Me, too. Hope I re-get blue after this round.

me too man

Me, too.

No score distribution for icpc style rounds

aight bro u can stop yapping now

your last attempted contest was 9 months ago...

hahaha

You are so funny bro! :))

i bet u can't. u know u have to be in top 150 to get 300point

GL!!!!

После раунда можно смотреть решения любого участника

Я тоже

What is ACMP?

Bro how ?

After 125 contests, I am still a Pupil and I believe in miracles.

YOU NOT GIVING UP GIVES ALL OF US HOPE!!

hope this round proof good for you

But E < D imo, so always be sure to read at least one more problem than the one you're stuck on.

you can do binary search instead of coming up a formula directly :D

For me B was trivial not C :((

there are 3 thing you can do in a day

1. attending a lesson
2. attending a lesson + one task
3. attending a lesson + two task

we have ((day -1) / 7 + 1) tasks unlocked.

so we will do 3. for tasks/2 days

then do 2. if tasks is odd

then 1. for remaining point

Binary search is probably an easier method, but I found a formula, so here it is:

Note that I'm pretty sure this constant-time solution is slower than binary search lmao.

235560283

this is my submission, i used math not binary search.

https://mirror.codeforces.com/contest/1902/submission/235665532

Your code gives WA on

2
1 2
1 2
1   
1 1 1

because in your dfs function, you say if you are the root node, don't initialize xor basis.

Can't we use Hashing in E?

D > F? Really? Idts.

here Submission

Can't that be hacked?

https://mirror.codeforces.com/contest/1902/submission/235606529

Hack welcomed :)

TBH the TL is too tight

My Hashing Submission Hacks are Welcome :)

You can brute force it asymptotically Submission

When solving the inequality, did you consider the case where the number of unlocked tasks is odd?

True One Piece fan

My trie solution passed and it was implemented using pointers.

wdym? I got AC with trie.

https://mirror.codeforces.com/contest/1902/submission/235578938

I got MLE(:/) with Trie too using PyPy3. After I converted the code to C++ after the contest, it passed magically :). But now that you say even in C++ trie won't always pass based on the implementation, I have a few things to learn.

You can brute force it asymptotically Submission

I built a trie of all strings where each node further keeps track of the number of strings in the subtrie, as well as the total length of the strings in the subtrie while excluding the common prefix that the node represents.

Then for each string, I read the characters in reverse-order and traverse the trie, with each step denoting a collapse step, while carefully accumulating the contributions to the desired sum of the other branches (concatenation cases) by utilizing the stored values.

My submission: 235592382 I used a 2D array for the trie, which was probably a terrible idea (almost broke the memory limit), but the logic should be sound. I can elaborate on the details further if you want.

it is simple to solve D in < 500ms if you use cpp

pypy isn't always faster than python. you can sometimes switch from pypy to regular python and won't tle

Take a look at Ticket 17179 from CF Stress for a counter example.

I'm the fool who couldn't solve E in 1 hour 40 minutes yet solved C in the first 11 minutes.

Nope. Well, it might always be mirrored around some line, but that line definitely isn't necessarily parallel to y = x. For example, if you keep going up, then reversing does nothing, and your mirror line is a vertical line, and similarly, going right only yields a horizontal mirror line.

Nope

I might be wrong with that thinking, but my base idea is that after flipping, the prefix and suffix of points remains unchanged. If so, we can somehow transform given point and check if it is within prefix, suffix, and given range.

You're on the right track. Let $$$p_i$$$ be the point we're on before executing the $$$i^{th}$$$ command. Now to reverse a segment $$$[l, r]$$$ of the path, shift the origin to $$$p_{r+1}$$$ and mirror all the points on this segment around $$$x=-y$$$, (the transform is $$$(x,y)\to(-x,-y)$$$). Then, shift the origin to $$$p_l$$$. If we treat the points as vectors, then the point before the $$$i^{th}$$$ command after mirroring (assuming $$$i$$$ is in the segment) is $$$p'_i = -(p_i - p_{r+1}) + p_l = p_l+p_{r+1}-p_i$$$

Why does this work? If the last command in the segment was R, then the second last point would be to the left of the ending point, so mirroring about the ending point will make the second point of the new segment to the right, which is what we want. Similarly this can be shown for any position and command on the segment. Finally we need to start at the original segment's starting point, so we shift our new points to have $$$p_l$$$ as the origin.

Pre-calculate times when you visit each position when moving forwards and backwards, plus some prefix sums to calculate the x and y offsets, then you're done!

Basically this, I have been heavily commenting my solution during the contest 235604170

After reversing, suppose :

x = s1 + s2 + s3...sl-1 + sr + sr-1 + sr-2.. s(r-z+1)

=> x = prefix_horizontal[l-1] + prefix_horizontal[r]-prefix_horizontal[r-z]

=> prefix_horizontal[r-z] = prefix_horizontal[l-1] + prefix_horizontal[r] — x

Similarly, prefix_vertical[r-z] = prefix_vertical[l-1] + prefix_vertical[r] — y

hence there's an index l-1<=z<=r such that,

prefix_horizontal[r-z] = prefix_horizontal[l-1] + prefix_horizontal[r] — x

prefix_vertical[r-z] = prefix_vertical[l-1] + prefix_vertical[r] — y

If you are choosing it inside the reversed part a[l,...,r]. Then it is choosing a[1..l-1] and a prefix from reversed(a[l...r]) which is suffix from a[l...r]

gcd should not be equal to 0 initlize it with A[1] — A[0]

If n = 1, then You are returning without reading the array.So, there will be misinput.

https://www.geeksforgeeks.org/abs-labs-llabs-functions-cc/

read this comment

When $$$n == 1$$$, your function returns without actually reading the array value, so the program would mistakenly treat this array value as the array length for the next test case.