检查数组是否代表斐波那契数列

给定一个由N 个整数组成的数组arr[] ，任务是检查是否可以使用所有数组元素形成斐波那契数列。如果可能，请打印“是”。否则，打印“否”。
例子：

Input: arr[] = { 8, 3, 5, 13 } 
Output: Yes 
Explanation: 
Rearrange given array as {3, 5, 8, 13} and these numbers form Fibonacci series.
Input: arr[] = { 2, 3, 5, 11 } 
Output: No 
Explanation: 
The given array elements do not form a Fibonacci series. 
 

方法：
为了解决上面提到的问题，主要思想是对给定的数组进行排序。排序后，检查每个元素是否等于前 2 个元素的总和。如果是，则数组元素形成斐波那契数列。
下面是上述方法的实现：

C++

// C++ program to check if the
// elements of a given array
// can form a Fibonacci Series
 
#include 
using namespace std;
 
// Returns true if a permutation
// of arr[0..n-1] can form a
// Fibonacci Series
bool checkIsFibonacci(int arr[], int n)
{
    if (n == 1 || n == 2)
        return true;
 
    // Sort array
    sort(arr, arr + n);
 
    // After sorting, check if every
    // element is equal to the
    // sum of previous 2 elements
 
    for (int i = 2; i < n; i++)
        if ((arr[i - 1] + arr[i - 2])
            != arr[i])
            return false;
 
    return true;
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 3, 5, 13 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (checkIsFibonacci(arr, n))
        cout << "Yes" << endl;
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to check if the elements of
// a given array can form a Fibonacci Series
import java. util. Arrays;
 
class GFG{
     
// Returns true if a permutation
// of arr[0..n-1] can form a
// Fibonacci Series
public static boolean checkIsFibonacci(int arr[],
                                       int n)
{
    if (n == 1 || n == 2)
        return true;
     
    // Sort array
    Arrays.sort(arr);
     
    // After sorting, check if every
    // element is equal to the sum
    // of previous 2 elements
    for(int i = 2; i < n; i++)
    {
       if ((arr[i - 1] + arr[i - 2]) != arr[i])
           return false;
    }
    return true;
}
     
// Driver code
public static void main(String[] args)
{
    int arr[] = { 8, 3, 5, 13 };
    int n = arr.length;
     
    if (checkIsFibonacci(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to check if the
# elements of a given array
# can form a Fibonacci Series
 
# Returns true if a permutation
# of arr[0..n-1] can form a
# Fibonacci Series
def checkIsFibonacci(arr, n) :
 
    if (n == 1 or n == 2) :
        return True;
 
    # Sort array
    arr.sort()
 
    # After sorting, check if every
    # element is equal to the
    # sum of previous 2 elements
 
    for i in range(2, n) :
        if ((arr[i - 1] +
             arr[i - 2])!= arr[i]) :
            return False;
 
    return True;
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 8, 3, 5, 13 ];
    n = len(arr);
 
    if (checkIsFibonacci(arr, n)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

C#

// C# program to check if the elements of
// a given array can form a fibonacci series
using System;
 
class GFG{
     
// Returns true if a permutation
// of arr[0..n-1] can form a
// fibonacci series
public static bool checkIsFibonacci(int []arr,
                                    int n)
{
    if (n == 1 || n == 2)
        return true;
         
    // Sort array
    Array.Sort(arr);
         
    // After sorting, check if every
    // element is equal to the sum
    // of previous 2 elements
    for(int i = 2; i < n; i++)
    {
       if ((arr[i - 1] + arr[i - 2]) != arr[i])
           return false;
    }
    return true;
}
         
// Driver code
public static void Main(string[] args)
{
    int []arr = { 8, 3, 5, 13 };
    int n = arr.Length;
         
    if (checkIsFibonacci(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript program to check if the elements of
// a given array can form a Fibonacci Series
 
    // Returns true if a permutation
    // of arr[0..n-1] can form a
    // Fibonacci Series
    function checkIsFibonacci(arr , n)
    {
        if (n == 1 || n == 2)
            return true;
 
        // Sort array
        arr.sort((a, b) => a - b);
 
        // After sorting, check if every
        // element is equal to the sum
        // of previous 2 elements
        for (i = 2; i < n; i++) {
            if ((arr[i - 1] + arr[i - 2]) != arr[i])
                return false;
        }
        return true;
    }
 
    // Driver code
     
        var arr = [ 8, 3, 5, 13 ];
        var n = arr.length;
 
        if (checkIsFibonacci(arr, n))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by umadevi9616
 

输出：

Yes

时间复杂度： O(N Log N)

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