快速加倍方法以找到第N个斐波那契数

给定整数N ，任务是找到第N个斐波那契数。
例子：

Input: N = 3 
Output: 2 
Explanation: 
F(1) = 1, F(2) = 1 
F(3) = F(1) + F(2) = 2 
Input: N = 6 
Output: 8 
 

方法：

• 矩阵求幂方法已经在前面讨论过了。尽管不使用矩阵乘法本身，但是加倍方法可以看作是对求第N个斐波那契数的矩阵求幂方法的一种改进。
• 斐波那契递归序列由下式给出

F(n+1) = F(n) + F(n-1)

• 矩阵幂方法使用以下公式

      $\[ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} \]$

• 该方法涉及昂贵的矩阵乘法，而且F _n被重复计算两次。
  另一方面，快速加倍方法基于两个基本公式：

F(2n) = F(n)[2F(n+1) – F(n)]
F(2n + 1) = F(n)2 + F(n+1)2

• 这是上述结果的简短说明：

Start with: 
F(n+1) = F(n) + F(n-1) & 
F(n) = F(n) 
It can be rewritten in the matrix form as: 

    $\[ \begin{bmatrix} F(n+1) \\ F(n) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} F(n) \\ F(n-1) \end{bmatrix} \] \[\quad\enspace= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^2 \begin{bmatrix} F(n-1) \\ F(n-2) \end{bmatrix} \] \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\enspace\enspace\thinspace......\\ \[= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \]$

For doubling, we just plug in “2n” into the formula:

    $\[ \begin{bmatrix} F(2n+1) \\ F(2n) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{2n} \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \] \[\quad\enspace= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \] \[\quad\enspace= \begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix} \begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix} \begin{bmatrix} F(1) \\ F(0) \end{bmatrix} \] \[\quad\enspace= \begin{bmatrix} F(n+1)^2 + F(n)^2 \\ F(n)F(n+1) + F(n)F(n-1) \end{bmatrix} \]$

Substituting F(n-1) = F(n+1)- F(n) and after simplification we get, 

    $\[ \begin{bmatrix} F(2n+1) \\ F(2n) \end{bmatrix} = \begin{bmatrix} F(n+1)^2 + F(n)^2 \\ 2F(n+1)F(n) - F(n)^2 \end{bmatrix} \]$

 

下面是上述方法的实现：

C++

// C++ program to find the Nth Fibonacci
// number using Fast Doubling Method
#include 
using namespace std;
 
int a, b, c, d;
#define MOD 1000000007
 
// Function calculate the N-th fibanacci
// number using fast doubling method
void FastDoubling(int n, int res[])
{
    // Base Condition
    if (n == 0) {
        res[0] = 0;
        res[1] = 1;
        return;
    }
    FastDoubling((n / 2), res);
 
    // Here a = F(n)
    a = res[0];
 
    // Here b = F(n+1)
    b = res[1];
 
    c = 2 * b - a;
 
    if (c < 0)
        c += MOD;
 
    // As F(2n) = F(n)[2F(n+1) – F(n)]
    // Here c  = F(2n)
    c = (a * c) % MOD;
 
    // As F(2n + 1) = F(n)^2 + F(n+1)^2
    // Here d = F(2n + 1)
    d = (a * a + b * b) % MOD;
 
    // Check if N is odd
    // or even
    if (n % 2 == 0) {
        res[0] = c;
        res[1] = d;
    }
    else {
        res[0] = d;
        res[1] = c + d;
    }
}
 
// Driver code
int main()
{
    int N = 6;
    int res[2] = { 0 };
 
    FastDoubling(N, res);
 
    cout << res[0] << "\n";
    return 0;
}

Java

// Java program to find the Nth Fibonacci
// number using Fast Doubling Method
class GFG{
 
// Function calculate the N-th fibanacci
// number using fast doubling method
static void FastDoubling(int n, int []res)
{
    int a, b, c, d;
    int MOD = 1000000007;
     
    // Base Condition
    if (n == 0)
    {
        res[0] = 0;
        res[1] = 1;
        return;
    }
    FastDoubling((n / 2), res);
 
    // Here a = F(n)
    a = res[0];
 
    // Here b = F(n+1)
    b = res[1];
 
    c = 2 * b - a;
 
    if (c < 0)
        c += MOD;
 
    // As F(2n) = F(n)[2F(n+1) – F(n)]
    // Here c = F(2n)
    c = (a * c) % MOD;
 
    // As F(2n + 1) = F(n)^2 + F(n+1)^2
    // Here d = F(2n + 1)
    d = (a * a + b * b) % MOD;
 
    // Check if N is odd
    // or even
    if (n % 2 == 0)
    {
        res[0] = c;
        res[1] = d;
    }
    else
    {
        res[0] = d;
        res[1] = c + d;
    }
}
 
// Driver code
public static void main(String []args)
{
    int N = 6;
    int res[] = new int[2];
 
    FastDoubling(N, res);
 
    System.out.print(res[0]);
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program to find the Nth Fibonacci
# number using Fast Doubling Method
MOD = 1000000007
 
# Function calculate the N-th fibanacci
# number using fast doubling method
def FastDoubling(n, res):
     
    # Base Condition
    if (n == 0):
        res[0] = 0
        res[1] = 1
        return
         
    FastDoubling((n // 2), res)
 
    # Here a = F(n)
    a = res[0]
 
    # Here b = F(n+1)
    b = res[1]
 
    c = 2 * b - a
 
    if (c < 0):
        c += MOD
 
    # As F(2n) = F(n)[2F(n+1) – F(n)]
    # Here c = F(2n)
    c = (a * c) % MOD
 
    # As F(2n + 1) = F(n)^2 + F(n+1)^2
    # Here d = F(2n + 1)
    d = (a * a + b * b) % MOD
 
    # Check if N is odd
    # or even
    if (n % 2 == 0):
        res[0] = c
        res[1] = d
    else :
        res[0] = d
        res[1] = c + d
     
# Driver code
N = 6
res = [0] * 2
 
FastDoubling(N, res)
 
print(res[0])
     
# This code is contributed by divyamohan123

C#

// C# program to find the Nth Fibonacci
// number using Fast Doubling Method
using System;
class GFG{
 
// Function calculate the N-th fibanacci
// number using fast doubling method
static void FastDoubling(int n, int []res)
{
    int a, b, c, d;
    int MOD = 1000000007;
     
    // Base Condition
    if (n == 0)
    {
        res[0] = 0;
        res[1] = 1;
        return;
    }
    FastDoubling((n / 2), res);
 
    // Here a = F(n)
    a = res[0];
 
    // Here b = F(n+1)
    b = res[1];
 
    c = 2 * b - a;
 
    if (c < 0)
        c += MOD;
 
    // As F(2n) = F(n)[2F(n+1) – F(n)]
    // Here c = F(2n)
    c = (a * c) % MOD;
 
    // As F(2n + 1) = F(n)^2 + F(n+1)^2
    // Here d = F(2n + 1)
    d = (a * a + b * b) % MOD;
 
    // Check if N is odd
    // or even
    if (n % 2 == 0)
    {
        res[0] = c;
        res[1] = d;
    }
    else
    {
        res[0] = d;
        res[1] = c + d;
    }
}
 
// Driver code
public static void Main()
{
    int N = 6;
    int []res = new int[2];
 
    FastDoubling(N, res);
 
    Console.Write(res[0]);
}
}
 
// This code is contributed by Code_Mech

Javascript

 
    // Javascript program to find the Nth Fibonacci
      // number using Fast Doubling Method
     
    let a, b, c, d;
    let MOD = 1000000007;
 
    // Function calculate the N-th fibanacci
    // number using fast doubling method
    function FastDoubling(n, res)
    {
        // Base Condition
        if (n == 0) {
            res[0] = 0;
            res[1] = 1;
            return;
        }
        FastDoubling(parseInt(n / 2, 10), res);
 
        // Here a = F(n)
        a = res[0];
 
        // Here b = F(n+1)
        b = res[1];
 
        c = 2 * b - a;
 
        if (c < 0)
            c += MOD;
 
        // As F(2n) = F(n)[2F(n+1) – F(n)]
        // Here c  = F(2n)
        c = (a * c) % MOD;
 
        // As F(2n + 1) = F(n)^2 + F(n+1)^2
        // Here d = F(2n + 1)
        d = (a * a + b * b) % MOD;
 
        // Check if N is odd
        // or even
        if (n % 2 == 0) {
            res[0] = c;
            res[1] = d;
        }
        else {
            res[0] = d;
            res[1] = c + d;
        }
    }
 
    let N = 6;
    let res = new Array(2);
    res.fill(0);
   
    FastDoubling(N, res);
   
    document.write(res[0]);
     

输出：

8

时间复杂度：重复平方减少了从线性到对数的时间。因此，使用恒定时间算法，时间复杂度为O(log n) 。

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