This is exactly the same as my sollution was (I was a tester)

The last tiebreaker isn't much of a deal right, for cells with the same max distance and manhattan distance from the center, we can choose any of them. I just sorted them in order of i and j indices.

As a side-note, this problem saved my ass, I almost thought my rating was cooked.

what does from 12 and 6 mean

How to do it I tried via recursion got stuck hard af then tried apply iterative approach still got fked af

yeah B was good

Actually!!

I did some calculation and used 2n-3 operations. I started from the bottom row and the gradually moved up to the top. [n, 1, n] --> [2, 1, 2] This will give us n-1 operations. After this I moved from top to bottom. [1, 2, 3] --> [n-2, n-1, n] This will be n-2 operations. Adding up we get a total of 2n-3 operations.

Hi, I am a bit confused about the case where $$$n=3$$$ and $$$m=3$$$ (the sample input). Are you doing the removals like this:

01 05 02
07 09 08
03 06 04

So your adding order is like:

09 05 08
03 01 02
07 04 06

And then the final penalty matrix becomes:

0 2 0
2 1 3
1 1 1

If I misunderstood your approach, sorry. Could you please provide a detailed example for the case when $$$n=3$$$ and $$$m=3$$$?

Yes, now it is working 🎉: 324418651

But I want to add something:

1. I don't know why the first if condition is needed.
2. I rotated your 3x3 matrix by 90 degrees.

actually strategy is flawed, there could be a solution sometimes when you have to set only a few bits of some number, for that notice that adding 1<<x, where x is the unset bit is the fastest way to set a bit, so simply keep track of positions of all unset bits, and then sort them and subtract one by one from k.

I got D1 after 2 minutes too. :(

they are same as light only turns when time modulo k = delay .. so if you reach it twice you will reach whentime modulo k = delay` only ..

i have another condition that didn't work: if we leave the traffic light in the same direction twice, then the answer is no. Would appreciate if someone how it's different from editorial condition

Aree just think it this way,.. u are proposing that we need to make sure we reached the traffic light both the times with same (t%k) value. But the point is, when u reach a red light, t%k should be d... Else it's green isn't it. So t%k always d, whenever u crash the same red light.

If you reach the light twice from the same direction. Then it means you have completed a whole loop which takes x*k time where, x>=1. As you reverse direction after visiting it the first time and then somehow make it back again. It can be concluded that you are stuck in an infinite loop.

My Submission

here t is time we reach that pos and dir is direction we are moving. we are checking what will be the light when we reach that pos.

324134942 I first precomputed for every traffic light for every time modulo k with direction if it can be left eventually with simple state transitions. Ignore all those parameters being passed for each function call.

dfs brute force — submission

324146023

What's an assert?

rip lol

Damn you have solved 1000 problems. Respect!!

couldn't even try d properly, only 20 mins were left.

for (auto j : ss)
    {
        sum += (1ll << j);
        if (sum <= k)
            c++;
        else
            break;
    }

I just modified this loop with break and it worked correctly.. but why do i need a break statement .. its quite obvious that if at the jth iteration sum exceeds k then at j+1th also it will exceed

324106167

Hey, I started coding 2-3 minutes into the competition. And then after that, because previously I was hitting like incorrect solutions and facing penalties on previous competitions, so I was building my own test cases, so by the time half an hour was complete I believe I had already coded up the first few problems which I thought would be easier for me to solve, and they were running on different test cases(mine). So once that was done I submitted them in the order the test cases I made stop running.

okay so my observation is that if we take the middle element and then go layer by layer, for any layer k, only the cells of the layer k-1 will be penalized (think about it).

now as we are going layer by layer we will compare the chessboard distance of that specific layer from the middle element and then go from in to out.

and as the problem itself suggests that the tiebreaker is manhattan distance, we will then choose those cells with the highest manhattan distance for adding the penalty.

let m = n = 3

initially we choose the middle element.

x x x
x 0 x 
x x x

from the middle elements all the cells of layer 1 are equidistant, but the closest are the adjacent cells and then the diagonal ones(bcs of manhattan distance)

now if we choose the the diagonal ones first then for every next coloring of element the penalty of any diagonal cell will exceed 3

x x 0      0 x 1      0 x 2      1 x 2      1 x 3     2 x 4 
x 1 x  =>  x 1 x  =>  x 1 x  =>  x 1 x  =>  0 1 x  => 0 1 x     
x x x      x x x      0 x x      0 x 0      0 x 1     0 0 1

as one can see that the top left cell exceeds 3. so the better approach is to go from in to out(from closer cells to farther according to the tiebreaker)

x 0 x      x 1 x      x 2 x      x 2 x      0 2 x      0 2 0      0 2 1      1 2 1
x 1 x  =>  x 1 0  =>  x 1 0  =>  0 1 1  =>  0 1 2  =>  1 1 2  =>  1 1 2  =>  1 1 2
x x x      x x x      x 0 x      x 0 x      x 1 x      x 2 x      0 2 x      0 2 0

thus we will extend this for multiple layers

now this can be easily done by sorting the coordinates with the first key as chessboard distance and the second key as manhattan distance.

Hi! Sorry for the delay, the issue should be fixed.

Your logic is okay, but you're not decreasing k, so it will give WA for for any 0 . Just do k -= wow3[i].ss while adding to the answer.

The main problem is that you're counting up to 60 bits, but it can never actually make all 60 bits high, because k can be at most 1e18 — which can set at most 59 bits. Your logic is doing now — pre, so the delta of 2^60 can end up being less than k.

so you code is adding 2^60 which is impossible to add

I mean when you use the integer with some ll quantity during multiplication, addition and all, it behaves like a long long number, otherwise using 1LL is the only way, it doesn't take any time also.

maybe checker need it.

Originally we used $$$n\le 2\cdot10^5$$$, but some $$$O(N\log^2{N})$$$ python solutions had a hard time passing, so we decided the lower the constraints a lot lower.

matrix size is 4 ... no. of operations used in the answer is 5

is not you, if you want to check some implementation ask here in comments, or wait writers fix that

All the code in Implementation seem to be linked to submissions in a private gym contest. So we just couldn't view any of them until someone fix it.

Firstly, let's imagine that the vertices of the trees have their identities tied to the identities of elements on the array. Let the vertex corresponding to index $$$i$$$ have label $$$a_i$$$.

We can show that the solution is correct by proving 2 things:

(Necessity) The cyclic sequence of the trees remains invariant under the operation. This is because no vertices can ever move to a different parent nor can the order of them change underneath the same parent.

(Sufficiency) This is the harder part — we want to show that $$$a$$$ can be transformed into any array that satisfies the same invariant. A nice trick to make this easier is to establish a "canonical" representation for the invariant, and proving that every array satisfying the invariant can be transformed into the canonical representation. Thankfully, there's a pretty natural one here: the pre-order traversals for all of the trees concatenated together in cyclic order, starting at an arbitrary tree; I'll call this $$$c$$$.

Consider any array $$$a$$$ that doesn't equal $$$c$$$. First, let's rotate $$$a$$$ so that $$$a_1$$$ corresponds to the root of the starting tree in $$$c$$$. Since preorder traversals start at the root, it follows that all occurrences of $$$m$$$ are in the correct place. If something else doesn't match, consider the first $$$i$$$ such that $$$a_i \neq c_i$$$. Let $$$j \gt i$$$ be the index of the first occurrence of $$$c_i$$$ in $$$a$$$ after index $$$i$$$. We'd like to swap $$$a_j$$$ to the left until it reaches $$$i$$$, but might run into something with value $$$a_j + 1$$$ or $$$a_j - 1$$$.

$$$a_j + 1$$$ is impossible — since the preorder traversal of the tree matched $$$a$$$ up until $$$i$$$, $$$a_j$$$'s parent must be in the correct place. Thus running into another $$$a_j + 1$$$ would mean that $$$a_j$$$ belonged to a different parent.

$$$a_j - 1$$$ is also impossible — since we know there's no occurrence of $$$a_j$$$ in $$$a_i, a_{i + 1}, ..., a_{j - 1}$$$, it must be true that its parent lies within $$$a_1, a_2, ..., a_{i - 1}$$$ and thus it must've occurred earlier in the traversal.

It follows that $$$a_j$$$ can be swapped to the left until it reaches $$$i$$$, and repeating this procedure would eventually yield $$$c$$$.

This blog may help you:

https://www.zhihu.com/question/3258240803/answer/27316958173