斐波那契数的总和

给定一个正数n，找到f ₀ + f ₁ + f ₂ +…的值。 + f _n其中，f _i表示第i个斐波那契数。请记住，f ₀ = 0，f ₁ = 1，f ₂ = 1，f ₃ = 2，f ₄ = 3，f ₅ = 5，…
例子 ：

Input  : n = 3
Output : 4
Explanation : 0 + 1 + 1 + 2  = 4

Input  :  n = 4
Output :  7
Explanation : 0 + 1 + 1 + 2 + 3  = 7

方法1(O(n))
蛮力法很简单，找到所有斐波那契数直到f(n)，然后将它们加起来。

C++

// C++ Program to find sum of Fibonacci numbers
#include
using namespace std;
 
// Computes value of first fibonacci numbers
int calculateSum(int n)
{
    if (n <= 0)
       return 0;
 
    int fibo[n+1];
    fibo[0] = 0, fibo[1] = 1;
 
    // Initialize result
    int sum = fibo[0] + fibo[1];
 
    // Add remaining terms
    for (int i=2; i<=n; i++)
    {
        fibo[i] = fibo[i-1]+fibo[i-2];
        sum += fibo[i];
    }
 
    return sum;
}
 
// Driver program to test above function
int main()
{
    int n = 4;
    cout << "Sum of Fibonacci numbers is : "
         << calculateSum(n) << endl;
    return 0;
}

Java

// Java Program to find
// sum of Fibonacci numbers
 
import java.io.*;
 
class GFG {
     
    // Computes value of first
    // fibonacci numbers
    static int calculateSum(int n)
    {
        if (n <= 0)
           return 0;
      
        int fibo[]=new int[n+1];
        fibo[0] = 0; fibo[1] = 1;
      
        // Initialize result
        int sum = fibo[0] + fibo[1];
      
        // Add remaining terms
        for (int i=2; i<=n; i++)
        {
            fibo[i] = fibo[i-1]+fibo[i-2];
            sum += fibo[i];
        }
      
        return sum;
    }
      
    // Driver program to test above function
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Sum of Fibonacci" +
        " numbers is : "+ calculateSum(n));
    }
}
 
// This code is contributed by Nikita tiwari.

Python3

# Python 3 Program to find
# sum of Fibonacci numbers
 
 
# Computes value of first
# fibonacci numbers
def calculateSum(n) :
    if (n <= 0) :
        return 0
  
    fibo =[0] * (n+1)
    fibo[1] = 1
  
    # Initialize result
    sm = fibo[0] + fibo[1]
  
    # Add remaining terms
    for i in range(2,n+1) :
        fibo[i] = fibo[i-1] + fibo[i-2]
        sm = sm + fibo[i]
         
    return sm
 
 
# Driver program to test
# above function
n = 4
print("Sum of Fibonacci numbers is : " ,
      calculateSum(n))
 
# This code is contributed
# by Nikita tiwari.

C#

// C# Program to find
// sum of Fibonacci numbers
using System;
 
class GFG
{
     
    // Computes value of first
    // fibonacci numbers
    static int calculateSum(int n)
    {
        if (n <= 0)
        return 0;
     
        int []fibo = new int[n + 1];
        fibo[0] = 0; fibo[1] = 1;
     
        // Initialize result
        int sum = fibo[0] + fibo[1];
     
        // Add remaining terms
        for (int i = 2; i <= n; i++)
        {
            fibo[i] = fibo[i - 1] + fibo[i - 2];
            sum += fibo[i];
        }
     
        return sum;
    }
     
    // Driver Code
    static void Main()
    {
        int n = 4;
        Console.WriteLine( "Sum of Fibonacci" +
                              " numbers is : "+
                              calculateSum(n));
    }
}
 
// This code is contributed by Anuj_67

Javascript

// Javascript Program to find sum
// of Fibonacci numbers
 
// Computes value of first
// fibonacci numbers
function calculateSum(n)
{
    let fibo = [];
    if (n <= 0)
    return 0;
 
    fibo[0] = 0;
    fibo[1] = 1;
 
    // Initialize result
    let sum = fibo[0] + fibo[1];
 
    // Add remaining terms
    for(let i = 2; i <= n; i++)
    {
        fibo[i] = fibo[i - 1] +
                    fibo[i - 2];
        sum += fibo[i];
    }
 
    return sum;
}
 
    // Driver Code
    let n = 4;
    document.write(`Sum of Fibonacci numbers is :
        ${calculateSum(n)} <br>`);
 
// This code is contributed by _saurabh_jaiswal

C++

// C++ Program to find sum of Fibonacci numbers in
// O(Log n) time.
#include 
using namespace std;
const int MAX = 1000;
 
// Create an array for memoization
int f[MAX] = {0};
 
// Returns n'th Fibonacci number using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1)? (n+1)/2 : n/2;
 
    // Applying above formula [Note value n&1 is 1
    // if n is odd, else 0].
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
 
    return f[n];
}
 
// Computes value of first Fibonacci numbers
int calculateSum(int n)
{
    return fib(n+2) - 1;
}
 
// Driver program to test above function
int main()
{
    int n = 4;
    cout << "Sum of Fibonacci numbers is : "
         << calculateSum(n) << endl;
    return 0;
}
Java] # Python 3 Program to find sum of 
# Fibonacci numbers in O(Log n) time.

MAX = 1000

# Create an array for memoization
f = [0] * MAX

# Returns n'th Fibonacci number
# using table f[]
def fib(n):
    
    n = int(n)

    # Base cases
    if (n == 0):
        return 0
    if (n == 1 or n == 2):
        return (1) 

    # If fib(n) is already computed
    if (f[n] == True):
        return f[n] 

    k = (n+1)/2 if (n & 1) else n/2

    # Applying above formula [Note value n&1 
    # is 1 if n is odd, else 0].
    f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1)) if (n & 1) else (2 * fib(k-1) + fib(k)) * fib(k)
    return f[n] 

# Computes value of first Fibonacci numbers
def calculateSum(n):

    return fib(n+2) - 1

# Driver program to test above function
n = 4
print("Sum of Fibonacci numbers is :", calculateSum(n)) 

# This code is contributed by
# Smitha Dinesh Semwal

输出 ：

Sum of Fibonacci numbers is : 7

方法2(O(Log n))
这个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。
F(i)是第i个斐波那契数。
S(i)指直到F(i)的斐波那契数的总和，

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)
-------------------------------

添加所有方程式，在左侧，我们有
F(0)+ F(1)+…F(n-1)，即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 —(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。
F(n)可以使用本文中的方法5或方法6在O(log n)的时间内求值(请参阅方法5和6)。
下面是基于此方法6的实现

C++

// C++ Program to find sum of Fibonacci numbers in
// O(Log n) time.
#include 
using namespace std;
const int MAX = 1000;
 
// Create an array for memoization
int f[MAX] = {0};
 
// Returns n'th Fibonacci number using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1)? (n+1)/2 : n/2;
 
    // Applying above formula [Note value n&1 is 1
    // if n is odd, else 0].
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
 
    return f[n];
}
 
// Computes value of first Fibonacci numbers
int calculateSum(int n)
{
    return fib(n+2) - 1;
}
 
// Driver program to test above function
int main()
{
    int n = 4;
    cout << "Sum of Fibonacci numbers is : "
         << calculateSum(n) << endl;
    return 0;
}

Java]

# Python 3 Program to find sum of 
# Fibonacci numbers in O(Log n) time.

MAX = 1000

# Create an array for memoization
f = [0] * MAX

# Returns n'th Fibonacci number
# using table f[]
def fib(n):
    
    n = int(n)

    # Base cases
    if (n == 0):
        return 0
    if (n == 1 or n == 2):
        return (1) 

    # If fib(n) is already computed
    if (f[n] == True):
        return f[n] 

    k = (n+1)/2 if (n & 1) else n/2

    # Applying above formula [Note value n&1 
    # is 1 if n is odd, else 0].
    f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1)) if (n & 1) else (2 * fib(k-1) + fib(k)) * fib(k)
    return f[n] 

# Computes value of first Fibonacci numbers
def calculateSum(n):

    return fib(n+2) - 1

# Driver program to test above function
n = 4
print("Sum of Fibonacci numbers is :", calculateSum(n)) 

# This code is contributed by
# Smitha Dinesh Semwal

C＃

// C# Program to find sum
// of Fibonacci numbers in
// O(Log n) time.
using System;
 
class GFG {
    static int MAX = 1000;
 
    // Create an array for memoization
    static int []f = new int[MAX];
     
    // Returns n'th Fibonacci
    // number using table f[]
    static int fib(int n)
    {
        for(int i = 0;i < MAX;i++)
        f[i] = 0;
         
        //Arrays.fill(f, 0);
        // Base cases
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return (f[n] = 1);
     
        // If fib(n) is
        // already computed
        if (f[n] == 1)
            return f[n];
            int k;
        if((n & 1) == 1)
            k = (n + 1) / 2 ;
        else
            k = n / 2;
     
        // Applying above formula
        // [Note value n&1 is 1
        // if n is odd, else 0].
        if((n & 1) == 1)
            f[n] = (fib(k) * fib(k) + fib(k - 1)
                                   * fib(k - 1));
        else
            f[n] = (2 * fib(k - 1) + fib(k)) *
                                       fib(k);
     
        return f[n];
    }
     
    // Computes value of first
    // Fibonacci numbers
    static int calculateSum(int n)
    {
        return fib(n + 2) - 1;
    }
     
    // Driver Code
    public static void Main()
    {
        int n = 4;
        Console.Write( "Sum of Fibonacci numbers is : "
                                    + calculateSum(n));
         
    }
}
 
// This code is contributed by nitin mittal.

输出 ：

Sum of Fibonacci numbers is : 7