題面

傳送門:http://codeforces.com/problemset/problem/691/E
E. Xor-sequences
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given n integers a1,  a2,  …,  an.

A sequence of integers x1,  x2,  …,  xk is called a “xor-sequence” if for every 1  ≤  i  ≤  k - 1 the number of ones in the binary representation of the number xi xi  +  1’s is a multiple of 3 and for all 1 ≤ i ≤ k. The symbol is used for the binary exclusive or operation.

How many “xor-sequences” of length k exist? Output the answer modulo 109 + 7.

Note if a = [1, 1] and k = 1 then the answer is 2, because you should consider the ones from a as different.

Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1018) — the number of given integers and the length of the “xor-sequences”.

The second line contains n integers ai (0 ≤ ai ≤ 1018).

Output
Print the only integer c — the number of “xor-sequences” of length k modulo 109 + 7.

Examples
inputCopy
5 2
15 1 2 4 8
outputCopy
13
inputCopy
5 1
15 1 2 4 8
outputCopy
5
題目大意：給定長度為n的序列，從序列中選擇k個數（可以重複選擇），使得得到的排列滿足xi與xi+1異或的二進位制中1的個數是3的倍數。問長度為k的滿足條件的序列有多少種？

分析：

1.DP方程的推導

設dp[i][j]$dp[i][j]$表示前i位以j結尾的方案數
則
dp[i][j]=∑ix=1dp[i−1][x]$dp\left[i\right]\left[j\right]=\sum _{x=1}^{i}dp\left[i-1\right]\left[x\right]$
(bitcount(a[x]$(bitcount(a[x]$ xor$xor$ a[i])mod$a[i])mod$ 3≡1)$3\equiv 1)$
bitcount(x)表示x的二進位制中1的個數

2.矩陣快速冪的優化

我們發現，狀態轉移方程的形式類似矩陣乘法
因為可以這樣變形：dp[i][j]=∑ix=1dp[i−1][x]×(bitcount(a[x]$dp\left[i\right]\left[j\right]=\sum _{x=1}^{i}dp\left[i-1\right]\left[x\right]\times (bitcount(a[x]$ xor$xor$ a[i])mod$a[i])mod$ 3==1)$3==1)$

所以狀態轉移方程可以寫成這樣
⎡⎣⎢⎢⎢⎢⎢dp[i][1]dp[i][2]⋮dp[i][n]⎤⎦⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢⎢101…01…⋱11⎤⎦⎥⎥⎥⎥⎥⎥⎥×⎡⎣⎢⎢⎢⎢⎢dp[i−1][1]dp[i−1][2]⋮dp[i−1][n]⎤⎦⎥⎥⎥⎥⎥$\left[\begin{array}{c}dp\left[i\right]\left[1\right]\\ dp\left[i\right]\left[2\right]\\ ⋮\\ dp\left[i\right][n]\end{array}\right]=\left[\begin{array}{ccccc}1& 0& 0& \dots & 1\\ 1\\ 1\\ \ddots \\ \dots & 1\end{array}\right]\times \left[\begin{array}{c}dp\left[i-1\right]\left[1\right]\\ dp\left[i-1\right]\left[2\right]\\ ⋮\\ dp\left[i-1\right][n]\end{array}\right]$

右邊那個很多1和0 的矩陣是n×n$n\times n$的狀態轉移矩陣，第i行第j列為1代表bitcount(a[x]$bitcount(a[x]$ xor$xor$ a[i])mod$a[i])mod$ 3≡1$3\equiv 1$，否則為0
顯然對角線全部為1（一個數異或它本身為0）
在DP之前我們可以預處理這個矩陣

由於最終答案為∑nj=1dp[k][j]$\sum _{j=1}^{n}dp[k][j]$,
狀態轉移方程可以改寫為