oops I misunderstood "in a row"
is "in a row" means cosecutive??

n == m

Well, there were quite a lot of LCA + segment tree problems by now, so by now people mostly should know what that is :)
As for me, i just took my solution to http://acm.timus.ru/problem.aspx?space=1&num=1471 and only had to modify it very slightly. Just, in there you had to operate with only minimums, and here you need to find a minimum in a certain range of your array of minimums, so there's that additional layer of minimums.

12222

your programm works only because of weak tests.
for example: 3 1
your answer is 0, but the correct one is 1.
1 _ _
1 _ 3
1 2 3 -> there is 1 person from both sides and the distance to number 2 is equal, so in the worst case the person will hit vasya.

Edited by author 13.01.2016 01:14

Edited by author 13.01.2016 01:14

Nope. The only inscribed rectangle with area=50 doesn't contain hole. Note, that there are only 4 possible rectangles in the solution.

Maybe I don't understand it clear, but I think it will get TL.
Difficult of this algorithm = O(n^2)

long long is better

There is information:"Если после очередного сгиба полоска стала полностью одноцветной".It's mean, both of sides needs to have one color

Yes, 6 is a subordinate of 1, 2, and 3.

Edited by author 19.11.2016 15:45

"An employee can send a memo only to another employee. The route of memo can consist of one employee."

So the answer should be 0 1 5.

I'll explain the test.
6
2 5 3 4 1 9
The first one is chosen as current candidate, and has 2 disadvantages. Then, he is compared with 2nd pirate (5 disadvantages), 3rd (3), 4th (4), 5th (1). When comparing with 5th pirate, we see that he has less disadvantages (1) than current candidate (2). So he becomes the current candidate. And then we finally compare him with last pirate (9 disadvantages).

In the end,
1st pirate with 2 disadvantages was compared to 2nd, 3rd, 4th, 5th — compared 4 times;
2nd pirate with 5 disadvantages was compared to 1st — compared 1 times;
3rd pirate with 3 disadvantages was compared to 1st — compared 1 times;
4th pirate with 4 disadvantages was compared to 1st — compared 1 times;
5th pirate with 1 disadvantages was compared to 1st, 6th — compared 2 times;
6th pirate with 9 disadvantages was compared to 5th — compared 1 times;

Out of all the pirates, the 1st one has the most compare times — 4. Thus, we output 1 — the index of a pirate who was compared most times.
But, if several pirates were compared the same amount of times — say, if 5th one was also compared 4 times (if there were 2 more pirates in input) — then both answers "1" or "4" would be correct in this case.

Is this any clearer?