在数组中查找划分所有数组元素的元素

给定n个非负整数的数组。在数组中找到这样的元素，所有数组元素都可以被数组整除。
例子 ：

Input : arr[] = {2, 2, 4}
Output : 2

Input : arr[] = {2, 1, 3, 1, 6}
Output : 1

Input: arr[] = {2, 3, 5}
Output : -1

方法是计算整个数组的GCD，然后检查是否存在等于数组GCD的元素。为了计算整个阵列的gcd，我们将使用欧几里得算法。

C++

// CPP program to find such number in the array
// that all array elements are divisible by it
#include 
using namespace std;
 
// Returns gcd of two numbers.
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to return the
// desired number if exists
int findNumber(int arr[], int n)
{
    // Find GCD of array
    int ans = arr[0];
    for (int i = 0; i < n; i++)
        ans = gcd(ans, arr[i]);
 
    // Check if GCD is present in array
    for (int i = 0; i < n; i++)
        if (arr[i] == ans)
            return ans;
 
    return -1;
}
 
// Driver Function
int main()
{
    int arr[] = { 2, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findNumber(arr, n) << endl;
    return 0;
}

Java

// JAVA program to find such number in
// the array that all array elements
// are divisible by it
import java.io.*;
 
class GFG {
 
    // Returns GCD of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Function to return the desired
    // number if exists
    static int findNumber(int arr[], int n)
    {
        // Find GCD of array
        int ans = arr[0];
        for (int i = 0; i < n; i++)
            ans = gcd(ans, arr[i]);
 
        // Check if GCD is present in array
        for (int i = 0; i < n; i++)
            if (arr[i] == ans)
                return ans;
 
        return -1;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 2, 2, 4 };
        int n = arr.length;
        System.out.println(findNumber(arr, n));
    }
}
 
// This code is contributed by Nikita Tiwari

Python3

# Python3 program to find such number
# in the array that all array
# elements are divisible by it
 
# Returns GCD of two numbers
def gcd (a, b) :
    if (a == 0) :
        return b
     
    return gcd (b % a, a)
     
# Function to return the desired
# number if exists
def findNumber (arr, n) :
 
    # Find GCD of array
    ans = arr[0]
    for i in range(0, n) :
        ans = gcd (ans, arr[i])
         
    # Check if GCD is present in array
    for i in range(0, n) :
        if (arr[i] == ans) :
            return ans
     
    return -1
     
# Driver Code
arr = [2, 2, 4];
n = len(arr)
print(findNumber(arr, n))
 
# This code is contributed by Nikita Tiwari

C#

// C# program to find such number in
// the array that all array elements
// are divisible by it
using System;
 
class GFG {
 
    // Returns GCD of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Function to return the desired
    // number if exists
    static int findNumber(int[] arr, int n)
    {
        // Find GCD of array
        int ans = arr[0];
        for (int i = 0; i < n; i++)
            ans = gcd(ans, arr[i]);
 
        // Check if GCD is present in array
        for (int i = 0; i < n; i++)
            if (arr[i] == ans)
                return ans;
 
        return -1;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, 2, 4 };
        int n = arr.Length;
        Console.WriteLine(findNumber(arr, n));
    }
}
 
// This code is contributed by vt_m

Javascript

 
    // Javascript program to find such number in the array
    // that all array elements are divisible by it
     
    // Returns gcd of two numbers.
    function gcd(a, b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Function to return the
    // desired number if exists
    function findNumber(arr, n)
    {
        // Find GCD of array
        let ans = arr[0];
        for (let i = 0; i < n; i++)
            ans = gcd(ans, arr[i]);
 
        // Check if GCD is present in array
        for (let i = 0; i < n; i++)
            if (arr[i] == ans)
                return ans;
 
        return -1;
    }
     
    let arr = [ 2, 2, 4 ];
    let n = arr.length;
    document.write(findNumber(arr, n));
 

输出 ：

2