通过从较小的元素中减去较大的元素来最小化数组总和

给定一个包含 n 个元素的数组，我们需要最小化数组总和。我们可以多次执行以下操作。

从数组中选择任意两个元素，比如 A 和 B，其中 A > B，然后从 A 中减去 B。

例子：

Input : 1
        arr[] = 1
Output : 1
There is no need to apply the above operation
because there is only a single number that is 1.
Hence, the minimum sum of the array is 1.

Input : 3
        arr[] = 2 4 6
Output : 6
Perform the following operations:-
subtract 2 from 4 then the array becomes 2 2 6
subtract 2 (at 2nd position) from 6 the array 
becomes 2 2 4
subtract 2 (at 2nd position) from 4 the array
becomes 2 2 2
Now the sum of the array will be 6.

方法：应用所有操作后，给定数组中的所有值都将相等，否则我们仍然可以选择两个数字 A 和 B 使得 A > B 并且可以进一步减少总和。应用所有操作后每个元素的值将等于数组的gcd，即ans。
因此，最小可能总和将等于 n * ans。
以下是上述方法的实现：

C++

// CPP program to find the minimum
// sum of the array.
#include 
using namespace std;
 
// returns gcd of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;   
    return gcd(b, a % b);
}
 
// returns the gcd of the array.
int gcdofArray(int arr[], int n)
{
    int ans = arr[0];
    for (int i = 1; i < n; i++)
        ans = gcd(ans, arr[i]);   
    return ans;
}
 
// Driver Function
int main()
{
    int arr[] = { 2, 4, 6 }, n;
    n = sizeof(arr) / sizeof(arr[0]);
    cout << n * gcdofArray(arr, n)
         << endl;
    return 0;
}

Java

// java program to find the minimum
// sum of the array.
import java.io.*;
 
class GFG {
     
    // returns gcd of two numbers
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
     
    // returns the gcd of the array.
    static int gcdofArray(int arr[], int n)
    {
        int ans = arr[0];
        for (int i = 1; i < n; i++)
            ans = gcd(ans, arr[i]);
        return ans;
    }
     
    // Driver Function
    public static void main (String[] args)
    {
        int arr[] = { 2, 4, 6 }, n;
        n = arr.length;
        System.out.println( n * gcdofArray(arr, n)) ;
         
    }
}
 
// This code is contributed by vt_m

Python3

# Python3 code to find the minimum
# sum of the array.
 
# returns gcd of two numbers
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)
     
# returns the gcd of the array.
def gcdofArray(arr, n):
    ans = arr[0]
    for i in range(n):
        ans = gcd(ans, arr[i])
    return ans
     
# Driver Code
arr = [ 2, 4, 6 ]
n = len(arr)
print(n * gcdofArray(arr, n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to find the minimum
// sum of the array.
using System;
 
class GFG
{
     
    // returns gcd of two numbers
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
     
    // returns the gcd of the array.
    static int gcdofArray(int []arr, int n)
    {
        int ans = arr[0];
        for (int i = 1; i < n; i++)
            ans = gcd(ans, arr[i]);
        return ans;
    }
     
    // Driver code
    public static void Main ()
    {
        int []arr = {2, 4, 6};
        int n;
        n = arr.Length;
        Console.WriteLine(n * gcdofArray(arr, n)) ;
         
    }
}
 
// This code is contributed by vt_m.

PHP

Javascript

输出：