Yes, the registration will be open soon.

Division 2 will contain the same set of problems as Technocup Round. Division 1 will contain harder problems.

The one actually is MikeMirzayanov...

Participate in the contest :)

Go to the exam :) If you don't participate on a div2, nothing will happen, You can do it virtual contest :) But You'll not have a virtual xm or something like that!

i would suggest to go for A, because contest comes first :p

Sure

HB

If you register for Technocup 2017 — Elimination Round 2, you will get both rating change and participation, if you register for Codeforces round 380, you will get just the rating change.

That you copy pasted this.

yes.

I think it'll be like 379 or a little bit harder. You can solve something I assume! :)

You can see problems from the Technocup 2017 Elimination Round 1.

yes

It happens; )))))

Same here... Had to use ios..

Strange..used cin but didn't get TLE, 22345324

t < s in A brought me down by 100+ ranks already :D

I solved D using greedy and C using binary search, dunno about E.

Find all the empty segments, and the max. number of ships placeable in the field is sum of (length of segment i / b). Let the number be x, and you need to shoot at least x-b+1 times to ensure a hit. Then just shoot at places where the number of placeable ships will decrease for a sure.

D — greedy — one should take every kth consecutive zero if a=1, dont take last a-1 such places.
E — consider height of tree to be 1..n-1, calculate answer as function of distinct values not present and values lying outside, take min overall.

Boring contest

Still top 100 so nice job anyways!

What exactly do you consider 'standard array length'? You have different constraints in each problem, it literally takes a second to look at them. Don't whine.

You may notice, that k <= 100 and -100 <= right - left <= 100

UPD: where 100 is surely = O(sqrt(n))

My solution with binary search didn't pass until I wrote .sync_with_stdio(0). How could O(n*k) pass then?

Now it's running!

UPD: uh-oh... It seems that queue is stopped?!

Use prefix sums.

They'll publish it tomorrow or later.

They noticed that systest today is a little bit fast, so they paused it a bit because CF traditions says it should be slow.

I think so.

They started div2 testing. Looks like they can't do both at the same time...

What does on fuel mean? If you mean tank capacity, my binary search works in 0.2s (it's 0..1e9 but it should not matter)

I am pretty sure that the reason for your TLE is integer overflow:

int lo = 1, hi = 2 * s;
.
.
.
int mi = (lo + hi) / 2;

Same thing.

It is not because of slow binsearch, but because of hanging binsearch. They have overflow or other mistakes in implementation.

Even solution with extra internal binary search (e.g. O(n·log²n)) passes the tests and fits in half of TL.

Pypy is much faster than python I think.

that's why there's pypy compiler

I think

1
o

This is where I got WA. -_-

Sure.

lol

Probably slow IO?

cin/cout http://codeforces.me/contest/737/submission/22362184

If the size of input/output is large enough and you want to use cin/cout, you have to call "std::ios_base::sync_with_stdio(false)" first.

Welcome to CodeForces buddy.

If you insist on using cin, cout make sure you are using c+14 and use ios::sync_with_stdio(false) .

Pretests are actually meant to be potato quality

You also output "?".

you outputed "a***b?", not "a***b"

That could be the case for div2; however, for div1 main reason of low participation seems to be a lot of other contests clashing with CF round — OpenCup stage, bunch of regionals (SWERC, CERC, NWERC); also some teams are on their road home from MIPT training camp.

Issue you mentioned has been discussed a lot of times already, and I don't think there are going to be some changes on it in near future. This system itself is really vulnerable (simply register another account to read problems I guess). It needs a lot of harsh actions — like checking IP addresses of contestants etc. Nowadays people aren't even banned for having multiple accounts or competing in teams using single account (when one account sometimes makes submissions from 2 or 3 cities during a contest). I'm personally fine with it in case we have to choose between platform improvement or having more contests — and improving "fair play" system :) I'm sure that I'm not at the top not because of cheaters but because of other contestants being much better than me :)

you print something strange at the end! I think it's '/n';

I was wondering why my solution to (div2)B wasn't working until I noticed that I'd erased the part which was reading the data and that I'd been processing uninitialized 2d array. :D

f doesn't return anything if it goes on until the end of the string

Hi, it's more convenient to post your whole submission. Let's say: 22363876. Also, as tfg has noted, your function f doesn't return any number if the control gets out of for loop without ever evaluating else clause. The behaviour of function f is therefore probably undefined and it just so happens that it returns correct value on your (and mine) computer, but not on the OJ. One way to modify this could be for example:

int f(int n)
{
	int i;
	for (i = n; i < N; i += 2)
	{
		if (d[i] == 'g'&&d[i + 1] == 'o')
			continue;
		else
			break;
	}
	return i;
}

That being said, it might not be the most beautiful solution to your problem. By the way, if you compile your program with "-Wall" flag, it gives you a warning:

g++ -Wall we.cpp
...
...
...
we.cpp:13:1: warning: control reaches end of non-void function [-Wreturn-type]

We have to change as few zeroes as possible to ones so that it is impossible to place a ships on the zeroes. Calculate for the initial configuration how many ships can be placed. After that, greedily remove the possibility of placing a ship, whenever there are b consecutive zeroes (i.e. change the b-th consecutive zero to one). Do it until fewer than a ships can be placed.

That's the right idea.

You check how many changes you need to make to create a tree with total depth from 1<=K<=N-1 (a tree with only a 0 has depth 0). All numbers from 1 to K must be present (there will be only one 0) so to do that you check if you can use the numbers > K to fill the gaps since you will have to change them anyway (to K or some other number lesse than K). If it still isn't enough, you will have to use the numbers from 1 to K to fill any missing gaps. If you consider the numbers that are 0 and aren't the chief as N, you will get the right answer since they are wrong and you need to change them.

Sort the cars by their volume of fuel and use binary search to find the first car that can arrive to the destination in time, lets say it's the k-th car. Then just itterate through all cars from k to n and pick the one which costs the least.

My code. Sorry but it's pretty ugly :D

We can binary search the least fuel f we need to get to the destination in time. And the answer is the cheapest car whose fuel limit is larger than or equal to f. Now the question is: how to check if we can get to the destination in time with a certain fuel limit.

Denote the distance between a pair of adjacent gas stations as d, the distance we need to drive in accelerate mode as a, the distance we need to drive in normal mode as b, the fuel limit as v, and the shortest time we need to get to the destination as t.

In order to get to the destination as fast as possible, we should use as much fuel as we can.

If 2d ≤ v we can always drive in accelerate mode, i.e. t += d;. If d > v then we don't have enough fuel even if we always drive in normal mode, so return false; in your check function.

Now we only need to consider the case when d ≤ v < 2d. Because we need to use as much fuel, so we have the following two equations:

2a + b = v and a + b = d.

Solve them, and we get a = v - d and b = 2d - v, i.e. t += (v-d) + 2*(2*d-v);

You can check my code for more details. Remember to use long long type or set the upper limit of the binary search to a little bit more than 1e9. I set the upper limit to 2e9 during the contest and it overflows :(

My code: 22361702

Not sure if this is the case here, but sometimes they have testcases that specifically catch unordered_maps. I think it's stupid, but it's happened in the past that you need to use the normal map (which is a bst) to pass the test-cases which are designed to specifically beat unordered_maps.

Looking for help in solving div1E.

I have drafted the minimum time should be max single machine/child play time, and I should rent second machines greedily by the length of play time queue. However, I have no idea how I should cope with providing a valid schedule.