这是竞争编程中最常用的技术之一。让我们首先考虑下面的简单问题。
查找第n个斐波那契数的最小时间复杂度是多少?
我们可以使用矩阵幂运算在O(Log n)时间中找到第n个斐波那契数。有关详细信息，请参见此方法4。在这篇文章中，讨论了矩阵幂运算的一般实现。
For solving the matrix exponentiation we are assuming a
linear recurrence equation like below:
F(n) = a*F(n-1) + b*F(n-2) + c*F(n-3) for n >= 3
. . . . . Equation (1)
where a, b and c are constants.
For this recurrence relation, it depends on three previous values.
Now we will try to represent Equation (1) in terms of the matrix.
[First Matrix] = [Second matrix] * [Third Matrix]
| F(n) | = Matrix 'C' * | F(n-1) |
| F(n-1) | | F(n-2) |
| F(n-2) | | F(n-3) |
Dimension of the first matrix is 3 x 1 .
Dimension of the third matrix is also 3 x 1.
So the dimension of the second matrix must be 3 x 3
[For multiplication rule to be satisfied.]
Now we need to fill the Matrix 'C'.
So according to our equation.
F(n) = a*F(n-1) + b*F(n-2) + c*F(n-3)
F(n-1) = F(n-1)
F(n-2) = F(n-2)
C = [a b c
1 0 0
0 1 0]
Now the relation between matrix becomes :
[First Matrix] [Second matrix] [Third Matrix]
| F(n) | = | a b c | * | F(n-1) |
| F(n-1) | | 1 0 0 | | F(n-2) |
| F(n-2) | | 0 1 0 | | F(n-3) |
Lets assume the initial values for this case :-
F(0) = 0
F(1) = 1
F(2) = 1
So, we need to get F(n) in terms of these values.
So, for n = 3 Equation (1) changes to
| F(3) | = | a b c | * | F(2) |
| F(2) | | 1 0 0 | | F(1) |
| F(1) | | 0 1 0 | | F(0) |
Now similarly for n = 4
| F(4) | = | a b c | * | F(3) |
| F(3) | | 1 0 0 | | F(2) |
| F(2) | | 0 1 0 | | F(1) |
- - - - 2 times - - -
| F(4) | = | a b c | * | a b c | * | F(2) |
| F(3) | | 1 0 0 | | 1 0 0 | | F(1) |
| F(2) | | 0 1 0 | | 0 1 0 | | F(0) |
So for n, the Equation (1) changes to
- - - - - - - - n -2 times - - - - -
| F(n) | = | a b c | * | a b c | * ... * | a b c | * | F(2) |
| F(n-1) | | 1 0 0 | | 1 0 0 | | 1 0 0 | | F(1) |
| F(n-2) | | 0 1 0 | | 0 1 0 | | 0 1 0 | | F(0) |
| F(n) | = [ | a b c | ] ^ (n-2) * | F(2) |
| F(n-1) | [ | 1 0 0 | ] | F(1) |
| F(n-2) | [ | 0 1 0 | ] | F(0) |因此，我们可以简单地将第二个矩阵乘以n-2倍，然后将其与第三个矩阵相乘以获得结果。使用除数和征服算法获得幂可以在(log n)时间内完成乘法(请参阅此内容)
让我们考虑以下问题：找到使用以下递归定义的系列的第n个项。
n'th term,
F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
Base Cases :
F(0) = 0, F(1) = 1, F(2) = 1我们可以使用以下命令找到第n个项：
Putting a = 1, b = 1 and c = 1 in above formula
| F(n) | = [ | 1 1 1 | ] ^ (n-2) * | F(2) |
| F(n-1) | [ | 1 0 0 | ] | F(1) |
| F(n-2) | [ | 0 1 0 | ] | F(0) |以下是上述想法的实现。
C++
// C++ program to find value of f(n) where f(n)
// is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
#include
using namespace std;
// A utility function to multiply two matrices
// a[][] and b[][]. Multiplication result is
// stored back in b[][]
void multiply(int a[3][3], int b[3][3])
{
// Creating an auxiliary matrix to store elements
// of the multiplication matrix
int mul[3][3];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
mul[i][j] = 0;
for (int k = 0; k < 3; k++)
mul[i][j] += a[i][k]*b[k][j];
}
}
// storing the multiplication result in a[][]
for (int i=0; i<3; i++)
for (int j=0; j<3; j++)
a[i][j] = mul[i][j]; // Updating our matrix
}
// Function to compute F raise to power n-2.
int power(int F[3][3], int n)
{
int M[3][3] = {{1,1,1}, {1,0,0}, {0,1,0}};
// Multiply it with initial values i.e with
// F(0) = 0, F(1) = 1, F(2) = 1
if (n==1)
return F[0][0] + F[0][1];
power(F, n/2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
// Multiply it with initial values i.e with
// F(0) = 0, F(1) = 1, F(2) = 1
return F[0][0] + F[0][1] ;
}
// Return n'th term of a series defined using below
// recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
int findNthTerm(int n)
{
int F[3][3] = {{1,1,1}, {1,0,0}, {0,1,0}} ;
//Base cases
if(n==0)
return 0;
if(n==1 || n==2)
return 1;
return power(F, n-2);
}
// Driver code
int main()
{
int n = 5;
cout << "F(5) is " << findNthTerm(n);
return 0;
} Java
// JAVA program to find value of f(n) where
// f(n) is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
import java.io.*;
class GFG {
// A utility function to multiply two
// matrices a[][] and b[][].
// Multiplication result is
// stored back in b[][]
static void multiply(int a[][], int b[][])
{
// Creating an auxiliary matrix to
// store elements of the
// multiplication matrix
int mul[][] = new int[3][3];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
mul[i][j] = 0;
for (int k = 0; k < 3; k++)
mul[i][j] += a[i][k]
* b[k][j];
}
}
// storing the multiplication
// result in a[][]
for (int i=0; i<3; i++)
for (int j=0; j<3; j++)
// Updating our matrix
a[i][j] = mul[i][j];
}
// Function to compute F raise to
// power n-2.
static int power(int F[][], int n)
{
int M[][] = {{1, 1, 1}, {1, 0, 0},
{0, 1, 0}};
// Multiply it with initial values
// i.e with F(0) = 0, F(1) = 1,
// F(2) = 1
if (n == 1)
return F[0][0] + F[0][1];
power(F, n / 2);
multiply(F, F);
if (n%2 != 0)
multiply(F, M);
// Multiply it with initial values
// i.e with F(0) = 0, F(1) = 1,
// F(2) = 1
return F[0][0] + F[0][1] ;
}
// Return n'th term of a series defined
// using below recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
static int findNthTerm(int n)
{
int F[][] = {{1, 1, 1}, {1, 0, 0},
{0, 1, 0}} ;
return power(F, n-2);
}
// Driver code
public static void main (String[] args) {
int n = 5;
System.out.println("F(5) is "
+ findNthTerm(n));
}
}
//This code is contributed by vt_m.Python3
# Python3 program to find value of f(n)
# where f(n) is defined as
# F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
# Base Cases :
# F(0) = 0, F(1) = 1, F(2) = 1
# A utility function to multiply two
# matrices a[][] and b[][]. Multiplication
# result is stored back in b[][]
def multiply(a, b):
# Creating an auxiliary matrix
# to store elements of the
# multiplication matrix
mul = [[0 for x in range(3)]
for y in range(3)];
for i in range(3):
for j in range(3):
mul[i][j] = 0;
for k in range(3):
mul[i][j] += a[i][k] * b[k][j];
# storing the multiplication
# result in a[][]
for i in range(3):
for j in range(3):
a[i][j] = mul[i][j]; # Updating our matrix
return a;
# Function to compute F raise
# to power n-2.
def power(F, n):
M = [[1, 1, 1], [1, 0, 0], [0, 1, 0]];
# Multiply it with initial values i.e
# with F(0) = 0, F(1) = 1, F(2) = 1
if (n == 1):
return F[0][0] + F[0][1];
power(F, int(n / 2));
F = multiply(F, F);
if (n % 2 != 0):
F = multiply(F, M);
# Multiply it with initial values i.e
# with F(0) = 0, F(1) = 1, F(2) = 1
return F[0][0] + F[0][1] ;
# Return n'th term of a series defined
# using below recurrence relation.
# f(n) is defined as
# f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
# Base Cases :
# f(0) = 0, f(1) = 1, f(2) = 1
def findNthTerm(n):
F = [[1, 1, 1], [1, 0, 0], [0, 1, 0]];
return power(F, n - 2);
# Driver code
n = 5;
print("F(5) is",
findNthTerm(n));
# This code is contributed by mitsC#
// C# program to find value of f(n) where
// f(n) is defined as
// F(n) = F(n-1) + F(n-2) + F(n-3), n >= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
using System;
class GFG {
// A utility function to multiply two
// matrices a[][] and b[][]. Multiplication
// result is stored back in b[][]
static void multiply(int[, ] a, int[, ] b)
{
// Creating an auxiliary matrix to store
// elements of the multiplication matrix
int[, ] mul = new int[3, 3];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
mul[i, j] = 0;
for (int k = 0; k < 3; k++)
mul[i, j] += a[i, k] * b[k, j];
}
}
// storing the multiplication result
// in a[][]
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
// Updating our matrix
a[i, j] = mul[i, j];
}
// Function to compute F raise to power n-2.
static int power(int[, ] F, int n)
{
int[, ] M = { { 1, 1, 1 }, { 1, 0, 0 },
{ 0, 1, 0 } };
// Multiply it with initial values i.e
// with F(0) = 0, F(1) = 1, F(2) = 1
if (n == 1)
return F[0, 0] + F[0, 1];
power(F, n / 2);
multiply(F, F);
if (n % 2 != 0)
multiply(F, M);
// Multiply it with initial values i.e
// with F(0) = 0, F(1) = 1, F(2) = 1
return F[0, 0] + F[0, 1];
}
// Return n'th term of a series defined
// using below recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
static int findNthTerm(int n)
{
int[, ] F = { { 1, 1, 1 }, { 1, 0, 0 },
{ 0, 1, 0 } };
return power(F, n - 2);
}
// Driver code
public static void Main()
{
int n = 5;
Console.WriteLine("F(5) is "
+ findNthTerm(n));
}
}
// This code is contributed by vt_m.PHP
= 3
// Base Cases :
// F(0) = 0, F(1) = 1, F(2) = 1
// A utility function to multiply two matrices
// a[][] and b[][]. Multiplication result is
// stored back in b[][]
function multiply(&$a, &$b)
{
// Creating an auxiliary matrix to store
// elements of the multiplication matrix
$mul = array_fill(0, 3,
array_fill(0, 3, 0));
for ($i = 0; $i < 3; $i++)
{
for ($j = 0; $j < 3; $j++)
{
$mul[$i][$j] = 0;
for ($k = 0; $k < 3; $k++)
$mul[$i][$j] += $a[$i][$k] *
$b[$k][$j];
}
}
// storing the multiplication result in a[][]
for ($i = 0; $i < 3; $i++)
for ($j = 0; $j < 3; $j++)
$a[$i][$j] = $mul[$i][$j]; // Updating our matrix
}
// Function to compute F raise to power n-2.
function power($F, $n)
{
$M = array(array(1, 1, 1),
array(1, 0, 0),
array(0, 1, 0));
// Multiply it with initial values i.e with
// F(0) = 0, F(1) = 1, F(2) = 1
if ($n == 1)
return $F[0][0] + $F[0][1];
power($F, (int)($n / 2));
multiply($F, $F);
if ($n % 2 != 0)
multiply($F, $M);
// Multiply it with initial values i.e with
// F(0) = 0, F(1) = 1, F(2) = 1
return $F[0][0] + $F[0][1] ;
}
// Return n'th term of a series defined
// using below recurrence relation.
// f(n) is defined as
// f(n) = f(n-1) + f(n-2) + f(n-3), n>=3
// Base Cases :
// f(0) = 0, f(1) = 1, f(2) = 1
function findNthTerm($n)
{
$F = array(array(1, 1, 1),
array(1, 0, 0),
array(0, 1, 0));
return power($F, $n - 2);
}
// Driver code
$n = 5;
echo "F(5) is " . findNthTerm($n);
// This code is contributed by mits
?>Javascript
输出 ：
F(5) is 7