good one!

mrweirdguy orz

Nice one, will def try it

Pretty useful, thanks!

Both the orderings are incorrect, my ordering fails the case given by chicubed

In your ordering, the first 6 queries would be:

(N-1) (N+1) ( N ) (N-2) ( N ) (N-1)

The position of X can be:

(N-3) (N-2) (N-1) ( N ) (N+1) (N+2)

After this we can keep moving X forward.

I think the problem just can't be solved using such orderings.

Damn we didn't thought about that, the interactor missed this as well leading to our incorrect solution getting AC

Submission link

Edit: The solution turned out to be wrong

An alternative solution to G:

While n>3, we can query {n , n-1 , n+1}, if all these are 0 then it implies that X in range [1,n-1]
When n==3, we can ask the following queries: {2, 4, 3, 2} and we will determine X.

The total number of queries is 3*(n-3) + 4

A lot of people cheated in D2 E as well, putting some of those here

Link 1

Link 2

Link 3

Link 4

Even there solutions are quite similar, looks like they both copied from the same source

Really enjoyed the contest, especially the C problem

I ran the editorial solution on the testcase in custom invocation in C++ 17, for that it ran in 3000ms, however for C++ 20 it ran in 800 ms, i don't really know why is that the case

For G the testcase

2

1 100

1 100000

This 1000 times would lead to TLE and is a valid testcase

orz

bilkul sleepyAdarsh bhai XD

heyyyankit orz!