = i;
	}
	for(int i = 0; i < n; i++)
	{
		cin >> arr[i];
	}
	for(int i = 0; i < n; i++)
	{
		int to = i;
		for(int j = 0; j < i; j++)
		{
			if(arr[i][j]=='0' && pos[i]>pos[j])
			{
				to = min(to, pos[j]);
			}
		}
		if(to==i)continue;
		int val = i;
		while(val-1>=to)
		{
			swap(ans[val], ans[val-1]);
			pos[ans[val]] = val-1;
			pos[ans[val-1]] = val;
			val--;
		}
	}
	for(auto &x : ans) cout << x+1 << " ";
	cout << endl;
}

hate speedforces and B ate all my damn time. 5 more minutes and could've done C whatever looking forward to better weekend contests

be me
spend almost 1 hour on problem B
can't solve it
reluctantly switch to problem C
solve it in 2 minutes flat
realize I only got 1000 points out of 1500 on C because of B
mfw B ruined everything

not even in my dreams would i expect seeing graphs in a div. 2 B lol

In problem C why we cannot do

int n;
cin>>n;
for(int i=0;i<n;i++)
cout<<1<<" ";

In problem D, I came up with the following conditions for subarrays with even length $$$l$$$ and median is $$$m$$$:

If $$$x$$$ is number of elements $$$<m$$$ and $$$y$$$ is number of elements $$$>m$$$, then $$$l-2x \geq 2$$$ and $$$l-2y \geq 2$$$.

Following this observation, $$$O(10n^2)$$$ solution is obvious to me. How can this be improved to $$$O(10n \log n)$$$? I was trying some $$$\texttt{ordered_set}$$$ and $$$\texttt{order_of_key()}$$$ stuff but could not come up with the solution. Can anyone help me with it? Here is my submission link.

Problem B was cool as it is the definition of topological sorting