广义斐波那契数

我们都知道斐波那契数(Fn)由递归关系定义

Fibonacci Numbers (Fn) = F(n-1) + F(n-2) 
with seed values 
F0 = 0 and F1 = 1 
 

同样，我们可以概括这些数字。这样的数字序列被称为广义斐波纳契数(G) 。
广义斐波那契数(G)由递归关系定义

Generalized Fibonacci Numbers (Gn) = (c * G(n-1)) + (d * G(n-2)) 
with seed values 
G0 = a and G1 = b 
 

寻找第N个词

给定广义斐波纳契数的四个常数值a，b，c和d，以及整数N ，任务是找到广义斐波纳契数的第N个项，即Gn 。

例子：

Input: N = 2, a = 0, b = 1, c = 2, d = 3 
Output: 2 
Explanation: 
As a = 0 -> G(0) = 0 
b = 1 -> G(1) = 1 
So, G(2) = 2 * G(1) + 3 * G(0) = 2

Input: N = 3, a = 0, b = 1, c = 2, d = 3 
Output: 7 
 

天真的方法：使用给定的值，找到序列中的每个项，直到第N个项，然后打印第N个项。
时间复杂度： O(2 ^N )

另一种方法：想法是使用DP列表查找直到第N个术语的所有术语，然后打印第N个术语。
时间复杂度： O(N)

高效的方法：使用矩阵乘法，我们可以解决log(N)时间中的给定问题。

$\small {Given\, G(0)=a\, and\, G(1)=b, \, therefore \, G(2)=c*a+d*b } \\ \small {} \\ \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} = \begin{bmatrix} d*b+c*a & b\\ b & a \end{bmatrix} \newline \text {Multiplying matrices on RHS will eventually give us our LHS }\\ \begin{bmatrix} G(3) & G(2)\\ G(2) & G(1) \end{bmatrix} = \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} \newline \text {One more step } \\ \begin{bmatrix} G(4) & G(3)\\ G(3) & G(2) \end{bmatrix} = \begin{bmatrix} G(3) & G(2)\\ G(2) & G(1) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} \newline \text {So by looking at the sequence above we can generalize the Nth term as... }\\ \begin{bmatrix} G(N) & G(N-1)\\ G(N-1) & G(N-2) \end{bmatrix} = \begin{bmatrix} G(2) & G(1)\\ G(1) & G(0) \end{bmatrix} * \begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} ^{\!N-2}\newline \text{Now }\begin{bmatrix} d & 1\\ c & 0 \end{bmatrix} ^{\!N-2}} \text{ can be solved in log(N) time complexity }$

下面是上述方法的实现：

C++

// C++ program to implement the
// Generalised Fibonacci numbers
#include 
using namespace std;
 
// Helper function that multiplies
// 2 matrices F and M of size 2*2,
// and puts the multiplication
// result back to F[][]
void multiply(int F[2][2], int M[2][2]);
 
// Helper function that calculates F[][]
// raised to the power N
// and puts the result in F[][]
void power(int F[2][2], int N, int m, int n);
 
// Function to find the Nth term
int F(int N, int a, int b, int m, int n)
{
    // m 1
    // n 0
    int F[2][2] = { { m, 1 }, { n, 0 } };
 
    if (N == 0)
        return a;
 
    if (N == 1)
        return b;
 
    if (N == 2)
        return m * b + n * a;
 
    int initial[2][2]
        = { { m * b + n * a, b },
            { b, a } };
 
    power(F, N - 2, m, n);
 
    // Discussed above
    multiply(initial, F);
 
    return F[0][0];
}
 
// Function that multiplies
// 2 matrices F and M of size 2*2,
// and puts the multiplication
// result back to F[][]
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Function that calculates F[][]
// raised to the power N
// and puts the result in F[][]
void power(int F[2][2], int N, int m, int n)
{
    int i;
    int M[2][2] = { { m, 1 }, { n, 0 } };
 
    for (i = 1; i <= N; i++)
        multiply(F, M);
}
 
// Driver code
int main()
{
    int N = 2, a = 0, b = 1, m = 2, n = 3;
    printf("%d\n", F(N, a, b, m, n));
 
    N = 3;
    printf("%d\n", F(N, a, b, m, n));
 
    N = 4;
    printf("%d\n", F(N, a, b, m, n));
 
    N = 5;
    printf("%d\n", F(N, a, b, m, n));
 
    return 0;
}

Java

// Java program to implement the
// Generalised Fibonacci numbers
import java.util.*;
 
class GFG{
 
// Function to find the Nth term
static int F(int N, int a, int b,
             int m, int n)
{
    // m 1
    // n 0
    int[][] F = { { m, 1 }, { n, 0 } };
 
    if (N == 0)
        return a;
    if (N == 1)
        return b;
    if (N == 2)
        return m * b + n * a;
 
    int[][] initial = { { m * b + n * a, b },
                        { b, a } };
                         
    power(F, N - 2, m, n);
 
    // Discussed below
    multiply(initial, F);
 
    return F[0][0];
}
 
// Function that multiplies
// 2 matrices F and M of size 2*2,
// and puts the multiplication
// result back to F[][]
static void multiply(int[][] F, int[][] M)
{
    int x = F[0][0] * M[0][0] +
            F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] +
            F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] +
            F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] +
            F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Function that calculates F[][]
// raised to the power N
// and puts the result in F[][]
static void power(int[][] F, int N,
                  int m, int n)
{
    int i;
    int[][] M = { { m, 1 }, { n, 0 } };
 
    for(i = 1; i <= N; i++)
        multiply(F, M);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 2, a = 0, b = 1, m = 2, n = 3;
    System.out.println(F(N, a, b, m, n));
     
    N = 3;
    System.out.println(F(N, a, b, m, n));
     
    N = 4;
    System.out.println(F(N, a, b, m, n));
     
    N = 5;
    System.out.println(F(N, a, b, m, n));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement the
# Generalised Fibonacci numbers
 
# Function to find the Nth term
def F(N, a, b, m, n):
     
    # m 1
    # n 0
    F = [[ m, 1 ], [ n, 0 ]]
 
    if(N == 0):
        return a
    if(N == 1):
        return b
    if(N == 2):
        return m * b + n * a
 
    initial = [[ m * b + n * b, b ],
                           [ b, a ]]
 
    power(F, N - 2, m, n)
 
    multiply(initial, F)
 
    return F[0][0]
 
# Function that multiplies
# 2 matrices F and M of size 2*2,
# and puts the multiplication
# result back to F[][]
def multiply(F, M):
     
    x = (F[0][0] * M[0][0] +
         F[0][1] * M[1][0])
 
    y = (F[0][0] * M[0][1] +
         F[0][1] * M[1][1])
 
    z = (F[1][0] * M[0][0] +
         F[1][1] * M[1][0])
 
    w = (F[1][0] * M[0][1] +
         F[1][1] * M[1][1])
 
    F[0][0] = x
    F[0][1] = y
    F[1][0] = z
    F[1][1] = w
 
# Function that calculates F[][]
# raised to the power N
# and puts the result in F[][]
def power(F, N, m, n):
 
    M = [[ m, 1 ], [ n, 0 ]]
    for i in range(1, N + 1):
        multiply(F, M)
 
# Driver code
if __name__ == '__main__':
 
    N, a, b, m, n = 2, 0, 1, 2, 3
    print(F(N, a, b, m, n))
 
    N = 3
    print(F(N, a, b, m, n))
 
    N = 4
    print(F(N, a, b, m, n))
 
    N = 5
    print(F(N, a, b, m, n))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement the
// Generalised Fibonacci numbers
using System;
class GFG{
  
// Function to find the Nth term
static int F(int N, int a, int b,
             int m, int n)
{
    // m 1
    // n 0
    int[,] F = { { m, 1 }, { n, 0 } };
    if (N == 0)
        return a;
    if (N == 1)
        return b;
    if (N == 2)
        return m * b + n * a;
    int[,] initial = { { m * b + n * a, b },
                        { b, a } };
    power(F, N - 2, m, n);
  
    // Discussed below
    multiply(initial, F);
    return F[0, 0];
}
  
// Function that multiplies
// 2 matrices F and M of size 2*2,
// and puts the multiplication
// result back to F[,]
static void multiply(int[,] F, int[,] M)
{
    int x = F[0, 0] * M[0, 0] +
            F[0, 1] * M[1, 0];
    int y = F[0, 0] * M[0, 1] +
            F[0, 1] * M[1, 1];
    int z = F[1, 0] * M[0, 0] +
            F[1, 1] * M[1, 0];
    int w = F[1, 0] * M[0, 1] +
            F[1, 1] * M[1, 1];
  
    F[0, 0] = x;
    F[0, 1] = y;
    F[1, 0] = z;
    F[1, 1] = w;
}
  
// Function that calculates F[,]
// raised to the power N
// and puts the result in F[,]
static void power(int[,] F, int N,
                  int m, int n)
{
    int i;
    int[,] M = { { m, 1 }, { n, 0 } };
    for(i = 1; i <= N; i++)
        multiply(F, M);
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 2, a = 0, b = 1, m = 2, n = 3;
    Console.WriteLine(F(N, a, b, m, n));
    N = 3;
    Console.WriteLine(F(N, a, b, m, n));
    N = 4;
    Console.WriteLine(F(N, a, b, m, n));
    N = 5;
    Console.WriteLine(F(N, a, b, m, n));
}
}
  
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript program to implement the
// Generalised Fibonacci numbers
 
// Function to find the Nth term
function F(N, a, b, m, n)
{
    var F = [ [ m, 1 ], [ n, 0 ] ]
     
    if (N == 0)
        return a;
    if (N == 1)
        return b;
    if (N == 2)
        return m * b + n * a;
     
    var initial = [ [ m * b + n * a, b ], [ b, a ] ]
     
    power(F, N - 2, m, n);
     
    // Discussed below
    multiply(initial, F);
     
    return F[0][0];
}
 
// Function that multiplies
// 2 matrices F and M of size 2*2,
// and puts the multiplication
// result back to F[][]
function multiply(F, M)
{
    var x = F[0][0] * M[0][0] +
            F[0][1] * M[1][0];
    var y = F[0][0] * M[0][1] +
            F[0][1] * M[1][1];
    var z = F[1][0] * M[0][0] +
            F[1][1] * M[1][0];
    var w = F[1][0] * M[0][1] +
            F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Function that calculates F[][]
// raised to the power N
// and puts the result in F[][]
function power(F, N, m, n)
{
    var i;
    var M = [ [ m, 1 ], [ n, 0 ] ];
 
    for(i = 1; i <= N; i++)
        multiply(F, M);
}
 
// Driver code
var N = 2, a = 0, b = 1, m = 2, n = 3;
document.write(F(N, a, b, m, n) + "</br>");
 
N = 3;
document.write(F(N, a, b, m, n) + "</br>");
 
N = 4;
document.write(F(N, a, b, m, n) + "</br>");
 
N = 5;
document.write(F(N, a, b, m, n));
     
// This code is contributed by Ankita saini
    

输出：

2
7
20
61