The only step costs 2 dollars, meaning the amount of money left will never exceed 1 dollar,

Why?

EDIT: You people have no respect for the Socratic Method.

Oh I see it now. The initial money is a real value.

NO

That's against the rules. We'd be disqualified for discussing the problems before the contest ends.

An idea for organizers: encrypt the input (ZIP should be ok I think), make tests downloadable at any moment and show password on request (which should also start countdown).

Same thing works for uploading: make user upload hash of their output (probably, with some salt so sharing hashes does not help) under 6 minutes and then allow more time to upload actual file.

why they have this one time 6 minutes rule ? it's so annoying for countries with crappy internet

it's fun because AFAIK there is special initiative for facebook employees "emulate crappy internet on your workplace", exactly for such cases. I suspect hackercup developers opted to avoid it

In sample inputs T < maxt.

My Outputs MD5:

a) 8ac11aa9fc477d369246edcfc70b9520

c) 7493cd269e214695e9492e3695b638f7

d) wrong((

My in/out files

Apparently, B is incorrect:)

You can grab the official I/O here: https://www.dropbox.com/sh/6f6wqzt8372azqo/AABAA8gWT44c_x4LBHnrTkCqa?dl=0

We'll be putting up a solutions post in a bit.

I completely agree. The test cases are extremely weak and certainly not fair to everyone. In problem B, a lot of people complained that the input contained N > 10000 (before they changed the constraints). The input file I received had N < 1000 for all cases. I know that the inputs are generated randomly and are different for every contestant, but what kind of randomly generated input contains N < 1000 for some and N < 10^5 for some others, considering all 50 cases? And how exactly do you call this fair?

Exact same problem repeats in C. It was mentioned that Ci <= 10^9, but in my input file, all Ci < 1000. Even apart from this, the test cases for C are very weak. Already a lot of contestants are mentioning the cases for which their solution fails in the comments. Not to mention those tricky cases where not using long double will lead to precision error for some contestants.

Please take care of these issues in the next round. I can accept somewhat weak test cases, but the input file of two contestants shouldn't differ so significantly in their constraints to ensure a fair competition.

Can you please explain why we need long double? I mean can't double handle 10^18? and only time here precision comes is when we multiply it segment values with 0.5, and finally divide by b-a. In the whole process,there is no extra division, so I'm confused why double won't work. I used only double, and it passes the cases you guys mentioned here :S

We want to finish washing the laundry as soon as possible. So we put the possible finishing times for the washing machines in a priority queue.

Then, we process the L items and assign each one of them to the earliest possible finishing time. This is just the minimum in the priority queue. Suppose this minimum time is X. Then, this washing machine is able to finish washing another item at time X + WashTimeOfThisMachine.

We can use the same reasoning for the driers, but it's not necessary. We will put an item to dry asap as well. This is X + D is there is a drier available, or D past the next available drier time. If we are processing item i, the next available drier is available at time WashTime[i-M] + D.

I solved it using binary search to find the minimum time to wash everything. If this minimum time is X then I divide X by wi for all w to get the maximum loads to be inserted in a washer and store these in a vector. Then I sort them, and choose the first L values after sorting. If p=(L-1)%m, then this is the dryer number which will have the last load. Thus, I calculate minimum drying time of last load by operating on p-dryer for L/m times using formula time=max(current dryer time, load's wash time).

I am just curious.
Does somebody find the self-documenting style of code clear and intuitive?
I try to write my code in that style, so it can serve a higher purpose and help someone struggling with thier's understanding (rather than just solve a problem). But recently I've heard an opinion that I am wrong and self-documenting code is more confusing than clear.

For example, what do you people think of the solution to the problem.
Do you find it clear or confusing?

Next Saturday at 10am PST.

Range 1 ≤ W_i ≤ 10⁹ is not specified in the Problem B.

Yeah, the data weakness was my fault. To be clear, the bounds weren't changed during the contest, in the sense that people were already receiving files with the higher bounds (I just typo'd the number of zeroes in the constraints). Additionally, the large cases weren't forcibly included in everyone's input files.

I'm sure there are a few people who wrote a slow solution but got lucky with a bad input file. Our precedent is to let those solutions stand, especially in a round with no fixed number of advancers. You can be sure I'll be quadruple-checking the Round 2 data.

My approach: find cumulative sum of the costs. then for each cumulative sums from last to first find the disjoint sets of modulos that can appear. Actually there can be 3 types of them call them L, M , R and they are set by the total cost of all the steps.

take the following test case:

2 9 20

8 2

the total cost is 10

the first number that is divisible by 10 after or equal to A=9 is 10

so L is [9,10)

the first number that is divisible by 10 before or equal to B=20 is 20

So R is [20,20]

and M is the range between L and R [10,20)

So if we modulo it by 10 we can get the following ranges:

1) From L: [9,10) 2) From M: [0,10) 3) From R: [0]

However notice that the modulo range those are above or equal to 8 can be spent by the first cost as the programmer greedily spends his money from first to last step

So the ranges are further edited.

1) From L: [1,2) 2) From M: [0,8)[0,2) 3) From R: [0]

From here you can calculate the expected value by calculating the length and midpoint of each ranges

That's really interesting :)

There aren't that many distinct tournament trees due to symmetry, so I guess your solution has a high probability to find the right answer in all tests.

If even in Google Codejam finals tomek ALMOST got away with using simulated annealing... Why not? :)

Even I did it in the same way! but my output failed.
For N = 1, 2, 4, 8 we can just iterate over all permutations. Which fits very well in the 6 minutes time.
But for N = 16, I used random_shuffle() doing around 70000 iterations. But sadly my output failed for 1 test case. Where as it worked for 80000 iterations. :(

This random has a bigger chances)))(отжиг)
http://mirror.codeforces.com/gym/100875/submission/15430079

Don't worry, you aren't alone ))): http://mirror.codeforces.com/gym/100875/submission/15422585

/*because of the absence of this condition (it should to be after the cycle ALSO, but not just in the cycle), my sol was wrong and i earned JUST 25 points*/ 
                if (cnt > 4)
                    cnt = 1;

Please tell me what is BPM?