Looks like it is back up!

Can't see problems yet though.

Yep, down for me too. Will Codeagon be extended or postponed or something?

I got two WA because of that :P

Man, it is nice to know that I got the idea right at least :D

How to solve E? I started with root at origin and then going 1e16 far for level 1 in 4 directions, 1e16 / 2 far for level 2 and so on. There was a bug in my implementation, so I don't know if it's correct or not.

Really great problems though, thanks for the contest!

you need to use a segment tree or a Fenwick tree to calculate sum from q[m].first to q[m].second in O(log n) time.

Your post is a bit too long to be funny :P

What is solution for "R!!Y!!B!!G!" test 17 for B?

For A: Pie Progress, wouldn't O(N**2 * M) DP be too slow? I see that it is mentioned in editorial as a solution. Did anybody get it to run?

What if there are two people with the same ranks?

In the problem setter's code for D, why don't we add the number represented by bi bi+1 ... bn to the mask before checking if the mask is valid or not?

Edit: NVM, I didn't read the question correctly (idiot).

I did it using two simultaneous hashes with different bases and got AC.

I will eat shoe if bangladesh ever place top 50 in world final

Literally nobody cares.

The online was indeed a sham. Problems took around 20 minutes to be judged and you couldn't submit anything for the last 5-10 minutes at all because the site was acting up. And it was all ignored.

The Chennai regionals however seemed better organised to me, with no judging problems at least.

I did the same thing!

That feeling when you find a mistake in your A solution just after you lock it. :'(

Ah well, you won't do it next time :D

I think overflow in sum1 and sum2 might be the problem here. Try using long instead of int.

How to solve Div 2 E / Div 1 C?

How?

How to solve Div 2 D and E?

D seemed like binary search to me but couldn't get anything concrete in time. :/

In your first code, the basic idea is correct but the graph construction is incorrect. It won't work on test cases like

4 2

1 1 1 1

1 2

1 4

3 4

because you're changing the adjacency list of only one vertex.

Here's an AC version of your code with the graph construction fixed and minor modifications to DFS: http://mirror.codeforces.com/contest/580/submission/22137984

How to solve E? I see some greedy solutions but don't know how they work correctly. :/

Internet points are important to people. :D

In my opinion, not giving anyone any points for Div 2 D and keeping it rated is a fine solution.

This is an awesome solution! Thanks.

Next one goes better hopefully. At least we'll have nothing to lose. :D

Man, I kept messing around with Linear Diophantine Equations for B, but couldn't get much from it. :/

I think I was doing the same for C but kept getting wrong answer. I think not getting B might have messed me up.

Ah well, next one I guess.

Welp, I'm fucked. Pupil status, here I come. :/

How to solve B and C?

Overflow in gpsum * b is the problem here. I made the same mistake :(

Hi, sorry if this is a stupid question, but what exactly are we doing in the BFS. I mean if we're doing BFS to find cells of the next layer wouldn't it still be wouldn't the BFS still be O(nm)? I read the editorial and it uses the same approach but I don't understand it.

That would still get a TLE as the number of edges in the graph is too high.

that feel when didn't even get to read E because I was stuck on D. :(

So uh, how to solve D? I was trying to think of DP but couldn't.