从较大的元素减去较小的元素后，数组元素的最小总和

给定数组arr ，任务是在应用以下操作后找到数组元素的最小和：
对于数组中的任何一对，如果a [i]> a [j]，则a [i] = a [i] – a [j] 。

例子：

Input: arr[] = {1, 2, 3}
Output: 3
modified array will be {1, 1, 1}

Input: a = {2, 4, 6}
Output: 6
modified array will be {2, 2, 2}

方法：在这里观察到，在每次操作之后，所有元素的GCD将保持不变。因此，最后，在应用了给定的操作之后，每个元素都将等于数组中所有元素的gcd。
因此，最终答案将是(n * gcd) 。

下面是上述方法的实现：

C++

// CPP program to Find the minimum sum
// of given array after applying given operation.
#include 
using namespace std;
  
// Function to Find the minimum sum
// of given array after applying given operation.
int MinSum(int a[], int n)
{
    // to store final gcd value
    int gcd = a[0];
  
    // get gcd of the whole array
    for (int i = 1; i < n; i++)
        gcd = __gcd(a[i], gcd);
  
    return n * gcd;
}
  
// Driver code
int main()
{
  
    int a[] = { 20, 14, 6, 8, 15 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    // function call
    cout << MinSum(a, n);
  
    return 0;
}

Java

// Java program to Find the minimum sum
// of given array after applying given operation.
  
import java.io.*;
  
class GFG {
     
// Recursive function to return gcd of a and b 
static int __gcd(int a, int b) 
{ 
    // Everything divides 0  
    if (a == 0) 
       return b; 
    if (b == 0) 
       return a; 
     
    // base case 
    if (a == b) 
        return a; 
     
    // a is greater 
    if (a > b) 
        return __gcd(a-b, b); 
    return __gcd(a, b-a); 
} 
// Function to Find the minimum sum
// of given array after applying given operation.
static int MinSum(int []a, int n)
{
    // to store final gcd value
    int gcd = a[0];
  
    // get gcd of the whole array
    for (int i = 1; i < n; i++)
        gcd = __gcd(a[i], gcd);
  
    return n * gcd;
}
  
// Driver code
  
    public static void main (String[] args) {
            int a[] = { 20, 14, 6, 8, 15 };
  
    int n = a.length;
  
    // function call
    System.out.println(MinSum(a, n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 program to Find the minimum 
# sum of given array after applying 
# given operation.
import math
  
# Function to Find the minimum sum
# of given array after applying 
# given operation.
def MinSum(a, n):
  
    # to store final gcd value
    gcd = a[0]
  
    # get gcd of the whole array
    for i in range(1, n):
        gcd = math.gcd(a[i], gcd)
  
    return n * gcd
  
# Driver code
if __name__ == "__main__":
  
    a = [20, 14, 6, 8, 15 ]
  
    n = len(a)
  
    # function call
    print(MinSum(a, n))
  
# This code is contributed by ita_c

C#

// C# program to Find the minimum sum
// of given array after applying given operation.
  
using System;
class GFG {
     
    // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0) 
           return b; 
        if (b == 0) 
           return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    } 
      
    // Function to Find the minimum sum
    // of given array after applying given operation.
    static int MinSum(int []a, int n)
    {
        // to store final gcd value
        int gcd = a[0];
      
        // get gcd of the whole array
        for (int i = 1; i < n; i++)
            gcd = __gcd(a[i], gcd);
      
        return n * gcd;
    }
      
      
    // Driver Program to test above function
    static void Main()
    {
        int []a = { 20, 14, 6, 8, 15 };
        int n = a.Length;
        Console.WriteLine(MinSum(a, n));
    }
      
    // This code is contributed by Ryuga.
}

输出：

5