How did you guys solve problem A so fast? I had to figure out primes for 1/p1 + 1/p2 + ... + 1/pn = 1, and after that combine them to form solutions for k > 1. It was not obvious to me...

Should I sleep before or after the contest?

haha you're right, I have just changed oset for a vector and got accepted

After this rotation, the query will be on the rectangle ((x+d, y+d), (x-d, y-d)). Why it's correct to use a rectangle that gets all positions of the grid that are at most 2 * d after the rotation?

https://code.google.com/codejam/contest/5304486/dashboard#s=a&a=2

Very weak test cases for B, I got AC large using a wrong max flow solution that allows to use a package multiple times.

My code gives 2 instead of 1 for this testcase:

1
3 2
10 10 10
10 10
10 1
10 10

Is it possible to change only the case of the handle?

ignore, same question above

You are right, I misunderstood the problem statement. Thanks a lot.

In the second test. Why can't I take each pupil on the bus per 3 meters?

let (p1, p2, p3) positions of the pupils i

t = 1.5 — (3, 1.5, 1.5) — p1 in the bus

t = 3 — (4.5, 4.5, 3) — p2 in the bus

t = 4.5 — (6, 6, 6) — p3 in the bus

total 4.5

Why this O(n²) solution http://codeforces.com/contest/699/submission/19241016 didn't exceeded the time limit? My hack attempt failed :(

Thanks!

What does log(10^15) mean? The upper bound is not clear to me.

What's the idea behind the topdown/recursive solution on problem C?

This new rating calculation is discouraging. It's need to work hard to get some points, it almost makes no difference.

Could someone give me a tip about how I can remove the "Memory limit exceeded" from my code for div.2 E? http://codeforces.com/contest/588/submission/13652236 I tried to make a LCA of sets...

I have used "News Feed Eradicator". This extension will be further useful, facebook is like a drug.

Div2 problem C — Can anyone explain me what "(a+b)/2*b" mean? Why can we divide by "2*b" ? Sorry, the tutorial was not clear to me :/

I get it now. Thanks for the explanation!

Could you show me a case when n == m the answer is 'no'? I understood for n > m, but why not n >= m?

You are using "cin" to read strings, it's very slow to read anything. It's better to use scanf instead of cin to reduce the runtime. You can also use "gets" to read string, it's faster than scanf. Take a look at: http://abhisharlives.blogspot.com.br/2012/06/really-fast-io-methods-for-programming.html

You could try to use fast I/O.

Take a look at:

http://ideone.com/Xec19L

it's better you to try to code it, isn't it?

I think you can do a BFS. Each number from set is a vertex, and for each number/vertex we can find out others ones with minimum operation, each number from set is an adjacent vertex of each vertex.

I think you could read some books and learn some basic algorithms to improve your understanding. The editorial often can be hard to understand because it considers you know something about it.

Take a look at http://www.quora.com/Computer-Programming/What-are-the-best-books-on-Competitive-programming-out-there