Sure.

such an aesthetic looking as a tester comment !! cool

practically and theoretically not possible

welcome

Yes Sir.

DP contest at Atcoder

here is a list of category wise dp problems : https://leetcode.com/discuss/general-discussion/1000929/solved-all-dynamic-programming-dp-problems-in-7-months

you can contact with the co-ordinator

You will.

Seeing by the graph you have been grinding lately when people were sleeping. That will surely pay off my Friend. See you after blue colour Sir.

You can't in div3 or edu rounds. They have their seperate period of time for hacking after the contest.

But for Div2 you can try that. Just lock the problems and go to room section. And start watching other people solution if you found some counter-testcase hack it.

When the contest ends in some of them there is a phase called: The Open Breaking Phase. And it is in this phase that you must enter someone's solution and press the "Hack!" Soon you are given a choice on how you will hack: with a generator or a manual hack? For the generator you write the code for the test. For a manual one, you write the test by hand. But you can't hack during the contest, only after it.

Welcome to the community on the behalf of specialist community Sir. Hope you have great time and learning.

Same here.

*positive change

This is a div 3 contest which is rated only for accounts having less than 1600 rating.Your rating is greater than 1600,so you're not an official participant.

Trick : number of factors of a number (10^9) that are perfect squares is less than 100

Somehow iterate over those factors...

$$$a[i] \cdot b \cdot b=a[k]$$$

if $$$a[i] \leq b$$$ then $$$a[i]^3 \leq a[k]$$$

if $$$a[i] \gt b$$$ then $$$b^3 \leq a[k]$$$

So you can solve problem in $$$O(N \cdot M^\frac{1}{3} \cdot logN)$$$

My solution works ~1200 ms.

You need no fancy algorithms. Casework on if the common ratio is greater than the first term. For each $$$a_i$$$, enumerate the common ratio from $$$2$$$ to $$$(a_i)^{\frac{1}{3}}$$$ to count the number of triples where the common ratio is less than the first term where $$$a_i$$$ is the greatest term, and then if $$$a_i\le 10^6$$$, enumerate the first term from $$$1$$$ to (and excluding) $$$(a_i)^{\frac{1}{2}}$$$ to count the number of triples where the common ratio is greater than the first term and $$$a_i$$$ is the middle term. Remember to add the case of three equal terms.

This runs in $$$O(MAX^{\frac{1}{3}}n)$$$ using a custom hash unordered map. Implementation at https://codeforces.me/contest/1822/submission/203344438

Congrats bro.

Need to learn and know: Is it legal (fair) to search in a browser something like this during the contest?

Thanks! We are glad that you enjoyed the contest!

Congratulations for the color change

for D also, I found the pattern, but couldn't solve it

Same i just copy pasted from oeis.

For C, I saw that:

• There are 4 edges with length $$$n$$$, which are the edges of the square as well.

• In the square, the length of each adjacency edges is decrease from $$$n - 1$$$ to 1, and $$$n - 2$$$ to 1.

• In the middle, there is an edge with length 1.

In conclusion, the formula I used without simplifying is: $$$4 * n + sum(n - 1) + sum(n - 2) + 1$$$ (Note that $$$sum(k)$$$ is the sum of the sequence from 1 to k).

n * (n + 1) + n + 2 I reached at this after a lot of rough work but its sad to see people directly guessed the test cases or used WolframAlpha/Oesis to guess the pattern.

There are two problems here:

1. a check like mp[v[j]*r] > 0 is relatively costly because it will insert an element into the map if it didn't exist before. It would be better to use map::count() or map::find() to check if an entry is present without modifying the map contents.

2. you should probably break out of the inner loop when v[j]*r*r is larger than the maximum value in v, otherwise you end up doing 1000×N = 100 million iterations in the worst case.