still waiting

Yes 18:15

will this be this final 5 mins or there's more to come. lol

An usually unusual timing lol :x... nx5mins too

Okay ,but what's the point?

Because we don't have a good set of problems for div. 1 round this time.

Thanks, fixed.

It is indeed sad, but I prefer having no Div. 1 rather than shitty Div. 1 where you only find out it's shitty after you've started participating.

Last time, there was a weak pretest. And this time, there's even not a div. 1.

This is competitive cheating

Whenever you see the words Codeforces Round #... it just means it is rated.

Yeah I thought they'll publish it at least 30 min before the round, but still nothing.

You heard? Don't the testers see the questions beforehand as they give a VC of the same contest? Just want to know.

hope i am able to solve 3 question

Looks like they don't want to reveal the score distribution for the technocup participants

i could finished my dinner

Don't worry so much about rating. Focus on learning

No, I Hope you will do good. Good luck.

oh no, I was waiting for youtube video from u)

maybe I am gonna drop down to specialist too

It is a sign to not attempt this contest for sure now.

another delay of 5 mins. maybe two negatives make a positive [ delta :) ]

I hope u reach Master in this one

Update: A huge negative delta is on the way. It was a sign xD.

mistakes were made(you didn't heed the signal, neither did I)

Okay this is not fun waiting (:

ok, again 5 more minutes to fix this I hope

Will this contest even start?

Using Inspect in you browser!!!

Hope is a good thing..

but may be you can give contest without thinking much about rating and surprise yourself..

"Ah shit, here we go again" (c)

why?

Ya!

The problems aren't accidentally visible during these delays so probably not Codechef :P

else downvote?? :)

Another One DJ Khalid

Ok Sir

I also want to report this. Is there a person manipulating different accounts in the contest? This should also be considered cheating. And since these accounts are newbies, it will make others rating drops much since you will be considered "defeated" by a bunch of newbies if these cheaters are not removed.

UPD: They finally appear on official ranking 22-26, 28-39 (except two CM and one expert)

WTF？？？

Actually, they were also cheated in Educational Codeforces Round #118, ranking 20~24. I expected them being caught cheating and became skipped then. Seems like they were not caught by Codeforces by using random space or different name of variables.

Arrayforces.

Oh my god !

Yeah, got first solve on F because of this. lol

Solution of D (90%): TRPLSRT

Ctrl + C
Ctrl + V

F = https://atcoder.jp/contests/arc115/tasks/arc115_e

somebody please help. why this is giving RTE?.link

It has undefined behaviour in line "ll x=a.size()-1, y=b.size()-1;". When a.size()=0 or b.size()=0.

For me C was horrible implementation problem, I fiddled like an hour on it.

Actually it is not so complecated. First observe that left and right side are independent of each other, each can be solved separate.

Then consider blocks of k elements, starting with the outermost (the longest distance) elements. For each block we need to walk the distance to the max one, and most likely back.

Do this for both sides. Then, after end, we can remove the one distance we not need to go back, that is max(abs(a[min]),a[max]). i.e. we go to the farthest element last, so can subtract that distance in the end.

please don't.

https://atcoder.jp/contests/arc115/tasks/arc115_e

I don't know how to prove it, but the ideia is that if inversions%2 = 0 or any number appears twice the answer is YES, otherwise is NO.

The cases where n = 1 and n = 2 are trivial.

Let's suppose n >= 3 and all elements are distinct. We can notice that by using a 3-cycle, the parity of the number of inversions in our vector remains the same. So, if all the elements are distinct, then we only have to check whether we have an even or odd number of inversions in our array.

If there are at least 2 equal elements, then we can use them to put all other elements in their place, so the answer is YES everytime in this case.

I did with counting inversion, such a triple swap of three distinct elements changes the inversion count by 0 or 2.

Also the array can be pre sorted. And if the array includes any element twice, then it is allways sortable because we can use the double element to simply swap elements.

Case 1: There are no duplicate values. In that case, we can decompose the permutation into cycles where $$$i \to a[i]$$$. After playing around with some cases on paper, you'll find that we can perform the following operations:

1. Merge two cycles of size $$$A$$$ and $$$B$$$ into a single cycle of size $$$A + B - 1$$$, or split one cycle into two.
2. Merge three cycles of size $$$A, B, C$$$ into a single cycle of size $$$A + B + C$$$, or split one cycle into three.
3. Increase of decrease the size of a cycle by $$$2$$$.

An array can be sorted if it can be split into cycles of size $$$1$$$.

So now we decompose the array into cycles, merge the cycles, and the answer is YES if the size of the resulting cycle is odd.

Case 2: There is a duplicate value. In that case, it's always possible. Say the value $$$x$$$ appears twice in the array. For the case where all our cycles merge into a cycle of even size, we can reduce the cycle to size $$$2$$$, then shift the cycle to one between two values $$$(x, y)$$$, then fix that swap with the help of the second $$$x$$$.

I solved D without using the ideas of inversions. First, for the case with duplicates, we can see that we can use any two of the same number and "move" a third number into sorted order.

So, for the case without duplicates, the array will be a permutation. The sorted array will have $$$a[i] = i$$$ for all $$$i$$$. Instead of using inversions, I tried, for every $$$i$$$, to swap $$$i$$$ into position $$$i$$$. If after all swaps the array is sorted, the answer is YES and otherwise NO.

To swap $$$i$$$ into position $$$i$$$, assume number $$$i$$$ is at position $$$x$$$ ($$$x \neq i$$$). Then, if $$$x \neq n$$$ we can select the positions $$$i$$$, $$$x$$$, and $$$n$$$ to move $$$i$$$ to position $$$i$$$. If $$$x = n$$$, we can use position $$$n - 1$$$ instead. This works because we only care if number $$$i$$$ is at position $$$i$$$, so we can arbitrarily use the last position to be our third number. All we have to do is keep track of the positions of each number when performing our operations and lastly check if the $$$a[i] = i$$$ for all $$$1 \leq i \leq n$$$.

Tried this during contest, but got ML. Not sure if there's a way to optimize.

Just use implicit segment tree and it works for ai <= 1e9

Yeah you can use a sparse segment tree.

You can use coordinate compression to build seg tree my submission