Older players fraud dead?
A,B
The cushions.
C - Prefix Enlightenment
This restriction is easy to see can be introduced up to each point comprises two sets. According to the routine, it is easy to think of these two sets even an edge, said their status to same / different.
Because ensures a solution, so take the time to sweep from left to right disjoint-set maintenance at a bipartite graph circumstances of each communication block, choose the smaller side.
If only one set covering, then it is equivalent to enforce this collection Check / uncheck, in doing disjoint-set time special sentenced click.
Code of goo.
D - Coffee Varieties (hard version)
As a Han Han, of course, practice and positive solutions are different. (Of course, a constant positive than Jiecha ......)
Each coffee is easy to think of statistics in the leftmost position. Consider violence, enumerate two locations \ (i <J \) , adding \ (i \) , adding \ (J \) , empty. This too \ (K = 1 \) case.
Consider each \ (K \) points (note that solution to a problem and not as \ (\ K FRAC 2 \) ) a sub-block. If we add \ (\ A) after the block plus \ (B \) block, then \ (B \) for each point in the block are the \ (B \) prefix block and \ (A \) block suffix compared. If we added backwards \ (A \) then backwards and then added to the block \ (B \) block, \ (B \) Each point block are the \ (B \) block and suffix \ (A \) prefix in the block were compared. Can be found at this time \ (B \) in each block are points and \ (A \) of each point of the block compared.
Is not considered \ (B \) block and will be the midpoint problems than later point in comparing your own, then for any two long blocks to do it again on the line. However, the constant is not OK.
We do first being a plus situation. It does not empty after considering the completion of two blocks, using the original information. Building a map, \ (I \) block to (J \) \ blocks connected side ( \ (I <J \) ), which take one side in the queue is equivalent to \ (I \) was added when the block \ ( J \) block. So we need every edge are gone again. Consider this figure to add some extra edge, making it the presence of Euler tour, so that you can once completed. Noting \ (I \) th block and the \ (\ frac n K -i + 1 \) showing the degree of match blocks can be additionally connected \ ((\ frac n K -i + 1, i , \ frac n K -2 * i + 1) \) side, so that the degree of each point are equal, and can run Euler. This time with the number of query \ (Km \) a, \ (m \) is about \ (\ FRAC with 3N ^ {2} {2} 4K ^ \) .
We hope that the opposite of the situation can be added in a similar manner to solve, but the opposite of added time \ (B \) for each point in the block are the \ (B \) block suffix to do a comparison, this is not what we want need.
Notes that the additional time being we have to put some points banning federal funding, the remaining points are different from each other in the same block, so you can add backwards is not the time to join banning federal funding of points, but add some the same number of blocks still alive charge point.
So he finished, almost against the upper bound of the number of inquiries, but do not need to empty the queue.
I did not tell you I did not understand it positive solution
Code: https://codeforces.com/contest/1290/submission/73847748
E - Cartesian Tree
This problem doing pre-knowledge: Definition \ (maxr (l) \) indicates the maximum \ (r \) , so that \ ([l, r] \ ) is a range of Cartesian tree, there is a \ (null \) . \ (minl (r) \) is similar. Then for a segment Cartesian tree \ ([L, R & lt] \) , as long as not the root, then the \ (maxr (l) = r \) and \ (minl (r) = l \) can be satisfied exactly a .
Back to this topic, consider to have been inserted into the determined position to fill in the number sequence is allowed to have vacancies.
Then \ ([l, r] \ ) of the sub-tree is the size of \ (\ [l, r] ) the number of non-null position, i.e. \ (FP (R & lt) -fp (L-. 1) \) .
Consider this thing apart with pre-knowledge, the answer is \ (\ sum \ limits_ {maxr (l) \ neq null} fp (maxr (l)) - \ sum \ limits_ {maxr (l) \ neq null} FP (. 1-L) + \ SUM \ {limits_ minL (R & lt) \} NEQ null FP (R & lt) - \ SUM \ {limits_ minL (R & lt) \} NEQ null FP (minL (R & lt) -1) \) , and then lose some of the constants on the root.
It is said that around is symmetric, we consider only the front two.
Found \ (maxr (l) \ neq null \) is equivalent to \ (P_. 1-L {}> P_L \) , then the second term to do well.
For the first item, after considering how to join a maximum maintenance \ (MAXR (L) \) . Because of \ (maxr (l) \) is \ ((l, n + 1 ] \) of a ratio \ (p_ {l-1} \) large position minus one, so that the maximum value will only be added to prefix take min, and a single point and modify their subsequent \ (MAXR \) .
You find that this thing can segment tree beats maintenance, and put a \ (fp \) is actually similar, with additional maintenance look outside Fenwick tree \ (fp \) just fine.
Probably so be it, the code temporarily cushions, and fill the afternoon.
F
The cushions.