They used to be in an investing company together. They were like best friends. They were the founders of the company, they worked together.

Snowden offended Barack Obama as well. He's got big balls.

Could you give me examples? We have some exclusions because of local finalists.

When I clicked on register button , it's showed me that I am not eligible. So that's not a problem at all.

No, some people just want to stay informed on what's happening with the contest

Ah yes, imaginary internet points. They truly are what make the world go round.

the only way is to submit wrong and wrong til it gets pretest passed !

Imagine it's a graph. You need to make it into a single cycle, and have an odd number of bs. If you have x > 1 cycles, you need exactly x operations to transform them into just 1 cycle, and then 1 more operation if the sum of all bi's is even.

The graph is combination of cycles, you need to make graph a single cycle. And if there is even number of 1's you should change something.

I counted the number of cycles in permutation and then calculated the number of changes we need to make in order to combine all of these cycles into a single cycle.
If all of the numbers in b were zeros I added 1 to the answer.

This approach is failing at test 8. Either my logic is completely wrong or I made a mistake in my implementation.

what was the right limit? (the left limit was 1, sure)

Can you explain your logic for div 2b ?

Binary Search was not necessary. You could brute-force by computing the number of pillows that you need for the answer to be at least 1, 2, 3... until p, where p = max(n-k+1, k), and then you could divide the number of remaining pillows (if there are some remaining) by n and add the result to p to get the answer.

I guess I need some extra hints :)

Also are you sure you aren't talking about Div1 B?

Div 1

Strangely, this gave TLE with C++11 and passed in 1263ms with C++14. Never knew C++14 had so much better optimizations.

C++11

C++14

Never would have thought there is such a big difference in speed between int and long long.

You can do the dp in O(N²) by using a prefix sum table.

We can also get rid of binary search by can starting from the root and going to the right child if it's maximum is positive. It's O(N log N) this way.

Um, I'm confused, Why do you need lazy propagation for this?

..... I came up with this idea and then forbade it.... that's too stupid

It is good when people complain only about problem F

At least problems A, B, C and D weren't bad which is not the case for many other contests

I think W4yneb0t's comment is sarcastic.

The problem is that arriving at the solution is straightforward and the difficulty lies entirely in coding it. I personally don't find that very fun.

Lol, I referred to problem E when telling about bignums xD. Should have mentioned it :D.

F is surely not about bignums, but I don't like F because it has very low ratio of "difficulty of getting idea right : difficulty of coding it" and that is more or less what I use to sort problems by how fun they are. We need to parse everything (hooray) and what is left to do is to make some rules producing (length of expression mod P-1, value mod P) for every expression. Only rule which I didn't figure out was those intervals, but I didn't think about it for longer than 15 seconds, but I guess it still can somehow be done without some brilliant ideas.

Btw, I do not predict your comment to get downvotes ;p.

I did that but got WA.http://mirror.codeforces.com/contest/760/submission/24051551

Well I got that we can transform the problem to counting different subsequences, but I do not understand why do we need to find this number for every length. Can you explain in a bit more details.

How does that works for this test :

7

2 3 4 7 6 1 6

0 1 0 0 0 0 0

I understood the cycle part during contest... however didn't get the part about even number of 1's ... please explain why even number of 1 won't work

I still don't understand Div2 E till now. Anyone care to explain?

Sure!

First of all, imagine that the time ranges on the tickets work backwards (ie: buying a 90 minute ticket at time 100 covers all trips from times 11 to 100). Then, for trip t_i, we would like to calculate the minimum cost c_i. There are three options:

1. Buy a ticket for one trip. This costs c_i - 1 + 20.
2. Buy a ticket for 90 minutes. Use a binary search to find the earliest trip t_m that would be covered by this ticket. This costs c_m + 50.
3. Buy a ticket for 1440 minutes. Again, find the earliest trip that happened in the past 1440 minutes. This costs c_m + 120.

Total time is because of the binary search at each step.

Another way is to do exactly what statement says.

	vi a(n);

	for (int & t : a) cin >> t;
	vi his(n, 0);

	his[0] = 20;
	cout << 20 << endl;

	for (int i = 1; i < n; ++i) {
		his[i] = 20;

		int paid1 = 0;
		for (int j = i - 1; j >= 0 && a[j] + 90 - 1 >= a[i]; --j) paid1 += his[j];

		if (paid1 + his[i] > 50) {
			his[i] = max(50 - paid1, 0);
		}

		int paid2 = 0;
		for (int j = i - 1; j >= 0 && a[j] + 1440 - 1 >= a[i]; --j) paid2 += his[j];

		if (paid2 + his[i] > 120) {
			his[i] = max(120 - paid2, 0);
		}

		cout << his[i] << endl;
	}

No DP, no binary search. 144kk operations worst case = 500 ms.