I hope it will be rated :D

There is wrote : " As they told me round will be rated ",how do you think ? it is rated or not?

So write the alphabet on the piano keys and write code while playing music. That will be the real art

Yes, It's first Turkish round.

I'm sure you will see.

!iralkaryab nisA

Bugun sinavvv erken kalkin cocuklar elimizde taze bilgisayarlari, ustumuzde yeni T-shitleri. (Turkish song) :)

2 hours

Onlar konusur Turkler yapar

Будут.

А почему может не быть?

I encountered the same situation last time.

I believe TLE is worst, because you probably need to write a completely different algorithm to get AC, whereas in the case of a WA, you may find a bug quickly and then solve the problem.

How about no.

I think you shouldn't ask for that during the contest...

I should say the same for Div2 :P 250-250-25000-25000-250000000

I just know div2 C didn't have test for n==1 cause I hacked someone....

I think many solutions of C will fail today.

OMG Nooooo I forgot. There can be only one bottle :( Now I realise why my solution was failing Div2 C on TestCase #3

Oh... that case fails me too. ;(

I have answer 4 for this test, but failed test 25.

UPD2: never mind, misread problem statement was sure that f(i,j) = gcd(... a[i — 1], a[j + 1] ... )

Even worse feeling, when you submit that solution after the contest and it says

"ALL TESTS PASSED"

Binary search the maximum height after removing k blocks, then binary search the minimum depth you can fill with k blocks. More or less.

Let's see if it passes the tests, but I just sorted the list and then searched from each end until I reached a value that was on the other side of the average (if I did this the difference was 0 or 1 depending on whether sum%n == 0) or you've removed too many blocks already, and then max-min

My solution is not using binary search. I am calculating final number where highest and lowest number end after k days, if they cross over answer is 0 or 1. 0 when total sum can be evenly distributed among n numbers, 1 otherwise.

For calculating final number where current highest ends, I keep on jumping to smaller non 1 frequency number if k is enough, else use up all of k and end up somewhere in middle (0 frequency number, this is breaking condition and at max happens only once). Maximum O(n) traverse.

Similarly calculated final state where lowest number end.

I hate hacking too :)) but I have to admit that understanding someone's code in such a relatively short time is really an admirable skill. Besides, it also helps careless coders realise their mistake. So, it is worth awarding.

The thing is that when you have a lot of people and only 5 problems, it's really hard to judge how well you perform (unless you performed exceptionally good/bad) so even though at times it might seem unfair, in the long run it is fair. And it really doesn't matter what kind of judgement system you use there will always be someone that will think it's unfair.

I did two binary searches one for best low value and one for best hi value. I think a ternary search will also work. Do not have the approach though.

I think it had only one bottle :

Testcase
0 1 0 2 0 0
1
0 3
Answer — 4

Try this one

UPD: Sorry, seems like I have seen some combination of your comment and the one above it and I thought you ask for Div2D :(

The test case was :

107 50 116 37 104 118
12
16 78
95 113
112 84
5 88
54 85
112 80
19 98
25 14
48 76
95 70
77 94
38 32

Participant's output
1579.43936611

Jury's answer
1576.895607473206

Dont know why I failed this http://codeforces.net/contest/672/submission/17863191

Use to_string() to convert integer to string in C++11.

And the easiest one for Div 2A http://www.codeforces.com/contest/672/submission/17859081 (no offence to the writer though)

Dynamic Programming..

state (index,adil,bera)

100000*2*2

adil and bera represent if they went first time or not..after that both start at bin cycle so no problem

Lets us consider there is only person.

After picking a bottle and throwing it in bin, then for all other bottles distances will be 2 * distance of that bottle from origin because you will start and end at origin after throwing each bottle. So first of all calculate sum of distance of all bottles and multiply it by 2.

Then find the bottle which at has this value minimum (distance of bottle from person — distance of person from origin). Add this to answer.

For two people, you can think of it in a similar way.

I solved it greedy...

Notice, that when Adil or Bera get to the recycling bin, they always walk the path to the bottle twice (it doesn't matter who is gathering bottles after the first recycled bottle).

That is, imagine these are distances from bottles to the recycling bin:
distFromBottleToBin[0] = ...
distFromBottleToBin[1] = ...
...
distFromBottleToBin[n - 1] = ...

If Adil or Bera started standing right on the bottles i and j, the total distance would be:
2·distFromBottleToBin[0] + 
2·distFromBottleToBin[1] + ... + 
1 ·distFromBottleToBin[i] + ... + 
1 ·distFromBottleToBin[j] + ... + 
2·distFromBottleToBin[n - 1].

Since they are not standing on them you should add up additional cost of traveling to the i'th and j'th (closest to them) bottles:
1·distFromBottleToBin[i] + distFromAdilToBottle[i] + 
1·distFromBottleToBin[j] + distFromBeraToBottle[j]

Notice also that if Adil and Bera are further from all the bottles than the recycling bin, then we have to choose the closest bottle to either one of them and one of them will be doing nothing.

Are you still using Windows XP?

I dint understand. Could please explain ?

Actually, most compilers are smart enough to return false, if nothing is returned!

maybe function bool in default version always return false;

study?

Well, maybe not the best way to do it, but you can just participate in contests, read editorials and implement solutions. Also for div1 problems, if you feel ok with it.

Thank you! It is very nice to hear this from you :D

Nope, this code gives 4.0000 and this is correct.

Bera takes the rubbish then goes to bin.

hi there! :o3

41 too as I see

Wrong answer on test 50 -_-

it is a risk to start contest from D.

You need to print 6 numbers after comma.

accepted code

You can see the test it failed at the bottom of the page. Precision problems.

NOOOOOOOOOOOOOOOOOOOO! why ;_; . just for it :(((((((((((((((.

thanks i have to kill myself

I didn't read your code. But your idea is O(n^2) right? For each point you check all other points. This was my innitial idea too, but there is a way to solve it in O(n), it's very similar to your idea, you just have to think a bit more.

No, they usually hide it :)

Actually my codes are ~100 lines for both. But as i see there are some quite long solutions.

102

My program returns 102 too and i got accepted in main tests

EDIT : solving by hand, answer is clearly 102, so we are both correct

OK, my program returns diffrent valuess running in Dev C++ and in online compiler so it's my mistake. Thanks.

According to your first hack based on previous code :-

The details of that hack

wrong answer 1st numbers differ — expected: '102.0000000', found: '200.0000000', error = '0.9607843'

Input: 100 0 0 100 0 0 2 0 1 1 0

Output: 200.0000000000

Answer: 102.000000000000

Time: 0

Memory: 4489216

[close]

I wasn't able to find any solution. I guess that once you solved the DP in N^2 or N^3 or something like that, the optimizations are not the big problem. What DP did you try to use?