D Dexterina’s Lab

题面

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Dexterina and Womandark have been arch-rivals since they’ve known each other. Since both are super-intelligent teenage girls, they’ve always been trying to solve their disputes in a peaceful and nonviolent way. After god knows how many different challenges they’ve given to one another, their score is equal and they’re both desperately trying to best the other in various games of wits. This time, Dexterina challenged Womandark to a game of Nim.

Nim is a two-player game in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects from a single heap. The player who can’t make a turn loses. By their agreement, the sizes of piles are selected randomly from the range [0, x]. Each pile’s size is taken independently from the same probability distribution that is known before the start of the game.

Womandark is coming up with a brand new and evil idea on how to thwart Dexterina’s plans, so she hasn’t got much spare time. She, however, offered you some tips on looking fabulous in exchange for helping her win in Nim. Your task is to tell her what is the probability that the first player to play wins, given the rules as above.

Input

The first line of the input contains two integers n (1 ≤ n ≤ $10^9$) and x (1 ≤ x ≤ 100) — the number of heaps and the maximum number of objects in a heap, respectively. The second line contains x + 1 real numbers, given with up to 6 decimal places each:P(0), P(1), … , P(X). Here, P(i) is the probability of a heap having exactly i objects in start of a game. It’s guaranteed that the sum of all P(i) is equal to 1.

Output

Output a single real number, the probability that the first player wins. The answer will be judged as correct if it differs from the correct answer by at most $10^{-6}$.

思路

​ 我们假设$A_{i,j}$表示前$i$堆石子，Xor和为$j$的概率

​ 那么得到如下的递推公式：

​ $A{i,j}=\sum{x\ xor\ y=j}A_{i-1,x}*P(y)$ （$x$和$y$之间有序）

​ 需要注意的是Xor和的上界，当$x$最大为100时，所有数的Xor和不会超过127（具体原因可以自己想想）

$A{i+1,j}=\sum{k=0}^tA{i,k}*M{k,j}​$（$t​$表示枚举上界）