Why am I getting so many downvotes? Does everyone else here just want math?

I do hope too. Constructive-only problems are lame in most of cases.

I think it is unreasonable to rate differently for the two rounds.

Possibly not. We aren't sure about scoring or time yet. Please do not feel any angush over this! When the contest rolls around, we are sure that you will be amoosed by our puns.

please read the blog post carefully once instead of asking so many questions and remaining confused all the time...this is not specifically for u.. for everybody in genereal...

I postponed registration to IndiaHacks 2016 to make it clear.

Totally agree.

In the last minutes in the qualification round jqdai0815 re submitted a wrong code in first problem which passed the pretest so that fail and not be qualified and he now registered in the unofficially version of the Elimination round, while the official round has 7 LGM and many other great coders.

I do not mean that jqdai0815 can not compete well in the official version, we all know who jqdai0815 is, but I think is it not fair to have 2 separate versions of the contest and 2 separate standing which will make one of them easier than the other one, and become in the top 10 ( in the unofficially ) much easier than the official.

I think this is affecting the number of participants since only around 2500 people have registered so far in both rounds.

I passed the qualification round and I registered for the Unofficial Edition, no problem at all

But your rating was increasing except the last one :/ :/ :/

sometimes sequences are more complex than you think

Yes, I am hoping too)

I guess Codeforces should then think about adding another title named "tourist" :D

Legend says one day tourist will cause an integer overflow!

Yes, many ways to solve a problem is a good stuff.

Thanks Zlobober! I'm glad you enjoyed our problems! :)

Same problem, 16795467.

Got this response from jury:

" We're really sorry, somehow the system broke in the last 5 or so minutes. It is accidentally running your code on G instead of F. Gleb is busy right now with other issues and we can't do anything about it.

We really apologize. :("

For me — RE: 16795457

The same.

Please, read UPD3. I'm sorry about it, it's my fault.

It seems there are some problem with the problem F.

Brb, posting that picture of a button with "START SYSTEST" for tons of contribution.

For 3rd question, I used a sliding window for O(N) time. I'm not sure how ternary search would work.

For the fourth, I binary searched until the graph was a valid one (that there was a valid ordering of nodes), but I think the way I checked graphs might be wrong. I found the winner of the graph (node without any parents) and continuously found the next best rapper.

I got stuck on D for ~20 minutes because of a bug in my binary search T_T.

4-th: no binsearch. Longest path + review edges and stop when all edges on the path viewed

You know, he's the master of last minute submissions

Hint: if you want to count the numbers of distinct subsequences, you should make something like . Where is number of distinct subsequences ending with character c which came last to the end of string.

See the second accepted answer here http://stackoverflow.com/questions/5151483/how-to-find-the-number-of-distinct-subsequences-of-a-string

Now, in our case, we had to apply a simple greedy for filling the extra characters. At each step, we will put the character from the alphabet size k, which had occurred earliest (i.e. if last[c] denotes last occurrence of character c, then the character c with minimum(last[c]).

Windows 10... You must be kidding...

Opera... You must be kidding...

"You must be kidding..." Yeah I must be kidding...

Thanks for sharing this screenshot. Through this I came to know about NBHEXT

you can find who is the winner with dsu but i dont think u can solve the whole problem

I don't think so, because dsu assumes undirected edges, and the problem reduces to whether for a subset of the edges E, for each pair i, j of vertices either path exists i → j or j → i. The trouble with this of course, is a set of edges like , where dsu will conclude ok, but actually we don't know which if b or c is better.

It can be solved with binary search+topological sort in NlogN.

Yeah, the same thing happened to me. I think there's a problem there... I got correct answer in custom invocation too.

At first I got TLE in Test 6, so I decided to optimize and submit again, only to get WA on pretest 1. I was like " What the...?! ". Maybe they have to recheck that...

+1, same problem. Pretests can theoretically be different from samples, but why then don't we have minuses in the scoreboard?

The same. Somebody should investigate it.

I think you forgot to delete blank line end of the output.

It is strange, but it seems like this problem appears only in the end of contest. Nobody passes this problem after 01:47.

I submitted "cout << 5 << endl; cout << 16 << endl;" in 01:56:56, but this code get WA on pretest1....

Please, read UPD3. I'm sorry about it, it's my fault.

I applied sliding window and then binary search. I think I did some overkill with my bs as I did it two times for odd length and four times for even length, but wanted to be absolutely sure... Lets see what happens after the system testing .. an overkill or an overwaste :v

I used binary search on covered range r. Then for each possible position i of the farmer in the array I checked if [i-r, i+r] could accommodate k cows and the farmer. With an array of sums it's possible to check each position in O(1).

How you used ternary search?

This is my approach-prefix sum of zeroes from left to right. Fill Closest_left_0_of_i array and Closest_right_0_of_i array, then use sliding window(or simply do an l=upper_bound() on zeroes array for each i where zeroes[i]>=k+1) to find the range [l,(r=i)] where they can fit, find midpoint of this range, and check the
min( max(|l-Closest_left_0_of_i[mid]|,|r-Closest_left_0_of_i[mid]|),max(|l-Closest_right_0_of_i[mid]|,|r-Closest_right_0_of_i[mid]|).
Take minimum over all i, and also for mid+1 in cases where l+r is odd.

ps: Formatting here is a little annoying. One can't simply get to a new line with return alone.

let "lf" be the left most room has been booked and "rh" right most and c the place of owner of cows.so the answer of problem( f(c) ) will be max( c — lf , rh — c).

next[i] means next 0 room for room i.(left to right)

first "lf" is the first 0 and "rh" is (k+1)th 0 and c is "lf" and do it until "rh" overflows:

while( f(c) <= f( next[c] ) )  set c next[c] and set ans min(ans , f(c) )

then set lf = next[lf] and rh = nex[rh]

so you can calculate the answer with O(N)

Yes.

In this case you get a 50 points penalty.

It is similar to v.pop_front(), so it does`nt waste linear time

Bet on #define vector deque magic

A similar thing happened a few months ago, where the checker was wrong for one single test case. It was then decided that said test case should be ignored, and all solutions that got correct answer in the remaining tests got Accepted in the end.

I think a similar course of action has to be taken here: All solutions submitted during the last 5 minutes should be run on Full Test Set for Problem F, and those who pass System Test should get the score for the problem. In case a user submitted many solutions with WA #1, I believe the first correct solution should be considered in order to minimize penalty.

I don't know if all this can be done, I'm not very familiar with Codeforces system inner working, but it would be great if you could do it. Let's hope for the best :)

Estimated time till start of system test?

this is the fourth time in a row for div1 as far as I remember

Binary search on the number of edges and do toposort on it!

Binary search the answer.

binary search on answer. To check whether an answer is posible or not , just add those edges and see if a unique topological sort exists or not.

Actually it's possible to solve it without binary search.

1. Check if all matches give you unique answer. If not, print -1.

2. If order is unique, let's assume that robot with index a_0 is better than one with index a_1, which is better than one with index a_2 etc.

3. Create set {{a0, a1}, {a1, a2}, ....}

4. Iterate through all matches and erase current match from set (if present).

5. As soon as set is empty, we know the order.

I didn't use binary search. I did a topological sort and then checked the maximum value of edges among consecutive vertices in toposort. Values of edges are index of match.

How are you finding k this way? You're basically checking if answer is not -1. Please explain, I think I misunderstood you.

"something like dijkstra or prim" should be topological sorting, I suppose.

As I can see, your isDist returns true iff there are no cycles? E.g. consider test 4 3 1 2 1 3 1 4 It looks to me your code will not output -1 though it should.

I think, jury should system test all WA#1 submissions and select the first passed (if there are such submits :D) as the final result.