Discrete Acceleration
Contest: Grakn Forces 2020
Difficulty rating: 1500
Tags: binary search, dp, implementation, math, two pointers

There is a road with length $$$l$$$ meters. The start of the road has coordinate $$$0$$$, the end of the road has coordinate $$$l$$$.

There are two cars, the first standing at the start of the road and the second standing at the end of the road. They will start driving simultaneously. The first car will drive from the start to the end and the second car will drive from the end to the start.

Initially, they will drive with a speed of $$$1$$$ meter per second. There are $$$n$$$ flags at different coordinates $$$a_1, a_2, \ldots, a_n$$$. Each time when any of two cars drives through a flag, the speed of that car increases by $$$1$$$ meter per second.

Find how long will it take for cars to meet (to reach the same coordinate).

Input format:
The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$): the number of test cases.

The first line of each test case contains two integers $$$n$$$, $$$l$$$ ($$$1 \leq n \leq 10^5$$$, $$$1 \leq l \leq 10^9$$$): the number of flags and the length of the road.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ in the increasing order ($$$1 \leq a_1 < a_2 < \ldots < a_n < l$$$).

It is guaranteed that the sum of $$$n$$$ among all test cases does not exceed $$$10^5$$$.

Output format:
For each test case print a single real number: the time required for cars to meet.

Your answer will be considered correct, if its absolute or relative error does not exceed $$$10^{-6}$$$. More formally, if your answer is $$$a$$$ and jury's answer is $$$b$$$, your answer will be considered correct if $$$\frac{|a-b|}{\max{(1, b)}} \leq 10^{-6}$$$.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Note:
In the first test case cars will meet in the coordinate $$$5$$$.

The first car will be in the coordinate $$$1$$$ in $$$1$$$ second and after that its speed will increase by $$$1$$$ and will be equal to $$$2$$$ meters per second. After $$$2$$$ more seconds it will be in the coordinate $$$5$$$. So, it will be in the coordinate $$$5$$$ in $$$3$$$ seconds.

The second car will be in the coordinate $$$9$$$ in $$$1$$$ second and after that its speed will increase by $$$1$$$ and will be equal to $$$2$$$ meters per second. After $$$2$$$ more seconds it will be in the coordinate $$$5$$$. So, it will be in the coordinate $$$5$$$ in $$$3$$$ seconds.

In the second test case after $$$1$$$ second the first car will be in the coordinate $$$1$$$ and will have the speed equal to $$$2$$$ meters per second, the second car will be in the coordinate $$$9$$$ and will have the speed equal to $$$1$$$ meter per second. So, they will meet after $$$\frac{9-1}{2+1} = \frac{8}{3}$$$ seconds. So, the answer is equal to $$$1 + \frac{8}{3} = \frac{11}{3}$$$.

Example input:
5
2 10
1 9
1 10
1
5 7
1 2 3 4 6
2 1000000000
413470354 982876160
9 478
1 10 25 33 239 445 453 468 477

Example output:
3.000000000000000
3.666666666666667
2.047619047619048
329737645.750000000000000
53.700000000000000

Editorial:
Solution $$$1$$$:
Let's make a binary search on the answer. If we have some time $$$t$$$ we can calculate the coordinate of each car after $$$t$$$ seconds. Let's define them as $$$x_1$$$ and $$$x_2$$$. If $$$x_1 \leq x_2$$$ let's move the left bound of the binary search, otherwise, let's move the right bound.
Time complexity: $$$O(n \log{\frac{1}{\epsilon}})$$$.
Solution $$$2$$$:
Let's calculate the time for each car and each flag, after which the car will reach this flag. We can find the first flag from left to right, on which the second car was before the first car. Using it we can find the segment between flags, on which the cars will meet. After that the answer can be found by some simple formula using their speed and times, on which they will reach the left and the right bound of this segment.
Time complexity: $$$O(n)$$$.