查找执行给定操作后获得的最终编号

给定正整数不同的数组arr [] ，任务是查找通过对数组的元素执行以下操作而获得的最终数字：
运算：取两个不相等的数字，并用它们之间的差替换较大的数字，直到所有数字相等为止。
例子：

Input: arr[] = {5, 2, 3} 
Output: 1 
5 – 3 = 2, arr[] = {2, 2, 3} 
3 – 2 = 1, arr[] = {2, 2, 1} 
2 – 1 = 1, arr[] = {2, 1, 1} 
2 – 1 = 1, arr[] = {1, 1, 1}
Input: arr[] = {3, 9, 6, 36} 
Output: 3 
 

天真的方法：由于最终答案始终是不同的，因此可以对数组进行排序，然后用两个最大元素的差值替换最大的项，然后重复该过程，直到所有数字相等为止。
高效的方法：从欧几里得算法中得知， gcd(a，b)= gcd(a – b，b) 。这可以扩展为gcd(A ₁ ，A ₂ ，A ₃ ，…，A _n )= gcd(A ₁ -A ₂ ，A ₂ ，A ₃ ，…，A _n ) 。
同样，假设在应用了给定的运算之后，最终获得的数字为K。因此，从扩展算法中，可以说gcd(A ₁ ，A ₂ ，A ₃ ，…，A _n )= gcd(K ，K，…，n次) 。由于gcd(K，K，…，n次)= K ，可以找到给定问题的解决方案
通过查找数组所有元素的gcd。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the final number
// obtained after performing the
// given operation
int finalNum(int arr[], int n)
{
 
    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++) {
        result = __gcd(result, arr[i]);
    }
    return result;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 9, 6, 36 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << finalNum(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the final number
// obtained after performing the
// given operation
static int finalNum(int arr[], int n)
{
 
    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++)
    {
        result = __gcd(result, arr[i]);
    }
    return result;
}
 
static int __gcd(int a, int b)
{
    return b == 0? a:__gcd(b, a % b);    
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 9, 6, 36 };
    int n = arr.length;
 
    System.out.print(finalNum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
from math import gcd as __gcd
 
# Function to return the final number
# obtained after performing the
# given operation
def finalNum(arr, n):
 
    # Find the gcd of the array elements
    result = arr[0]
    for i in arr:
        result = __gcd(result, i)
    return result
 
# Driver code
arr = [3, 9, 6, 36]
n = len(arr)
 
print(finalNum(arr, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the readonly number
// obtained after performing the
// given operation
static int finalNum(int []arr, int n)
{
 
    // Find the gcd of the array elements
    int result = 0;
    for (int i = 0; i < n; i++)
    {
        result = __gcd(result, arr[i]);
    }
    return result;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 9, 6, 36 };
    int n = arr.Length;
 
    Console.Write(finalNum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript implementation of the approach   
// Function to return the final number
    // obtained after performing the
    // given operation
    function finalNum(arr , n) {
 
        // Find the gcd of the array elements
        var result = 0;
        for (i = 0; i < n; i++) {
            result = __gcd(result, arr[i]);
        }
        return result;
    }
 
    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Driver code
     
        var arr = [ 3, 9, 6, 36 ];
        var n = arr.length;
 
        document.write(finalNum(arr, n));
 
// This code contributed by aashish1995

输出：

3

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