总和为斐波那契数的子数组总数

给定一个由N 个整数组成的数组arr[] ，任务是计算总和为斐波那契数的子数组的总数。
例子：

Input: arr[] = {6, 7, 8, 9} 
Output: 3 
Explanation: 
The subarray whose sum is fibonacci numbers are: 
1. {6, 7}, sum = 13 (5 + 8) 
2. {6, 7, 8}, sum = 21 (8 + 13) 
3. {8}, sum = 8 (3 + 5)
Input: arr[] = {1, 1, 1, 1} 
Output: 4 
Explanation: 
The subarray whose sum is fibonacci numbers are: 
1. {4, 2, 2}, sum = 8 (3 + 5) 
2. {2}, sum = 2 (1 + 1) 
3. {2}, sum = 2 (1 + 1) 
4. {2}, sum = 2 (1 + 1) 
 

方法：这个想法是生成所有可能的子数组并找到每个子数组的总和。现在，通过使用本文中讨论的方法检查每个和是否是斐波那契数。如果是，则计算所有这些总和并打印总计数。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check whether a number
// is perfect square or not
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}
 
// Function to check whether a number
// is fibonacci number or not
bool isFibonacci(int n)
{
    // If 5*n*n + 4 or 5*n*n - 5 is a
    // perfect square, then the number
    // is Fibonacci
    return isPerfectSquare(5 * n * n + 4)
           || isPerfectSquare(5 * n * n - 4);
}
 
// Function to count the subarray with
// sum fibonacci number
void fibonacciSubarrays(int arr[], int n)
{
    int count = 0;
 
    // Traverse the array arr[] to find
    // the sum of each subarray
    for (int i = 0; i < n; ++i) {
 
        // To store the sum
        int sum = 0;
 
        for (int j = i; j < n; ++j) {
            sum += arr[j];
 
            // Check whether sum of subarray
            // between [i, j] is fibonacci
            // or not
            if (isFibonacci(sum)) {
                ++count;
            }
        }
    }
 
    cout << count;
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    fibonacciSubarrays(arr, n);
    return 0;
}

Java

// Java program for the above approach
 
class GFG{
 
// Function to check whether a number
// is perfect square or not
static boolean isPerfectSquare(int x)
{
    int s = (int) Math.sqrt(x);
    return (s * s == x);
}
 
// Function to check whether a number
// is fibonacci number or not
static boolean isFibonacci(int n)
{
    // If 5*n*n + 4 or 5*n*n - 5 is a
    // perfect square, then the number
    // is Fibonacci
    return isPerfectSquare(5 * n * n + 4)
        || isPerfectSquare(5 * n * n - 4);
}
 
// Function to count the subarray
// with sum fibonacci number
static void fibonacciSubarrays(int arr[], int n)
{
    int count = 0;
 
    // Traverse the array arr[] to find
    // the sum of each subarray
    for (int i = 0; i < n; ++i) {
 
        // To store the sum
        int sum = 0;
 
        for (int j = i; j < n; ++j) {
            sum += arr[j];
 
            // Check whether sum of subarray
            // between [i, j] is fibonacci
            // or not
            if (isFibonacci(sum)) {
                ++count;
            }
        }
    }
 
    System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 6, 7, 8, 9 };
    int n = arr.length;
 
    // Function Call
    fibonacciSubarrays(arr, n);
}
}
 
// This code contributed by PrinciRaj1992

Python3

# Python3 program for the above approach
import math
 
# Function to check whether a number
# is perfect square or not
def isPerfectSquare(x):
     
    s = int(math.sqrt(x))
    if s * s == x:
        return True
    return False
 
# Function to check whether a number
# is fibonacci number or not
def isFibonacci(n):
     
    # If 5*n*n + 4 or 5*n*n - 5 is a
    # perfect square, then the number
    # is Fibonacci
    return (isPerfectSquare(5 * n * n + 4) or
            isPerfectSquare(5 * n * n - 4))
 
# Function to count the subarray with
# sum fibonacci number
def fibonacciSubarrays(arr, n):
     
    count = 0
     
    # Traverse the array arr[] to find
    # the sum of each subarray
    for i in range(n):
         
        # To store the sum
        sum = 0
         
        for j in range(i, n):
            sum += arr[j]
             
            # Check whether sum of subarray
            # between [i, j] is fibonacci
            # or not
            if (isFibonacci(sum)):
                count += 1
                 
    print(count)
 
# Driver Code
arr = [ 6, 7, 8, 9 ]
n = len(arr)
 
# Function Call
fibonacciSubarrays(arr, n)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check whether a number
// is perfect square or not
static bool isPerfectSquare(int x)
{
    int s = (int) Math.Sqrt(x);
    return (s * s == x);
}
 
// Function to check whether a number
// is fibonacci number or not
static bool isFibonacci(int n)
{
    // If 5*n*n + 4 or 5*n*n - 5 is a
    // perfect square, then the number
    // is Fibonacci
    return isPerfectSquare(5 * n * n + 4) ||
           isPerfectSquare(5 * n * n - 4);
}
 
// Function to count the subarray
// with sum fibonacci number
static void fibonacciSubarrays(int []arr, int n)
{
    int count = 0;
 
    // Traverse the array []arr to find
    // the sum of each subarray
    for(int i = 0; i < n; ++i)
    {
        
       // To store the sum
       int sum = 0;
       for(int j = i; j < n; ++j)
       {
          sum += arr[j];
           
          // Check whether sum of subarray
          // between [i, j] is fibonacci
          // or not
          if (isFibonacci(sum))
          {
              ++count;
          }
       }
    }
    Console.Write(count);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 6, 7, 8, 9 };
    int n = arr.Length;
 
    // Function Call
    fibonacciSubarrays(arr, n);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program for the above approach
 
// Function to check whether a number
// is perfect square or not
function isPerfectSquare(x)
{
    var s = parseInt(Math.sqrt(x));
    return (s * s == x);
}
 
// Function to check whether a number
// is fibonacci number or not
function isFibonacci(n)
{
    // If 5*n*n + 4 or 5*n*n - 5 is a
    // perfect square, then the number
    // is Fibonacci
    return isPerfectSquare(5 * n * n + 4)
           || isPerfectSquare(5 * n * n - 4);
}
 
// Function to count the subarray with
// sum fibonacci number
function fibonacciSubarrays(arr, n)
{
    var count = 0;
 
    // Traverse the array arr[] to find
    // the sum of each subarray
    for (var i = 0; i < n; ++i) {
 
        // To store the sum
        var sum = 0;
 
        for (var j = i; j < n; ++j) {
            sum += arr[j];
 
            // Check whether sum of subarray
            // between [i, j] is fibonacci
            // or not
            if (isFibonacci(sum)) {
                ++count;
            }
        }
    }
 
    document.write( count);
}
 
// Driver Code
 
var arr = [ 6, 7, 8, 9 ];
var n = arr.length;
 
// Function Call
fibonacciSubarrays(arr, n);
 

输出：

3

时间复杂度： O(N ² ) ，其中 N 是元素的数量。

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