Watch me win this contest ...

I hope for +pos delta too!
I only AC six problems(ABCDEF)...T_T

as an author i can confirm that he can say that he is a tester

As a Codeforces user I can say I am a jerk

Other authors deserve the chance too

We got FLT memes before GTA VI :(

all the best

good luck

gl bro

same

hhhhh, enfin quelqu’un qui comprend la vraie douleur du competitive programming ! Entre les erreurs bêtes, les tests cachés et les cas tordus, ce round promet d’être un grand moment de réflexion… ou de désespoir.

Mais bon, tant qu’on ne finit pas avec un score de zéro, c’est une victoire. Bonne chance, que ton code compile du premier coup et que les WA t’oublient !

Seems like the use of LLMs has impacted comments as well. Implementing defensive measures against it is necessary, but not an easy task. Let's see how things unfold.

Yes, you get money for creating rounds on codeforce. Unfortunately, I don’t know how much they pay for creating rounds

he added very good problem about kinder pingui

As a former groupmate of some authors I can say there's no Om Nom unfortunately.

please dont

Click on the number of registrations

I hope to get +5 this time

Us! But today I'll

so what?

SO you used GPT?

Bruh seriously, just stop complaining about GPT under every single Div3/Div4 post. The staff knows about it. Everybody knows about it. Just let the staff do their job, they are already trying very hard to enforce the rules.

It's nothing, but if you want to get higher rating, I think study more algorithms is better than participating in CF contests.

Never heard of it, but I like triangles

there is a simple trick to compute whether ncr is even or odd you can find it in usaco guide combinatorics section link : https://usaco.guide/problems/cses-2419-xor-pyramid/solution

for me it's triangles + recursive

Here

This

notice how pascal triangle and magic triangle are related odd==K and even==0 ~~~~~

~~~~~
  //16
    //16 16
    //16 0 16
    //16 16 16 16
    //16 0 0 0 16
    //1
    //1 1
    //1 2 1
    //1 3 3 1
    //1 4 6 4 1

~~~~~

~~~~~

kk...kk->k0...00k and those 2 ks "initialize" 2 new triangles https://ibb.co/Cpc13fdL a row has a form of R + 00...00 + R, where R is the corresponding row in the smaller triangle, so its possible to solve this problem recursively

The problem is a roundabout way of saying number of pairs where Manhattan distance is the same as regular distance. Or in other words, for every pair of staves that share one X or Y coordinate.

For a given x or y, the number of pairs is sum from 0 to (number of staves on that X/Y minus 1). For instance, if there's 2 staves with x = 3, then on x = 3 there's 1 pair (sum from 1 to 1 is 1). If there were 6 staves, there'd be 1 + 2 + 3 + 4 + 5 = 15 pairs. You see?

Very easy to make the needed amount of pairs with 500 staves because of how it grows. Sum of all numbers 1 to n is about half n squared, 500 squared is ginormous so 500 is more than enough staves to fit the problem requirements no matter what level spell.

Trick is to just make sure you keep them separate. Pick a y coordinate, stack staves on that row going up by 1 x every time until adding an extra one would take you over number of pairs. Then move up a row (change y), but don't reset x to not add any extra pairs. This solution makes all pairs have the same row, NEVER the same column so no two staves have the same x. (In my actual solution I reverse X and Y here but same thing). Works no matter what, just have to iterate through staves and move up by one, simple addition.

Add numbers 0 to n until their bitwise OR contains a bit that is not contained in X, then add X in the remaining positions. If the total bitwise OR in the end is not equal to X replace last number with X

try to add current mex till you cant( means if((mex|x)!=x)) break here for last element check if adding mex will make OR ==x if not just add X in the end else add mex

Hint: be greedy to maximize mex first. If failed, to get the k, assign the last element = k.

You need n-1 numbers that has same set bits as number x. And after getting those n-1 numbers, the last number will be a number which makes total OR value equals to x. So keep track of made value and required value. To maximize the MEX value take numbers starting from 0, 1, ... and so on, untill for some number some bits are on where x has off bits.

E<B,C,D for me

should have tried D that was easier

I thought so at first but look carefully at the constrains

in D , n<=2000 shifting left means you will take one element to the back and nothing else will change , no lets see how this will affect inversions all the elements that are bigger than lets X will add + inversions , all elements that are smaller than X will subtract inversions. so you just needed to check how numbers between ith element to jth element will affect inversion change. you can check my code

For D, the operation is basically shifting one element to the right some number of times. Every time it passes over a larger number the number of inversions INCREASES by 1, a smaller number causes it to decrease. Trick is the input size is actually small enough to just brute force it from there. So you just go through each index, count from one after til the end number of biggers and smallers. At any index here the minimized inversions is the number of smallers minus number of biggers. Keep track of max every time, O(n^2) answer fits the time constraints.

if i'm not wrong all the titles are references. thought i'm not weeb enough to remember all of them, they just sound like trash-mid anime titles.