Plus, It thanked MikeMirzayanov for his awesome Codeforces and Polygon platforms. :)

How many did youb solve in the icpc preli?

It has just 5 problems, so yeah, it's gonna be hard.

That's quite not possible.

Yes, I can not understand the statement of problem B. Kind of shit!

I have the same problem.

2 -10000000 1 10000000 1

This solution does not fit the time limit restriction.

I solved it by binary search on radius R: then for a fixed R for each point you can calculate the interval of the x coordinate of the center of the circle (by pen, paper, math). Then you need to check whether all n these intervals have an intersection.

what was ur greedy after the binsrch on answer?!

I also got WA on pretest 4. But I overcame it after increasing the number of iterations of the binary search (unfortunately I got TLE on later tests and didn't have time to optimize the algorithm :( )

edit: nvm, my algorithm was not the same as yours

HINT: Remove numbers at an odd position's, until there more than 3 elements.

As I guess: 6

Answer would be:

1 1 1 2 2 6

Problem D was one of the best problem I have ever tried but missed to submit, i just needed 10 seconds, found a little bug in last moment (-_-).I tried only problem D in 2 hours.

my pretest passed solution:

initially current_gcd = 1, then remove every even index (1-indexing) if n is even, or remove every odd index if n is odd from the original sequence so we can get larger gcd faster, except when n = 3 we just remove index 1 and 2 because its better.

The idea was the following: Initially the gcd is 1 and this needs to be increased to some g>1. Consider the smallest prime factor of g (say p). To increase the gcd from 1 to g we need to remove at least all the numbers which are not divisible by p. We need to minimize the number of numbers not divisible by p, which is same as maximizing the number of numbers divisible by p.

It can be observed that except for n=3 in all other cases, p=2 (We can hard code the solution for n<3). After removal of "bad" numbers, all the remaining numbers will be the first floor(n/2) multiples of 2. Now it is possible to recursively build the solution, by solving for n=floor(n/2) and multiplying the resulting sequence by 2 elementwise.

Problem C was not Math, just IQ test. Don't be tricked by GCD !

You got E for dp today.

I solved E in 200 lines using 3 segment trees and dp.

B find every cell which is surrounded by ‘#’， when found one，copy these '#'s to another graph(initialized by ‘.’). compare two graphes，judge whether they match.

I don't know how to solve C，(((QAQ ORZ

I didn't even read E during the contest because all the other bullshit took away my time, but after reading it now, I think it can be solved with prefix jumps over ancestors and binary search, but I still have to test my solution.

I'll see if I'm right once the turtle finishes system testing all solutions for this sorry excuse for a contest.

You thought THIS was good?

Well, yeah bro..... you're either too easy-going or you're a math freak. Choose whichever you like.

I actually really liked the problems, especially B and C, and I think I have an idea for E...

Problems are more enjoyable to one when one has managed to solve them.

Well I got AC with that idea (at least on pretests).

1e9 pure computations only take about 0.2-0.4s for C/C++ solutions. Other languages (Java, Python, etc.) will surely get TLE for that.

Looking through all of your comments so far... you seem to be a very, very angry man

My approach = Question says center of square 3*3 will not be painted rest all elements of 3*3 grid will be painted,so center of matrix can be (2,2) to (n-1,m-1). So for every possible center check for all 8 side boxes and if none of the neighboring cell contains '.' then assign a certain fixed value to all neighboring cells. Repeat this process for whole matrix and at last check this matrix with actual matrix ( i.e if matrix[i][j]=='#' && value not assigned). If both are same then "Yes" else "No".

I think it will be more fair to rejudge E with stronger data.

The idea of starting from leaves and going as far as possible each time is correct. It could time out with strong tests if done naively (i.e. simple dfs), but the answer should be correct.

The key is to try to go up as far as possible for all leaves, even if there's a middle vertex that's already included in another path from another leaf. For the sample test you provided, a correct greedy solution would do the following...

• Process node 3 and make path (3,2) with total weight 6.
• Process node 4 and make path (4,2,1) with total weight 6.
• Skip node 2 because it's already included in a path.
• Skip node 1 because it's already included in a path.

Well... if you look at the correct answer and compare it to the your program's answer, you'll actually find that it's pretty far off. Quite a bad approximation (even the accepted version barely passes the limit). 1e-6 is a standard precision requirement.

I'd say the implementation could use some work. Most people who got accepted had way better (tighter) approximations. I'd say it's LUCKY that your second submission passed (and not unlucky that the first one didn't).

Also... it doesn't really have anything to do with luck. This particular test is easy to think about, it's obvious that it has the highest probability of introducing precision errors and is also very easily to compute by hand (with pen and paper). Therefore, it should be one of the first cases you'd want to check your program on. Not stress-testing your code is your fault, not anybody else's.

TL;DR: try to analyze your mistakes, improve yourself and work to get better instead of blaming the system :)

I think task D was a valid problem, since dealing with real numbers' precision is key in competitive programming. However, I think this is a valid suggestion as well. If the problemsetter's intention was for contestants to figure out a thoroughly precise solution, Why does a weak program pass by fixing small details? As far as I am aware of, a good problem should be able to break any solution that is not the intended one, and this one didn't. Not to mention that this is a two-hour contest, where penalty for wrong submissions and delay is significant. Therefore I do believe it is a bit unfair to leave ambiguous constraints, since people will waste too many points on a solution that isn't even supposed to pass.

System test for E is so weak. Many O(n²)(worst cases) approaches pass the test, which is unfair for those who employ a correct algorithm of O(n log n).