斐波纳契数的总和|套装2

给定正数N ，任务是找到第一个(N + 1)斐波那契数之和。

例子：

Input: N = 3
Output: 4
Explanation:
The first 4 terms of the Fibonacci Series is {0, 1, 1, 2}. Therefore, the required sum = 0 + 1 + 1 + 2 = 4.

Input: N = 4
Output: 7

天真的方法：有关解决问题的最简单方法，请参阅本文的上一篇文章。
时间复杂度： O(N)
辅助空间： O(1)

高效的方法：可以通过以下观察和计算来优化上述方法：

令S(N)代表斐波那契数列的前N个项的和。现在，为了简单地找到S(N) ，计算第(N + 2)^个斐波那契数，并从结果中减去1 。该系列的第N^个项可以通过以下公式计算：

$F_N = \frac{(\frac{1 + \sqrt{5}}{2})^{N}}{\sqrt{5}}$

现在，可以通过(F _{N + 2} – 1)来计算S(N)的值。

因此，我们的想法是使用上式计算S _N的值：

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the sum of
// first N + 1 fibonacci numbers
void sumFib(int N)
{
   
    // Apply the formula
    long num = (long)round(pow((sqrt(5) + 1)
                               / 2.0, N + 2)
                           / sqrt(5));
 
    // Print the result
    cout << (num - 1);
}
 
// Driver Code
int main()
{
    int N = 3;
    sumFib(N);
    return 0;
}
 
// This code is contributed by Dharanendra L V.

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to find the sum of
    // first N + 1 fibonacci numbers
    public static void sumFib(int N)
    {
        // Apply the formula
        long num = (long)Math.round(
            Math.pow((Math.sqrt(5) + 1)
                         / 2.0,
                     N + 2)
            / Math.sqrt(5));
 
        // Print the result
        System.out.println(num - 1);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;
        sumFib(N);
    }
}

Python3

# Python program for the above approach
import math
 
# Function to find the sum of
# first N + 1 fibonacci numbers
def sumFib(N):
   
    # Apply the formula
    num = round(pow((pow(5,1/2) + 1) \
                    / 2.0, N + 2) \
                / pow(5,1/2));
 
    # Prthe result
    print(num - 1);
 
# Driver Code
if __name__ == '__main__':
    N = 3;
    sumFib(N);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
class GFG
{
 
    // Function to find the sum of
    // first N + 1 fibonacci numbers
    public static void sumFib(int N)
    {
        // Apply the formula
        long num = (long)Math.Round(
            Math.Pow((Math.Sqrt(5) + 1)
                         / 2.0,
                     N + 2)
            / Math.Sqrt(5));
 
        // Print the result
        Console.WriteLine(num - 1);
    }
 
// Driver Code
static public void Main()
{
        int N = 3;
        sumFib(N);
}
}
 
// This code is contributed by jana_sayantan.

---

输出：

4

时间复杂度： O(1)
辅助空间： O(1)

替代方法：请按照以下步骤解决问题：

• 第N^个斐波那契数也可以使用生成函数来计算。
• 第N^个斐波那契数的生成函数由下式给出：

$G(X) = \frac{X}{(1 - X - X^{2})}$

• 因此，斐波纳契数之和的生成函数由下式给出：

$G'(X) = \frac{X}{((1 - X - X^{2}) * (1 - X))}$

• G’(X)的部分分解后， G’(X)的值由下式给出：

$G'(X) = \frac{a}{(a + 1) * (a - b) * (X + a)} + \frac{b}{(b + 1) * (b - a) * (X + b)} + \frac{1}{(a + 1) * (b + 1) * (X - 1)}$

where, 
$a = \frac{(1 + \sqrt5)}{2}, b = \frac{(1 - \sqrt5)}{2}$

• 因此，斐波那契数之和的公式为：

$S_{N} = \mid \frac{1}{b^{N + 1} + b^{N} + b^{N - 1} + b^{N - 2} } \mid - 1$

下面是上述方法的实现：

C++

// C++  program for the above approach
#include 
using namespace std;
 
// Function to find the sum of
// first N + 1 fibonacci numbers
void sumFib(int N)
{
   
  // Apply the formula
  double num = (1 - sqrt(5)) / 2;
 
  long val = round(
    abs(1
        / (pow(num, N + 2)
           + pow(num, N + 1)
           + pow(num, N)
           + pow(num, N - 1)))
    - 1);
 
  // Print the result
  cout<

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to find the sum of
    // first N + 1 fibonacci numbers
    public static void sumFib(int N)
    {
        // Apply the formula
        double num = (1 - Math.sqrt(5)) / 2;
 
        long val = Math.round(
            Math.abs(1
                     / (Math.pow(num, N + 2)
                        + Math.pow(num, N + 1)
                        + Math.pow(num, N)
                        + Math.pow(num, N - 1)))
            - 1);
 
        // Print the result
        System.out.println(val);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 3;
 
        // Function Call
        sumFib(N);
    }
}

C＃

// C# program for the above approach
 
 
using System;
 
public class GFG {
 
    // Function to find the sum of
    // first N + 1 fibonacci numbers
    public static void sumFib(int N)
    {
        // Apply the formula
        double num = (1 - Math.Sqrt(5)) / 2;
 
        double val = Math.Round(
            Math.Abs(1
                     / (Math.Pow(num, N + 2)
                        + Math.Pow(num, N + 1)
                        + Math.Pow(num, N)
                        + Math.Pow(num, N - 1)))
            - 1);
 
        // Print the result
        Console.WriteLine(val);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 3;
 
        // Function Call
        sumFib(N);
    }
}
 
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度： O(1)
辅助空间： O(1)