Hahaha you are so funny.

You must be new here...

Great way to farm karma eh?

I agree with you :)

You are absolutely right!

is div2 harder to raise rating for CM than div1 or div1+2 because of new accounts?

same to you...

but round is unrated now :(

Hint: If there exists segment of length x, where b<=x<a. B can always win.

it's easy to see that each element should increase at most 2 times

but u need to use FAST IO, it was my mistake

Case $$$1$$$: When there's a segment with length $$$x$$$ where $$$b\leq x \lt a$$$, Bob always wins.

Case $$$2$$$: When there's at least two segments with length at least $$$2b$$$, Bob can always divide one fitting into case $$$1$$$ and thus win.

Case $$$3$$$: When there's no segment with length at least $$$2b$$$, the winner depends on the parity of number of segment with length $$$\geq a$$$.

Case $$$4$$$: When there's exactly one segment with length at least $$$2b$$$, enumerate over all possible moves for Alice with this segment, then it reduces to case $$$3$$$.

Check wheter there exists segment of length x, where b<=x<a. If yes then B can always win. Otherwise let cnt be # of segments with length >=2*b and now we only take care of segments >=a. If( cnt>=2) Then B always win because no matter what A does he can split segment of length x, where b<=x<a. If(cnt==0) then all segments have length x, where a<=x<2*b, so only parity determines the winnner. If(cnt==1) A have to move in this segment first, so we can consider all his moves on this segment, and if any of them provides him a win, he can win, otherwise he cant.

My observation: in B move if there is some segment of size $$$ \gt =2*b$$$ he can win.

At first consider only blocks with no less than b letters. If there is a block with at least b but less than a letters then second player wins. If it is a turn of a second player and there is a block with at least 2 * b letters then second player can create a block of length b and win. So, check at first if there is a block of length between b and a — 1, if there is, then second player wins. Then count amount of blocks of length at least 2 * b. If it is 0, then in every other block each player can make only one turn, so total number of turns is amount of blocks. If it is at least two, then second player wins. If it is one, bruteforce the move of the first player, and check if he can win(after his move all the blocks should be the length less than 2 * b so number of moves will be the number of blocks of length at least b).

Same thoughts, but can't figure out how to obtain the answer from those 4 values.

1
3 3 0

answer is 2.
1 team: c c m
2 team: c m m

The answer was simple min(a,b,(a+b+c)/3))

never mind... sorry

Input: 100 Apparently, the checker was buggy, so many wrong solutions passed during the contest and now they re-judged the solutions.

dude...

I feel the same :(

Test case 1 1 0 No output in your code

I think this will be while(ans >= 0)

1
18 13 0

answer: 10

try 10 10 0

You can find an optimal solution where no border is increased more than $$$2$$$ units, so it can be solved by $$$dp$$$ using this observation.

Proof: If adjusting border $$$i$$$ to be finally of length $$$a_i$$$ will make it equal to some adjacent border, and adjusting it to be of length $$$a_i+1$$$ will make it equal to the other adjacent border, so for sure adjusting it to be $$$a_i+2$$$ will make it different from both adjacent borders, and there is no point in increasing it more than this.

Each elements in $$$a$$$ should increase at most $$$2$$$.
Let $$$dp(i, j), j \in [0;2]$$$ is the minimal cost to make the fence from $$$1$$$ to $$$i$$$ (inclusive) great.
Remember keep tracking the last element in previous state.
Let $$$prev[j], j \in [0;2]$$$ is the last element of the great fence after we increase $$$a_i$$$ with $$$j$$$ units.
Base case:
- $$$dp(1, 0) = 0$$$
- $$$dp(1, 1) = b[1]$$$
- $$$dp(1,2) = 2*b[1]$$$
and
- $$$prev[0] = a_1$$$
- $$$prev[1] = a_1 + 1$$$
- $$$prev[2] = a_1 + 2$$$

Recurrence:

Check my submission for more detail.

deleted

No, because game isn't impartial.

No, you also need to take into account the number of bipartitions.

Let $$$f(C)$$$ be the number of arrangements such that none of the numbers in set $$$C$$$ appears. Then our answer is $$$2^n - f({0}) - f({1}) - f({2}) + f({0,1}) + f({0,2}) + f({1,2}) - f({0,1,2})$$$

$$$f({0})$$$ and $$$f({2})$$$ are the number of independent sets.

$$$f({1})$$$ is $$$2^{(\text{number of components})}$$$

$$$f({0,2})$$$ is the number of bipartitions.

$$$f({0,1})$$$ and $$$f({1,2})$$$ is $$$2^{ \text{number of isolated vertices}}$$$

finally $$$f({0,1,2})$$$ is $$$2^n$$$ if the graph has no edges and $$$0$$$ otherwise.

Actually, because it's guaranteed that all elements of the multiset are powers of two, you can solve it simply and directly just like this.

The only long code is SOS DP though, so I don't really see much of a problem.

I think $$$s$$$ should be $$$c$$$ right?