Tutorial is loading...

Author: arsijo

Developer: stanislav.bezkorovainyi

Tutorial is loading...

Author: arsijo

Developer: stanislav.bezkorovainyi

Tutorial is loading...

Author: stanislav.bezkorovainyi

Developer: stanislav.bezkorovainyi

Tutorial is loading...

Author: Lewin

Developer: Lewin

Tutorial is loading...

Author: Noam527

Developer: Noam527

Tutorial is loading...

Author: Noam527

Developer: Noam527

Tutorial is loading...

Author: Lewin

Developer: Lewin

Tutorial is loading...

Author: _h_

Div2 C. Why my solution failed on test no. 17 I used Difference array and Binary search.? http://codeforces.com/contest/1075/submission/45302142

Not sure if this is your case, but I also used the same technique and got WA on test no.17 Maybe this test case might help you?

My code outputs 2 but now I see that it should be 1...T_T

Overall complexity of Div.2C should be

O(nlog(n)) because of sorting.NSA wants to know your location

That's true. Thank you.

Thank you for this excellent contest.

NSA here, open up

Dad?

aboAdnan, I am your father. Open up right now. Our family has been working for the NSA for generations under farmersrice and the codeforces triumvirate. Welcome to the underground world of cf.

Anyone know the tag for problem C Div 2? Solution presented above is greedy?

Binary Search/Two Pointers. But I think it is more helpful to view it as adhoc where you have to make observations. In this case main observation based on vertical lines span the whole grid, and no horiztonal touching.

Div2C: I did everything right, but forgot to write sort()... I am feeling so bad about this alexa play despacito

Div1C: i did everything right but didnt pay attention that border size can be in exponent 18, which is not supported by default by javascript which i use... And it is the second contest at row... can you imaging how bad is my feeling =)

Nice !

Div1D is not original.

Here is the same problem in one of the Indian Regionals — https://www.codechef.com/KGP17ROL/problems/XORQUERY

E and F aren't great either. My understanding is that CF wants to encourage problem writers and the bar for problem quality isn't set too high, so unless contest organizers want high quality original problems (which wasn't the case this time but in the past for this reason some companies had problems set by their own engineers and not outsourced) you should not have high expectations.

Hey,

I did not intend to copy/steal a problem by any means. Neither me nor the testers knew of this problem (and also no other task that is similar to div1D), otherwise it would be brought up. I'm sure you understand that we want to have tasks with as high quality as we can, and also being original. Still, I'll try to look deeper for tasks similar to mine, but honestly, I think it's impossible to assure my problems are original.

Anyway, I hope you still managed to enjoy some other tasks from the round :).

Hi, Can someone tell me whats wrong with Code??

Interactive Problem

Thanks in Advance

The problemis that you need to run a bfs to find the closest vertex instead of finding any vertex like you do with your dfs.

Isn't W(3,5)⊕W(5,2) equal W(2,2), not W(3,2)?

Because [3⊕4⊕5]⊕[2⊕3⊕4⊕5] is equal to 2.

And that would mean that W(a,b)⊕W(b,c)=W(a,c)⊕b.

In the presented solution, we look at the subarray by its end borders. So,

Can anyone help me out with Div. 2 F? I'm getting WA on test 14. I'm using dsu and merge small-to-large, and but I cannot tell what my error is and this problem is quite hard to debug.

45341296

update: I fixed my error, a pretty big one, actually. I was updating the subtrees with p[l]^p[r] instead of x^p[l]^p[r].

Thanks a lot for detailed explanations!

Can anyone explain the dp part of Div1-C, Also what are c1,c2..,c6 wrt this question?

We need to find 3 points, such that |x1 — x2| + |y1 — y2| + |x2 — x3| + |y2 — y3| + |x3 — x1| + |y3 — y1| is maximal. We are going to try all possibilities of assigning + or — to each of these pairs. For example if we have +--+-+, we are going to have (x1 — x2) — (y1 — y2) — (x2 — x3) + (y2 — y3) — (x3 — x1) + (y3 — y1), which evaluates to 2*x1 — 2*y1 — 2*x2 + 2*y2 + 0*x3 + 0*y3. c1 here is the coefficient next to x1, c2 is the coefficient next to y1...c6 is the coefficient next to y3. So we calculate arrays P, Q, R as in the editorial and then do dp to get the maximum for such combination. The dp state is [current point][taken points]. You can also check out my implementation 45365331.

Thanks a lot :) , I got majorly confused on how taking all possible combinations wont give wrong answer because we are taking the cases are not possible too, For those who are confused on this part -> max value of the expression can be |x1 — x2| + |y1 — y2| + |x2 — x3| + |y2 — y3| + |x3 — x1| + |y3 — y1|, any other case will be less than or equal to this, Hence this wont affect our maximum. With similar reason we can prove why this wont work for minimum.

Thanks, stanislav.bezkorovainyi for writing a descriptive and well explained tutorial for problem 1074A.

Thank you :)

DIV2-B any better approach??

can someone explain A part.... Dont understand why is z=x