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📜  快速加倍方法以找到第N个斐波那契数

📅  最后修改于: 2021-05-04 07:47:43             🧑  作者: Mango

给定整数N ,任务是找到第N个斐波那契数。
例子:

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:

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


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)