Good Luck to you!!!

Will Good Bye 2017 be rated????

The idea is really nice! I think here on CF problem sets and groups for which rating is calculated are bound too tight. We could have several groups(Div 1.2, Div 1.4, Div 1.6 etc) competing on same problemset. Currently getting to div.1 you most likely will be back in div.2 after one of next contests. Only Div1+Div2 is ok for 1900-2000 people :)

They have already thanked him.

At least he greeted codeforces.

LoL...they are changed to 7.

From the past contests, no.

Nope.

But you still can rank up by hacking people higher than you. :D

Difficulty is the same as Div2 problems

...

Hacking doesn't affect your score in any way.

1. Testing of Educational round (after the end of Goodbye)
2. Testing of Goodbye round
3. Rating for Educational (probably somewhere during the testing)
4. Rating for Goodbye

It is ACM style contest, so problem A worth same as F.

Isn't that what virtual is for?

where can i use the predictor

:(

Why is that? :( I think it is educational

We have "question to jury" option btw. Right below the problem list in the main page of the contest.

Hacks don't affect anyone's rating, whether they've participated or not.

Inversion number parity always changes whenever you swap two arbitrary elements in a permutation.

swapping any two numbers changes the parity of number of inversions.. so just calc the initial number of inversions and how many pairs are being swapped in every query

The parity of the number of inversions of a composition of two permutations is the sum of the parities of the permutations, mod 2. A permutation that reverses a segment of length d has inversions, hence it suffices to add this after each query and consider the amount mod 2.

try sth like 3 2

1 2

It's random case with n = 200000, k = 2 and some permutation in answer

200000 2

102991 8046

2 4 4

2 4 4 it seems

Try this:
9 3
9 5 3

warning: suggest parentheses around comparison in operand of '==' [-Wparentheses] if (N==a==b)

that's what -Wall, -Wextra etc is for

You can count the initial number of inversions (we will call it x thenceforth) through brute force.

For each query, it is certain that inside the segment, the inverse pair will become non-inverse and vice-versa. Therefore, the oddity of x after each query is decided by the oddity of nC2 (the number of distinct pair in the query's segment).

If nC2 is odd, then if x was even back then, it will be odd, and vice-versa. Simply because the numbers of non-inverse and inverse pairs are not both odd or both even, therefore, the state of x will be changed.

Similarly, if nC2 is even, x keeps the same state it was before the query.

You can check out this comment.

Also, since Good Bye 2017 seems to have no rating restrictions (Div.1 and Div.2 users can join equally), then I guess there is no need to calculate rating before this round.

I think you mean inversions, in problem D you could have just done two loops to check for one, and for an better algorithm read this.

And if by any chance you wanted the number of permutations, there are n! of them and you can read more about them here.

Or you are trolling (no offence).

edit : no trolling.

no need to use dp its just a simple math

int n, a, b;
cin >> n >> a >> b;

int sm = min(a, b);
int mxs = max(a, b);
int ct = 1,mx = 0;
while (ct < n) {
	int cx = n - ct;
	int smc = sm/ct;
	int mxc = mxs/cx;
	mx = max(mx,min(smc,mxc));
	ct++;
}

cout << mx;

Some of the n plates are going to be of type 1 and the others of type 2. If you know how much plates of each type you will have it's easy to get the optimal answer, Then just try every possible distribution of the number of plates, there are only O(n) of them.

awoo's words

1.Testing of Educational round (after the end of Goodbye)

2.Testing of Goodbye round

3.Rating for Educational (probably somewhere during the testing)

4.Rating for Goodbye

that's the order

if you understand the solution now then please explain it. It will be a great help :)

no

NO

I am not sure if it can be solved in something under .

Here is such a problem but with swaps instead of segment reversals.

When performing the stack sort, you can never push a number onto the stack that is bigger than a number already on the stack, otherwise they will come off the stack out of order. So a successful stack-sort, if it exists will always be: pop the stack until the top is bigger than the front of A (or empty), then pop from A onto the stack. You can simulate that for the first k values to check if it is valid (as in, numbers popped were 1, 2, 3, ... up to some value). Note that the stack will always be sorted (smallest at the top).

To complete the sequence, suppose that the top of the stack is currently X. A must then contain all missing number less than X before any value greater than X. To get a lexicographically maximal solution, they should appear in A in decreasing order. It's also easy to verify that this will give a valid solution.

It's strange it doesn't auto-refresh for me when I open someone's code.

segment tree. but i've tried to solve it by UFS && block_division

Try to solve it with several segment trees :D

I've seen two solutions:

1. O((Q + H)\sqrt{N} + N), where H = 100. Divide the array into blocks of size . For each block and value, store a linked list of all the places that value occurs in that block (in no particular order). To process a query, process the two partial blocks at the ends iteratively, but for the full blocks, just do an O(1) splice of one linked list onto another.

2. O(QH\log Q + N). Store a segment tree over the updates, where each element in the segment tree is a transition table. At the leaves, an transition table is either the identity (for an "inactive" update) or an identity but with f(x)=y (for an "active" query). The root of the segment tree gives the composition of all the active updates. Now do a sweep through the array, activating updates when reaching l and deactivating them when reaching r, and looking up each a[i] through the root of the segment tree. Composing two transition tables takes O(H) time, so each activation/deactivation takes O(H\log Q) time.

EDIT: took the big-O times out of maths mode since the images seemed to be broken.

Build a segment tree, where each node has an array to[1...100](to[i] refers to what the number i will become after all modifications on the interval).

Yet actually, I solved the problem using 100 bitsets(S[1...100]) of length n, where S[i] consists of all appearances of number i. Despite the O(nq) complexity, the constant is so small that I passed all the test cases. 33723537

Can you link one of this submissions ?

You can use test generators.

Find a diameter of the tree first, then delete all the points outside the diameter, finally delete the points on it one by one..

Can you explain the preprocessing part of your code? How to count initial no. Of inversions

((1,3)(2)(4))for every 4 numbers.

We can start the garland with time interval = 2 on an odd number(let's say we start from 1). So according to the problem statement 1,3,5,7,......,2*n + 1 are covered by this garland. With odd time intervals covered, now we need to cover only the garlands that occur at even time intervals.

Now lets start another garland with time interval = 4 from an even number( say 2). Now this garland covers time interval 2,6,10,14,....4*n — 2. Now we see the only time intervals left are multiples of 4. 4,8,12,....,4*n. Thus starting the next garland with time interval = 4 from 4 will cover all the time intervals. Hence we get the solution.

Hope this helped!