I will also give this contest today

#include<bits/stdc++.h>
using namespace std;
int main(){
   cout<<"I will also give this contest :)"<<"\n";
}

I think because they don't have a final problemset yet, maybe because they have 8 or more problems made but they are not sure which problem to put on 1-2 places, and they will decide later

I am Very Bad, contest.*

Skill issue,copium.

Masterpiece. I laughed sooo hard when I saw this :D

I spend 30 minutes to debug F)))

I think D is shit and unnatural.

what was your approach? I tried like this but failed — the pairwise mins in array A must have an increasing order. for odd n the pairing excludes the first element, but it has a constraint to be the smallest integer.

Orz

Can you explain it please?I think it's the longest path in the graph after deleting some edges but there are some cases that I choose edges that I shouldn't

I got wa on test case 50

There are a number of situations you must think of.

WA on test 20 three times and 32 twice.

for problem F, Test 32 had a situation where the entire grid was filled with '*' for sure.

Failing testcase: Ticket 6329

Try this, it helped me pass #32. Correct answer is 11.

2
20 13

You can register for practice and still submit.

Nope, note that you can only change the relative order between two elements after sorting ($$$a_{n} \leftrightarrow a_{n - 1}, a_{n - 2} \leftrightarrow a_{n - 3}, ...$$$)

you don't really need to do the operation.

remember the total number N of current icons on the screen, and the number of icons in the first N cells. The result is the difference between them. Use fenwick tree to find out the second value.

You can consider unraveling the matrix resulting in a vector (i.e. taking the first column, then second, then third, and stick them to make a new vector). Then, the apps need to be in a prefix of this vector. when adding a image, you extend the prefix, otherwise you need to shrink the prefix (where you need to move the icons). Then, to find how many images you need to move is to calculate the number of images you have in the prefix you currently are in. If this number is X, and you have Y images on the desktop, then $$$Y - X$$$ is the answer. To calculate X, you can do some if's when erasing/inserting images (to see if they spawn/despawn directly from your prefix)

You had to count gaps in the icon arrangement using segment/fenwick trees.

Video Solutions: A-F

The statement is a bit hard to understand, but anyway it's still OK to me.

Hey, thanks for the problem, and for the years of service!

I think the statement was okay, but when reading it for the first time, I felt like an image with the statement would be perfect. Or maybe, explanation of one test case.

Bringing statements back to reality is a good goal. I appreciate the effort.

CP-english is unfortunately a low-trust highly-synthetic dialect that's really hard to get used to even as a native speaker of normal human english.

Therefore, as a moron who had to ask in abject paranoia for the clarification 'is a palindrome a palindrome today, sir?' in that contest, it's maybe unsurprising that today's F was incomprehensible to me despite the real-world context... I'd reckon it's because we're routinely asked to ignore/overwrite it for synthetic predicates despite collisions ('alphabetical' to mean something besides sorted in the ordinary sense), distortions ('local maximum' when what was really meant was 'peak' or 'width-1 local maximum'), or sign-flips ('distance' and 'closeness' are opposites, but the problem happily defines one to be the sum of the other).

For what it's worth, I was envisioning desks in a classroom (being lenient on 'desktop' as if it were just sloppy wording for 'desk') and just couldn't make any sense of why there'd be icons etc. Cp itself is 'command line' more than desk icons...? Any mention of 'computer', 'screen', or even 'resolution' might've helped pull me back out of that dead end, but it's still, as always, up to the reader to read. My mistake and certainly not the last...

At least in the palindrome case, there was the option to code my way out '())(' as 'occo' etc. For today's F, I just bailed out of respect/fear of its position (plus I was still mulling over the nested 1538G in E and badly tempted by an ultimately incorrect thought on G).

So... nothing especially toxic, just... something we can all keep working at (better than retreating to formalisms, imo)

Otherwise, usual thanks and orz's, good stuff.

edit: 'local maximum' matches 'peak' better... don't remember which the problem asked for whenever it was tho...

As a tester, I agree with vovuh. I also thought everyone encountered desktop arranging at least once and would understand the problem perfectly. To me, it seems like a very nice life-based problem so I didn't suggest any changes to the statement. Will definitely keep such things in mind from now on though.

For problem F, why did the question allow a time limit of 3 seconds? It'd be reasonable to make the participants to use a time limit of 1 second instead.

"But I can't get how people are not familiar with the order of icons appearance on the desktop."

On mac, by default it is right to left:

Educational rounds and div3s are meant to have weak pretest. Why do you think there is a 12 hour hacking period?

Dear contestant...

Please don't worry if you had to hack yourself in order to see if your solution is good. In a programming competition, the tests used can be useless, you should have assured your code works. Sorry if this does not fit in your view.

Author of xem sdod

try 5 1000 7

I think $$$1\space8\space100$$$ is a good choice.

9 hacks with 100 1

I spent about 30 minutes thinking of a solution for this problem before I realize It was not the real problem

hacked. thank you

me too!!!And I also used "#define int long long" which caused me to miss the problem until the last few minutes when I changed it to (int)pow(2,a.length())

The condition is that, after removing any number of edges, the final graph MUST have in-degree and out-degree of EVERY vertex to be strictly less than the original graph (except if the in/out degree was 0 for any vertex in original graph).

For example, if there is a graph 1->2->3->4, then we can't just remove 1->2, because in the resulting graph (2->3->4), the in-degree and out-degree of 3 are the same as that in the original graph.

I'm trying to explain my solution while my solution is hacked...so I deleted all my notes...

The idea is to make any two element to zero by three ways.

• If you shoot directly to two lowest valued position. You can get one of the answer.

• If you shoot at two consecutive position. You can get one of the answer. If maximum of the two number is greater than or equal to the twice of smaller one then shooting larger one will give the answer, else you need to shoot both to get optimal answer.

    int a = max(v1[i], v1[i-1]);
    int b = min(v1[i], v1[i-1]);
    if(b*2<=a) {
         int x = a;
         int y = ceil((double)(x)/2.0);
         ans = min(ans, y);
    } else {
         int x = v1[i]+v1[i-1];
         int y = ceil((double)(x)/3.0);
         ans = min(ans, y);
    }

• If you shoot at at one position and it's effect may be seen on both left and right side and can contribute to answer.

    int x = v1[i]+v1[i+2];
    int y = ceil((double)(x)/2.0);
    ans = min(ans, y);

• Minimum of all above cases may give the correct answer. Till now this solution is not hacked.
• Correct me if I am wrong.

and you have to give Div3 to cross the newbie stage, DEADLOCK :p!

i did it

Once my rank was about 200 higher than everyone else I knew I would get hacked. Luckily there were so many that they did not end up affecting my actual position all that much (before I was around 700, now I'm around 700)

Test: 10 3 10 9 8 7 2 5 10 10 10 Answer is 3, but you have 2(

There are more than 50 pages of successful hacking to E. Never imagined to this...

There are no pretests in div.3, div.4 or educational rounds. It's full feedback meaning submissions are judged on all tests. Although, the tests on E were weak no doubt. The solves went from ~3000 to ~550 after system testing was over.

skill issue

My dear friend, you are absolutely wrong jury always have a proof to defend their solutions

The hacks are tested against jury's solution, that means the jury solution gives different output from the hacked solutions. Which means it's correct.

And then get a WA coz your proof was wrong

There are 3 ways to locate two broken sections:
Case 1. Adjacent
Case 2. Through one
Case 3. Different locations
We need calculate 3 minimums independently and take a total minimum. Let section's durabilities are x, y. Without reducing generality x<=y.
Case 1. The number of onager's shots is:
- if 2*x<y then (y+1)/2
while Monocarp shoots more durable section, the adjacent section will be broken
- else (x+y+2)/3
Monocarp can decrease total durability of both sections by 3 every shot may be except the last one
Case 2. The number of onager's shots is:
x+(y-x+1)/2
At first Monocarp shoots x times right between the sections. Then Monocarp shoots second section as target (y-x+1)/2 times.
Case 3. The number of onager's shots is calculated independently for x and y:
(x+1)/2+(y+1)/2