Sorry, the implementation might be unauthorized, we're solving it...

https://codeforces.me/contest/2224/submission/373788886

The java implementation for 2224D solution 2 got TLE with Fenwick tree, but it passed with merge sort. Wondering why...

The solution for Problem $$$\textbf{Div2}$$$ $$$C$$$ is not clear!!

I created video editorial for D. Zhily and Barknights

I created video editorial for E. Zhily and Signpost.

can someone pls point out the mistake in this? (the function full of comments is taken from gfg)

Specifically, we have Mv=lcm(Mu,du). Then m≡Ru (mod Mu) implies m≡kMu+Ru (mod Mv), k∈[0,gcd(Mu,du)−1]. By enumerating all k and computing the corresponding r, we can obtain the congruence equation for su,r. If some r is not obtained, then the congruence equation corresponding to su,r has no solution.

Should not $$$k$$$ range over $$$[0, \frac{d_u}{\gcd(M_u, d_u)} - 1] $$$ ?

EDIT: I got it. $$$k \in [0, \frac{d_u}{\gcd(M_u, d_u)} - 1]$$$ if the modulus is $$$d_u$$$

I'm new here. Can somebody tell me why the source in the implementation shows "N/A"? Is my viewing permission sufficient?

can someone please help me figure out why my submission resulted in rte for preblem C? i used a binary searching like technique on centroid

https://ideone.com/OYXng2

slowEditorialForces

Is problem (div2)D solvable in Python without getting TLE?

Solution of problem B can be implemented easily by using mathematics if we observe pattern in the optimal answer when $$$max(a_1, a_2, ..., a_n) \geq mex(a_1, a_2, ..., a_n)$$$ and when it's not.

Check out this submission

The fastest task analysis this year!!!

Anyone else tried solving 2E/1C using binary search?

Hi, I am not able to see the code in the submission given under implementation. Does it have anything to do with this being my first contest? How can I fix this? Appreciate anyone's help.

Also, can someone tell me where I went wrong in my solution for Problem B, Div 2?

Edit: Got it, LL conversion error.

I think Div2 problem C is much simpler than Div2 B.

weird contest

C is easier than B...

For problem div2 E/div1 C, i suggest an alternative solution using binary search on centroid decomposition tree. Though i will not go much into explaining it, the way it works is check whether a node u in the centroid tree is reachable from 1, if it is then the answer must lies inside the subtree of u, then compute the next normally. If not then it can be seen that the answer is not inside the subtree of u, so we find go to the parent of the node u in the original tree and its centroid, note that we always move to a centroid with higher depths in the centroid decomposition tree.

I wonder if this technique caa be extended even more though, as centroid decomposition is almost like DNC on tree, so maybe we can extend it even further?

in question A there is written we have to find number of positive number in array, is 0 not a positive number??

any help, my submission submission got different wrong answers for different c++ versions , i.e : in c++23 it got WA on test2 case 9 , in c++17 got WA test2 case 6 , in c++20 got WA test2 case 11, what's the issue despite i submit the SAME code && the "failed" testcases as codeforces saif got the correct output Locally but different output on codeforces

Can anybody explain whats the proof of greedy in div2 C?Like why is it that balancing the difference in A & B is optimal?What if A has more open brackets so we can have its diff less than B and give more open brackets to B before while ensuring that no one has diff<0 at any point?I have the intution on why greedy is working but i just cant prove it

i had a different solution to 2224C (zhily and bracket swapping). the basic idea is to alternate between setting $$$a_i$$$ as '(' and $$$b_i$$$ as '(' at all entries $$$i$$$ such that $$$a_i \neq b_i$$$. then, just check if the resultant strings are valid. it is easy to prove this is correct by noticing the max discrepancy in open and closed parentheses this creates is one.

for an example, if indices 4, 6, 9, and 10 differ between strings $$$a$$$ and $$$b$$$, set $$$a_4$$$ as '(', $$$a_6$$$ as ')', $$$a_9$$$ as '(' and $$$a_10$$$ as ')'. then, check if $$$a$$$ and $$$b$$$ are valid.

in $$$D$$$ let $$$d = a * b$$$ then answer is just $$$\frac{inv(d) - n * (inv(a) + inv(b))}{n (n - 1)}$$$ where $$$inv$$$ is the number of inversions.

$$$\; \; a * b = [a_0 \times b_0, ..., a_0 \times b_{n- 1}, a_1 \times b_0, ... ]$$$

Div 1F: What exactly is multivariate Lagrange inversion and where can I find a proof (in English)?

Has anyone done 2224D using binary search in Java, I would like to see your solution, I am getting TLE with the same code in java but it runs in cpp.

Overdid the first problem by lazy seg tree LOL !!!

why is Div2 D not accepting O(N^2 * log(N)) solution? Is my solution too expensive because of constant factor? Did anyone solve this using pb_ds ?

My submission: 382022983

From D1E editorial,

In fact, during the problem preparation, this conclusion is naturally obtained from a more complex but powerful structure; for simplicity, we now consider the direct proof above.

What's this structure?

help https://codeforces.me/contest/2224/submission/394401365