what even is pinely?

thanks, fixed

What a coincidence... Me too! I would like to request a refund...

Also the round is nice, you can participate if you want

practise more problems in 1200-1300 range

• Solve A and B
• If this doesn't help, Solve A and B faster
• If this still doesn't help, something's wrong in the process

I don't know what happened to Pinely Round but it was canceled and replaced by common Div2.

It is rated for people with less than $$$2100$$$ rating.

can you please tell me whats is the meaning of rollback what is the meaning of this line " Rating changes for last rounds are temporarily rolled back. They will be returned soon."

same goal :)

Actually B is a double problem.

B1 is understanding B, then B2 is solving B.

congrats!

How do you see your approximate performance and rating before rating canculations? Is it a browser extension or it could be unlocked in settings?

"No human is limited". Believe in yourself, maybe one day you can become a LGM!

it requires absolutely no algorithmic skills and even worse, it is easy to pass with an intuitive solution without proof

I have no idea how to solve D, though spent a lot of time on it

But for B it was a clear ternary search from the start

why

cout<<123456789.5<<endl; and you will get 1.23457e+08 in the output I believe this is the reason

I think you made the same mistake I kept on making for an hour. Basically, (a+b)/c, where c is a double and a, b are ints, returns a double in scientific notation. You need to convert that scientific notation to a decimal.

The idea is simple first of all, imagine two strings, Reverse of $$$s_1$$$ and orginal $$$s_2$$$. Now everytime you perform an operation, you select a suffix of length $$$k$$$ from both the strings, reverse them and swap them. It simplifies the problem (makes it easier to visualize).

So, first reverse $$$s_1$$$. i.e $$$s_1 <= reverse(s_1)$$$ Now, with a bit of playing around you can prove that you can

1. Swap any letters at position $$$i$$$, $$$s_1[i]$$$ and $$$s_2[i]$$$. (i.e exchange the letters at position $$$i$$$ of both the strings. Note that $$$s_1$$$ is not the original $$$s_1$$$, it is the reversed string).
2. Swap any pairs $$$(s_1[i] , s_2[i])$$$ and $$$(s_1[j] , s_2[j])$$$, for any index $$$i$$$ and $$$j$$$.

Now its easy to solve!

no. it requires ternary search!

ternary search, but what was testcase 3 :((

yep

You could binary search, yes.

But you could also just calculate the minimum $$$x - t$$$ and the maximum $$$x + t$$$ and take the midpoint as the answer.

Proof: If the same person has both the minimum $$$x - t$$$ AND the maximum $$$x + t$$$, then this person is taking a crapton of time to get dressed, and everybody else can reach this location by the time this person is done.

Otherwise, suppose person $$$P$$$ has the minimum $$$x - t$$$ and person $$$Q$$$ has the maximum $$$x + t$$$. Since $$$P \neq Q$$$, this means $$$x_P < x_Q$$$ (otherwise, whoever has the higher $$$t$$$ would have claimed the other's rank). The claim is that the ideal midpoint is $$$\frac{x_p - t_p + x_q + t_q}{2}$$$. It's easy to see that $$$P$$$ and $$$Q$$$ will spend the longest time to reach this point, compared to everyone else, since anyone who spends longer time will either have a lower $$$x - t$$$ than $$$P$$$ (if they were moving right) or a higher $$$x + t$$$ than $$$Q$$$ (if they were moving left). Here, $$$P$$$ and $$$Q$$$ take the same amount of time, and it's the optimal time, because changing the position in any direction will cause one of $$$P$$$ or $$$Q$$$ to spend more time to get there.

I used binary search,but got TLE at point 4.

Find the minimum value of $$$x_i - t_i$$$ and the maximum value of $$$x_i + t_i$$$ (they can overlap). The answer is the midpoint of these two values.

You shouldn't feel stupid though, because even though it's actually not that hard to prove this result, I think it's very difficult to actually come up with this idea with enough confidence that it might work, before one can consider trying to prove it.

use ternary search

An alternative solution to binary/ternary search:

For each person, we create two new locations, the location that is t[i] units to the left of the person and the location that is t[i] units to the right of the person. Now we find the location among these locations that is furthest to the right and then the location among the locations that is furthest to the left and pick the location exactly in the middle between them.

However, I did not prove this solution and it might not work after the system tests, but it just felt right. The main reason for this idea is that in an optimal solution there will be a latest person that comes from the left (and arrives at the same time as if he was t[i] units to the left of where he originally started), and there will be a latest person from the right. If these people do not arrive at the same time it is probably not the optimal location because you can move the location closer to the person that comes later.

A cool thing about this solution though is that it can run in O(n) and not O(nlogn) like most other solutions.

Edit: This solution does in fact work since it passed system tests

"Mom, can we have subtractions in absolute value?"

"No, we have subtractions in absolute value at home"

Subtractions in absolute value at home:

Geometry

Yes :(

Link

Are you for real? Where?

Ok that explains why c has thousands of AC while B doesn’t

https://www.youtube.com/watch?v=QWYxr-IEwE0

so it would seem

Hint for solution without any search: Find their equivalent starting points

Use binary search to get the answer. In fact, here's a parabolic and you should find the x of the lowest point. Implementation

Was it really about floating point precision??? I did a solution where I sorted the $$$x[i]$$$, could obtain the minimum time assuming the meeting point is in between each segment($$$x[i]$$$ and $$$x[i+1]$$$) in $$$O(1)$$$, checked the time taken if meeting was at each $$$x[i]$$$, and still got it wrong... I am extremely demoralized, normally I'd walk away from a contest thinking I'd learn something but that was just terrible, the difficult D and E didn't help either

Oh nice, very creative idea!

What if there are many person with same max_dressing time?

I think you misunderstand the problem. Applying operation with k = 5 on initial strings results in s1 = daabaa and s2 = aabada.

Yea seems like a lot of people copied C, because C wasn't that easy, these incidents keep ruining the performances of those who participate fairly.

It's ok. We all have some bad times, look at my plot for example

I even thought of solving three problems... 😂😂😂

You got to be kidding me, I just edited my contest solution, adding "set_precision" and got AC, I assumed that just defining all variables with long double will work, I have never felt worse after a CP contest ever

No neko-tyan and no bowl of rice for the authors

#include <bits/stdc++.h>
using namespace std;

int main() {
 
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T; cin >> T;
    while (T--) {

    }
    return 0;
}  

You should put them first in main().

[deleted]

actually he's right. just skip (but better ban) those whose solution looks like:

I understand, that a lot of people pings you in comments, or you might already know about this situation, but MikeMirzayanov, geranazavr555, can you, please take this into account, while searching for cheaters? If needed, I can personally provide handles of few people with solutions like the one from above

p.s. link for video if needed:

By the way, I found 150+ cheaters using my submission parser. It looks for weird chars like '10' or '20'.

UPD. Thanks to Codeforces team for removing these cheaters!