It's tourist Round.

Why are you getting downvoted?

you are right there.. hattrick done...

how can this normal comment have negative vote?

If I am saying something wrong then feel free to rectify it.

There is no obligation to participate.

This round was the Vk cup final. And all the contestants of this round were GM or above, also they're very well known in CF.

It's highly unlikely that they would leak problems or cheat.

First of all, most of the participants of the final round of this VK cup have written contests for CODEFORCES before. So I think they will respect other contests, if not then the same thing could happen to their contest(by testers or whatever else) and that, of course, is not what an author would want.

What you mentioned is correct, but that round was not based on the final round of another contest, if I'm not wrong it was based on an elimination round.

What you have seen are virtual participants, which are not rated(they joined the contests after it finished).

How do you propose to win using this information?

Download done!

Thought that was gonna be a rick roll lmao

neither, it is for the offical contest.

I think its a lesson for us that we should never_giveup. :p

Congrats never_giveup !!

Amen

Rickroll):

but why

Hi beluga :}

Me too :(

Notice that the insertions put the following restriction on the final sorted array: for some positions $$$k$$$ and $$$k+1$$$ the values strictly differ $$$a_{k} \lt a_{k+1}$$$. For all other positions, the values might be equal. You have to count the number of such positions because if several insertions happen in the same place, they all contribute to a single restriction. To do so, you can track the current "strict" positions with a treap (it will allow you to increment positions for a suffix).

Given the number of restrictions, let's look at the differences between subsequent elements. Add a 1 to the very beginning of the array and $$$n$$$ at the very end for simplicity. It's easy to see that for positions with restrictions the differences have to be at least 1, for other positions -- at least 0. In total, they should sum up to $$$n-1$$$, as the final element is now $$$n$$$ and the first is $$$1$$$. So, you have to divide $$$n-1-N_{restrictions}$$$ 1-point-differences between $$$n+1$$$ positions. By subtracting $$$N_{restrictions}$$$ we account for the "strict" positions. It is just a binomial coefficient.

Find the minimum number of breaks (let this be $$$x$$$). Now, every swap between A winning and B winning in will increase the number of breaks by 2 (because in the min. number of breaks, either all of A's wins are B's serves or all of B's wins are A's serves). So, we can reach everything in the sequence $$$x, x + 2, ... x + 2k$$$ breaks, where $$$k$$$ is the number of swaps we can make. You need to repeat this for both serving sequences ($$$ABABA$$$ or $$$BABAB$$$).

Fix a K = total number of breaks._ Let K1 = number of breaks for Alice and K2 = number of breaks for Bob. Naturally, we have K1 + K2 = K. Now consider two cases: 1) The total number of games (a.k.a, a + b) is even: we know that the number of games where Alice serves the ball (call it HomeA) should be equal to the number of games where Bob serves the ball (call it HomeB)

HomeA = HomeB
a - k1 + k2 = b - k2 + k1
a - b = 2 x k1 - 2 x k2
k1 - k2 = (a - b) / 2

and you also have k1 + k2 = K, so you could solve these. If k1 or k2 is negative or greater than a and b respectively, then the fixed value of K is invalid. 2) When a + b is odd, you could do the same work but this time, HomeA + j = HomeB where j = {1, -1}

That is how I solved it not sure if there is an easier way to be fair.

Consider a table

             A serve  B serve  Total win
A win        aa       ab       a (given)
B win        ba       bb       b (given)
Total serve  count_a  count_b       

if a+b is even:
   count_a = count_b = (a+b)/2

if a+b is odd: consider two cases
   count_a = (a+b+1)/2, count_b = (a+b-1)/2 
   count_a = (a+b-1)/2, count_b = (a+b+1)/2

• You consider all possible values of aa which is between 0 to count_a inclusive.
• For each possible value of aa, you can fill out the rest of the matrix like how you fill up a Sudoku.
• The assignment is valid if all four entries are non-negative.
• For each valid assignment, add k = ab + ba to the result
• If a+b is odd, you consider the two possible set of values count_a and count_b could have.

At least you solve B.

+1

i did binary search

First of all, you have to find maximum one for every group. While finding maximum our heroes power should be at least:

armor of monster — index of monster.

after this, you should sort values of maximums and start from minimum one. if maximum in this group is bigger than answer increase answer