矩阵二分快速幂优化dp动态规划
矩阵快速幂代码:
int n; // 所有矩阵都是 n * n 的矩阵
struct matrix {
int a[100][100];
};
matrix matrix_mul(matrix A, matrix B, int mod) {
// 2 个矩阵相乘
matrix C;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
C.a[i][j] = 0;
for (int k = 0; k < n; ++k) {
C.a[i][j] += A.a[i][k] * B.a[k][j] % mod;
C.a[i][j] %= mod;
}
}
}
return C;
}
matrix unit() {
// 返回一个单位矩阵
matrix res;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j) {
res.a[i][j] = 1;
} else {
res.a[i][j] = 0;
}
}
}
return res;
}
matrix matrix_pow(matrix A, int n, int mod) {
// 快速求矩阵 A 的 n 次方
matrix res = unit(), temp = A;
for (; n; n /= 2) {
if (n & 1) {
res = matrix_mul(res, temp, mod);
}
temp = matrix_mul(temp, temp, mod);
}
return res;
}
矩阵快速幂模板
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 110;
const int MOD = 1e9 + 7;
#define mod(x) ((x)%MOD)
//https://www.51nod.com/Challenge/Problem.html#!#problemId=1113
int n;
struct mat{
int m[maxn][maxn];
}unit;
//重载矩阵的乘法
mat operator * (mat a,mat b){
mat ret;
ll x;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
x = 0;
for(int k=0;k<n;k++){
x += mod((ll)a.m[i][k] * b.m[k][j]);
}
ret.m[i][j] = mod(x);
}
}
return ret;
}
//初始化矩阵
void init_unit(){
for(int i=0;i<maxn;i++)
unit.m[i][i] = 1;
return;
}
//矩阵快速幂 前面代码已经重载过乘法运算 并取模 :就等于二分快速幂 + 矩阵乘法
mat pow_mat(mat a,ll n){
mat ret = unit;
while(n){
if(n&1) ret = ret*a;
a = a*a;
n >>= 1;
}
return ret;
}
int main(){
ll x;
init_unit();
cin>>n>>x;
mat a;
//输入数据
for(int i = 0;i < n;i++){
for(int j=0;j<n;j++){
cin>>a.m[i][j];
}
}
a = pow_mat(a,x);//计算a矩阵的x次幂
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(j+1==n) cout<<a.m[i][j]<<endl;
else cout<<a.m[i][j]<<" ";
}
}
return 0;
}