传送门：https://nanti.jisuanke.com/t/31453
本题是一个组合数学（DP，滑稽）题。
一个环上有N个位置，标号为1~N。设第i（1≤i≤N）个位置上的数为x[i]，限制条件为：0≤x[i]<2k。另有限制条件：当位置i和位置j相邻时，x[i]⊕x[j]≠2k-1。求满足限制条件的环的状态数。
可以考虑将环切割成链，以分析问题。设：
①a[n]：长度为n，且首尾相同的满足上述条件的链的状态数；
②b[n]：长度为n，且首尾相反的满足上述条件的链的状态数；
③c[n]：长度为n，其他的满足上述条件的链的状态数。
于是，有以下转移方程：
$\begin{cases}
\ a_n=a_{n-1}+c_{n-1} \\
\ b_n=b_{n-1}+c_{n-1} \\
\ c_n=(2^k-3)c_{n-1}
\end{cases}$
可以写成矩阵乘法的形式：
$\begin{pmatrix}
a_i\\
b_i\\
c_i
\end{pmatrix}
=\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
2^k-2 & 2^k-2 & 2^k-3
\end{pmatrix}
\begin{pmatrix}
a_{i-1}\\
b_{i-1}\\
c_{i-1}
\end{pmatrix}$
初始条件：
$\begin{pmatrix}
a_1\\
b_1\\
c_1
\end{pmatrix}
=\begin{pmatrix}
2^k\\
0\\
0
\end{pmatrix}$
可以通过矩阵快速幂在O(logN)时间内求解。
答案为$ans(N,k)=(2^k-1)a_{N-1}+(2^k-2)(b_{N-1}+c_{N-1})$。
参考程序如下（实现上引用了本人编写的矩阵类的部分代码）：
#include <bits/stdc++.h>
using namespace std;
const int64_t mod = 1e9 + 7;
int64_t pwr(int64_t x, int p)
{
if (p == 0) return 1;
if (p & 1) return x * pwr(x, p ^ 1) % mod;
return pwr(x * x % mod, p >> 1);
}
int64_t a[3][3];
#define MAX_N 3
/***
* Matrix is a type of 2D-array.
* Here implements Basic Operators of Matrix.
*/
struct Matrix_t {
int row, col;
int64_t elem[MAX_N][MAX_N];
Matrix_t() {}
Matrix_t(int row, int col) : row(row), col(col) {}
Matrix_t(int row, int col, int64_t elem[][MAX_N]) {
this->row = row;
this->col = col;
if (elem != nullptr) {
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
this->elem[i][j] = elem[i][j];
}
}
}
}
Matrix_t(const Matrix_t& obj) {
this->row = obj.row;
this->col = obj.col;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
this->elem[i][j] = obj.elem[i][j];
}
}
}
virtual ~Matrix_t() {
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
elem[i][j] = 0;
}
}
row = 0;
col = 0;
}
Matrix_t& operator =(const Matrix_t& rhs) {
row = rhs.row;
col = rhs.col;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
elem[i][j] = rhs.elem[i][j];
}
}
return *this;
}
//Override: Matrix Multiplication.
Matrix_t operator *(const Matrix_t& rhs) {
int64_t res[MAX_N][MAX_N];
for (int i = 0; i < row; i++) {
for (int j = 0; j < rhs.col; j++) {
res[i][j] = 0;
for (int k = 0; k < col; k++) {
res[i][j] += elem[i][k] * rhs.elem[k][j] % mod;
res[i][j] %= mod;
}
}
}
return Matrix_t(row, rhs.col, res);
}
};
/***
* Square Mairtx is a type of Matrix.
* Here implements Integer Power of Square Matrix.
*/
struct SquareMatrix_t : Matrix_t {
int sz;
SquareMatrix_t() {}
SquareMatrix_t(int sz) : Matrix_t(sz, sz), sz(sz) {}
SquareMatrix_t(int sz, int64_t elem[][MAX_N]) : Matrix_t(sz, sz, elem), sz(sz) {}
virtual ~SquareMatrix_t() {
for (int i = 0; i < sz; i++) {
for (int j = 0; j < sz; j++) {
elem[i][j] = 0;
}
}
sz = 0;
}
void init() {
for (int i = 0; i < sz; i++) {
elem[i][i] = 1;
}
}
SquareMatrix_t& operator =(const Matrix_t& rhs) {
sz = rhs.row;
for (int i = 0; i < sz; i++) {
for (int j = 0; j < sz; j++) {
elem[i][j] = rhs.elem[i][j];
}
}
return *this;
}
SquareMatrix_t operator ^(int p) {
SquareMatrix_t res(sz);
SquareMatrix_t tmp(*this);
res.init();
while (p) {
if (p & 1) res = res * tmp;
tmp = tmp * tmp;
p >>= 1;
}
return res;
}
};
int main(void)
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
memset(a, 0, sizeof(a));
int n, k;
cin >> n >> k;
int64_t p = pwr(2, k);
if (n == 1) {
cout << p << endl;
continue;
}
a[0][0] = 1;
a[0][2] = 1;
a[1][1] = 1;
a[1][2] = 1;
a[2][0] = (p - 2 + mod) % mod;
a[2][1] = (p - 2 + mod) % mod;
a[2][2] = (p - 3 + mod) % mod;
SquareMatrix_t mat(3, a);
SquareMatrix_t res = mat ^ (n - 2);
int64_t a = res.elem[0][0] * p % mod;
int64_t b = res.elem[1][0] * p % mod;
int64_t c = res.elem[2][0] * p % mod;
int64_t ans = (a * ((p - 1 + mod) % mod) % mod + (b + c) % mod * ((p - 2 + mod) % mod) % mod) % mod;
cout << ans << endl;
}
}
原文地址：https://www.cnblogs.com/siuginhung/p/9616106.html