|Codeforces Round #493 (Div. 2)|
There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.
There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.
Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $$$[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$$$ $$$\to$$$ two cuts $$$\to$$$ $$$[4, 1 | 2, 3, 4, 5 | 4, 4, 5, 5]$$$. On each segment the number of even elements should be equal to the number of odd elements.
The cost of the cut between $$$x$$$ and $$$y$$$ numbers is $$$|x - y|$$$ bitcoins. Find the maximum possible number of cuts that can be made while spending no more than $$$B$$$ bitcoins.
First line of the input contains an integer $$$n$$$ ($$$2 \le n \le 100$$$) and an integer $$$B$$$ ($$$1 \le B \le 100$$$) — the number of elements in the sequence and the number of bitcoins you have.
Second line contains $$$n$$$ integers: $$$a_1$$$, $$$a_2$$$, ..., $$$a_n$$$ ($$$1 \le a_i \le 100$$$) — elements of the sequence, which contains the equal number of even and odd numbers
Print the maximum possible number of cuts which can be made while spending no more than $$$B$$$ bitcoins.
1 2 5 10 15 20
1 3 2 4
1 2 3 4 5 6
In the first sample the optimal answer is to split sequence between $$$2$$$ and $$$5$$$. Price of this cut is equal to $$$3$$$ bitcoins.
In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.
In the third sample the sequence should be cut between $$$2$$$ and $$$3$$$, and between $$$4$$$ and $$$5$$$. The total price of the cuts is $$$1 + 1 = 2$$$ bitcoins.