范围内的斐波那契数之和

给定范围[l，r] ，任务是找到和fib(l)+ fib(l + 1)+ fib(l + 2)+….. + fib(r) ，其中fib(n)是^第n^个斐波那契数。
例子：

Input: l = 2, r = 5 
Output: 11 
fib(2) + fib(3) + fib(4) + fib(5) = 1 + 2 + 3 + 5 = 11
Input: l = 4, r = 8 
Output: 50 
 

天真的方法：只需以O(r – l)时间复杂度来计算fib(l)+ fib(l + 1)+ fib(l + 2)+….. + fib(r) 。
为了在O(1)中找到fib(n) ，我们将借助黄金分割率。
斐波那契使用Binet公式的计算

fib(n) = phiⁿ – psiⁿ) / ?5 
Where, 
phi = (1 + sqrt(5)) / 2 which is roughly equal to 1.61803398875 
psi = 1 – phi = (1 – sqrt(5)) / 2 which is roughly equal to 0.61803398875 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
 
// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}
 
// Function to return the required sum
ll calculateSum(int l, int r)
{
 
    // To store the sum
    ll sum = 0;
 
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
 
    return sum;
}
 
// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);
 
    return 0;
}

Java

// Java implementation of the approach
import java.lang.Math;
class GFG
{
     
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int)Math.round(Math.pow(phi, n) / Math.sqrt(5));
}
 
// Function to return the required sum
static int calculateSum(int l, int r)
{
 
    // To store the sum
    int sum = 0;
 
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));
 
}
}
 
//This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach
 
# Function to return the nth
# Fibonacci number
def fib(n):
    phi = ((1 + (5 ** (1 / 2))) / 2);
    return round((phi ** n) / (5 ** (1 / 2)));
 
# Function to return the required sum
def calculateSum(l, r):
     
    # To store the sum
    sum = 0;
 
    # Calculate the sum
    for i in range(l, r + 1):
        sum += fib(i);
 
    return sum;
 
# Driver Code
if __name__ == '__main__':
    l, r = 4, 8;
    print(calculateSum(l, r));
 
# This code contributed by Rajput-Ji

C#

// C# implemenatation of above approach
using System;
 
class GFG
{
     
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int)Math.Round(Math.Pow(phi, n) / Math.Sqrt(5));
}
 
// Function to return the required sum
static int calculateSum(int l, int r)
{
 
    // To store the sum
    int sum = 0;
 
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
 
    return sum;
}
 
// Driver code
public static void Main()
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// Javascript implementation of the approach
 
    // Function to return the nth Fibonacci number
    function fib(n) {
        var phi = (1 + Math.sqrt(5)) / 2;
        return parseInt( Math.round
        (Math.pow(phi, n)/ Math.sqrt(5))
        );
    }
 
    // Function to return the required sum
    function calculateSum(l , r) {
 
        // To store the sum
        var sum = 0;
 
        // Calculate the sum
        for (i = l; i <= r; i++)
            sum += fib(i);
 
        return sum;
    }
 
    // Driver code
     
        var l = 4, r = 8;
        document.write(calculateSum(l, r));
 
 
// This code contributed by gauravrajput1
 

输出：

50

高效的方法：想法是找到斐波那契数之和与^第n^个斐波那契数之间的关系，并使用Binet公式计算其值。
关系推论

1. F(i)是第i^个斐波那契数。
2. S(i)指直到F(i)的斐波纳契数之和。

We can rewrite the relation F(n + 1) = F(n) + F(n – 1) as below: 
F(n – 1) = F(n + 1) – F(n) 
Similarly, 
F(n – 2) = F(n) – F(n – 1) 
… 
… 
… 
F(0) = F(2) – F(1)
Adding all the equations, on left side, we have 
F(0) + F(1) + … + F(n – 1) which is S(n – 1) 
 

所以，
S(n – 1)= F(n + 1)– F(1)
S(n – 1)= F(n + 1)– 1
S(n)= F(n + 2)– 1
为了找到S(n) ，只需计算第(n + 2)^个斐波那契数，然后从结果中减去1 。
所以，
S(l，r)= S(r)– S(l – 1)
S(l，r)= F(r + 2)– 1 –(F(l + 1)– 1)
S(l，r)= F(r + 2)– F(l + 1)

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}
 
// Function to return the required sum
int calculateSum(int l, int r)
{
 
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
 
// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int) Math.round(Math.pow(phi, n) / Math.sqrt(5));
}
 
// Function to return the required sum
static int calculateSum(int l, int r)
{
 
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
import math
 
# Function to return the nth
# Fibonacci number
def fib(n):
 
    phi = (1 + math.sqrt(5)) / 2;
    return int(round(pow(phi, n) /
                         math.sqrt(5)));
 
# Function to return the required sum
def calculateSum(l, r):
 
    # Using our deduced result
    sum = fib(r + 2) - fib(l + 1);
    return sum;
 
# Driver code
l = 4;
r = 8;
print(calculateSum(l, r));
 
# This code is contributed by mits

C＃

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int) Math.Round(Math.Pow(phi, n) /
                            Math.Sqrt(5));
}
 
// Function to return the required sum
static int calculateSum(int l, int r)
{
 
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
 
// Driver code
public static void Main()
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}
 
// This code is contributed
// by Akanksha Rai

输出：

50

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