Game
Contest: Codeforces Round 492 (Div. 1) [Thanks, uDebug!]
Difficulty rating: 2500
Tags: math

Allen and Bessie are playing a simple number game. They both know a function $$$f: \{0, 1\}^n \to \mathbb{R}$$$, i. e. the function takes $$$n$$$ binary arguments and returns a real value. At the start of the game, the variables $$$x_1, x_2, \dots, x_n$$$ are all set to $$$-1$$$. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an $$$i$$$ such that $$$x_i = -1$$$ and either setting $$$x_i \to 0$$$ or $$$x_i \to 1$$$.

After $$$n$$$ rounds all variables are set, and the game value resolves to $$$f(x_1, x_2, \dots, x_n)$$$. Allen wants to maximize the game value, and Bessie wants to minimize it.

Your goal is to help Allen and Bessie find the expected game value! They will play $$$r+1$$$ times though, so between each game, exactly one value of $$$f$$$ changes. In other words, between rounds $$$i$$$ and $$$i+1$$$ for $$$1 \le i \le r$$$, $$$f(z_1, \dots, z_n) \to g_i$$$ for some $$$(z_1, \dots, z_n) \in \{0, 1\}^n$$$. You are to find the expected game value in the beginning and after each change.

Input format:
The first line contains two integers $$$n$$$ and $$$r$$$ ($$$1 \le n \le 18$$$, $$$0 \le r \le 2^{18}$$$).

The next line contains $$$2^n$$$ integers $$$c_0, c_1, \dots, c_{2^n-1}$$$ ($$$0 \le c_i \le 10^9$$$), denoting the initial values of $$$f$$$. More specifically, $$$f(x_0, x_1, \dots, x_{n-1}) = c_x$$$, if $$$x = \overline{x_{n-1} \ldots x_0}$$$ in binary.

Each of the next $$$r$$$ lines contains two integers $$$z$$$ and $$$g$$$ ($$$0 \le z \le 2^n - 1$$$, $$$0 \le g \le 10^9$$$). If $$$z = \overline{z_{n-1} \dots z_0}$$$ in binary, then this means to set $$$f(z_0, \dots, z_{n-1}) \to g$$$.

Output format:
Print $$$r+1$$$ lines, the $$$i$$$-th of which denotes the value of the game $$$f$$$ during the $$$i$$$-th round. Your answer must have absolute or relative error within $$$10^{-6}$$$.

Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. Your answer is considered correct if $$$\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$$$.

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

Note:
Consider the second test case. If Allen goes first, he will set $$$x_1 \to 1$$$, so the final value will be $$$3$$$. If Bessie goes first, then she will set $$$x_1 \to 0$$$ so the final value will be $$$2$$$. Thus the answer is $$$2.5$$$.

In the third test case, the game value will always be $$$1$$$ regardless of Allen and Bessie's play.

Example input:
2 2
0 1 2 3
2 5
0 4

Example output:
1.500000
2.250000
3.250000

Example input:
1 0
2 3

Example output:
2.500000

Editorial:
One can show by induction that the expected value of the game is $$$\mathbb{E}[f] = 2^{-n} \sum_{x \in \{0, 1\}^n} f(x)$$$.
Consider the first turn. For notation, let $$$v_{i, 0}$$$ be the expected value of the game when $$$x_i$$$ is set to $$$0$$$, and let $$$v_{i, 1}$$$ be the expected value of the game when $$$x_i$$$ is set to $$$1$$$. By induction, it is easy to see that $$$\frac{v_{i, 0} + v_{i, 1}}{2} = \mathbb{E}[f].$$$
Consider Allen's strategy. If it is Allen's turn, he will set $$$x_s = t$$$, where $$$0 \le s < n, 0 \le t \le 1$$$ are such that $$$v_{s, t}$$$ is maximal. As $$$\frac{v_{i, 0} + v_{i, 1}}{2} = \mathbb{E}[f]$$$ for all $$$i$$$, it is clear that $$$v_{s, 1-t}$$$ is actually minimal among all the $$$v_{i, j}$$$. This means that Bessie would have chosen to set $$$x_s = 1-t$$$ if it were her turn. Therefore, the expected game value is $$$\frac{v_{s, t} + v_{s, 1-t}}{2} = \mathbb{E}[f].$$$