Impossible bro

You will get back to PUPIL

same goal in this contest

Congrats on becoming a blue

.

As a participant i am at a loss how this information is relevant to this contest

Wishing my next one to be the same!

Good luck

Same :)

how?

Both have the same meaning.

онатдыр

Me too ^_~

but "do not have a point of 1900 or higher in the rating."

Transfer it to me

You get points like you submitted it 10 minutes later.

No, that's not what they meant. You should be aware that if you got a accepted after(for example) two failed submission and someone got it from the first time (consider that both of u solved it at same time) he will get penalty less than u .Also,He rank higher than you.

No, if you do not submit any problems, then it will be unrated for you.

is there any chance to rated this contest

Bruh, you solved upto F but couldn't solve C :/

Why? Because of the problem’s statement? I don't see any problems and it's a good problem.

i had the same thing, try virtual contests, they bring less stress gl

your profile picture reminded me of Sam Hogan. Good old days

nvm, looks like it's just checking all the cases. I thought it's some classic algorithm.

You can check out my solution for e, it's just maths :)

You can just precompute prefix sums for d <= sqrt(n), and for d > sqrt(n) just brute force the answer. In the first case a query takes O(1) and in the second case its O(sqrt(n)). Preprocessing takes O(n sqrt(n)).

Basic idea is precomputing answers for all d <= $$$\sqrt{N}$$$, which can be done by prefix sums in O(N*$$$\sqrt{N}$$$). For d > $$$\sqrt{N}$$$, simply iterate and get your answer.

why have it unrated for all, just the cheaters should be eliminated

play optimally means that if there is a winnig state for player A, then player A will play in such a way that will win.

No need to check every possible move if you find a general case for wich A wins

#define bin bitset<64>

Your bitsets only store $$$64$$$ bits, not enough for $$$10^5$$$ bits.

can you explain your approach for the problem F ?

Same lol :)

I kept tripping over the math for B and C.

Then for D, I was thinking it was a heap/priority queue solution, but I couldn't figure it out.

Was hoping to make Pupil this round, but oh well :_(

just need to participate in as many contests as you can

I found that D is extremely likely to be fakesolved for me. I cannot find the bug since my code pass all examples and got WA on test 2.

With 1 solve and your default expected performance being 1400 you should expect a huge drop in rating. However since you're unrated, there's a +500 bonus on the first round, so your rating will still increase.

You would need to wait for the system test to end, which would happen about 18 hours from now, since there is open hacking phase.

Do continue practice. It will improve your problem solving skills and logic building skills. I am also a amateur coder .But now i taking it seriously I was able to solve two problems. Earlier i was not able solve any problem. But with continuous practice i have some improve skills

F can be solved in O((n+q)*sqrt(n)) , by dividing in two cases d <= sqrt(n) {for which you can use suffix sums) and d > sqrt(n) for which just sum directly. However , I believe taking d <= sqrt(q) , will be more better.

you can just simulate it !

Single if is enough to compare two intervals where two players could meet.

241806826 241811094 241818999 241814972 this particular method was posted online

using a while loop seems to be hackable. I would write some test case but meh

Maybe this can help

My solution (241838398) precalculated prefix sum of a[i] and prefix sum of a[i]*i, for all $$$d\leq\sqrt{n}$$$. It's $$$O(n^{3/2})$$$ time and $$$O(n^{3/2})$$$ space, which seems to just fit the time and memory limit.

Make diagonal prefix sums, like 2d prefix sum and try each of 4 possible directions in O(1) for n*m shooting squares.

I failed to implement it in time

Bruteforce every possible starting location. To move the starting position by one cell you need to add/remove subsegments of a row, column, or diagonal. From here it's down to your implementation skills.

There is a simple solution in $$$O(nm*min(n,m))$$$.

Assume WLOG $$$n \le m$$$, The solution is: iterate over every $$$(i,j)$$$ as the start location, iterate over every row $$$x$$$ that the $$$(i,j)$$$ will reach with the operation, find the sum of the elements of the row that is in the range of the operation with simple prefix sum.

This work in O(n^2 m), if $$$m < n$$$, just rotate the input. This is code: submission

I had a slightly different solution:

Solve the problem for the down/right diagonal only. Rotate the grid four times and print the best answer over each rotation.

To solve for the down/right diagonal, notice that a shot from point (r, c) of size k is equivalent to a shot from point (0, 0) of size k + r + c, minus the combined number of targets in the first r-1 rows of that shot and the number of targets in the first c-1 columns of that shot. This can be found using inclusion/exclusion and prefix sums.

So, we can enumerate the targets in order of their Manhattan distance from (0, 0) and check starting points for the shot using a sliding window technique.

Yes, we will fix it and rejudge all affected submissions.

Perhaps overflow?

Also, vector erase method is $$$O(n)$$$, so your code is $$$O(n^2)$$$ which doesn't fit the time limit. Instead, try to use a pointer int L = 0, and increment it to simulate erasing the first position.

You are using erase, I did the same and got tle. Either use a dequeue or two pointers.