根据给定条件最小化数组的总和

给定整数数组A。任务是使用以下规则使数组元素的总和最小化：
选择两个索引i和j以及任意整数x ，以使x为A [i]的除数，然后将它们更改为A [i] = A [i] / x和A [j] = A [j] * X。

例子：

Input: A = { 1, 2, 3, 4, 5 }
Output: 14

Divide A[3] by 2 then
A[3] = 4/2 = 2,

Multiply A[0] by 2 then
A[0] = 1*2 = 2

Updated array A = { 2, 2, 3, 2, 5 }
Hence sum = 14

Input: A = { 2, 4, 2, 3, 7 }
Output: 18

方法：如果将任何数字除以x，则最好将x与数组中存在的最小数字相乘。

这个想法是获得数组的最小值，找到特定元素的除数，并不断检查总和减少了多少。

下面是上述方法的实现：

C++

// C++ implementation
#include 
using namespace std;
  
// Function to return the minimum sum
void findMin(int arr[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // sort the array to find the
    // minimum element
    sort(arr, arr + n);
  
    int min = arr[0];
    int max = 0;
  
    for (int i = n - 1; i >= 1; i--) {
        int num = arr[i];
        int total = num + min;
        int j;
  
        // finding the number to
        // divide
        for (j = 2; j <= num; j++) {
            if (num % j == 0) {
                int d = j;
                int now = (num / d)
                          + (min * d);
  
                // Checking to what
                // instance the sum
                // has decreased
                int reduce = total - now;
  
                // getting the max
                // difference
                if (reduce > max)
                    max = reduce;
            }
        }
    }
    cout << (sum - max);
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findMin(arr, n);
}

Java

// Java implementation of the above approach
import java.util.*;
  
class GFG 
{
      
    // Function to return the minimum sum 
    static void findMin(int arr[], int n) 
    { 
        int sum = 0; 
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
      
        // sort the array to find the 
        // minimum element 
        Arrays.sort(arr); 
      
        int min = arr[0]; 
        int max = 0; 
      
        for (int i = n - 1; i >= 1; i--) 
        { 
            int num = arr[i]; 
            int total = num + min; 
            int j; 
      
            // finding the number to 
            // divide 
            for (j = 2; j <= num; j++) 
            { 
                if (num % j == 0) 
                { 
                    int d = j; 
                    int now = (num / d) + 
                              (min * d); 
      
                    // Checking to what 
                    // instance the sum 
                    // has decreased 
                    int reduce = total - now; 
      
                    // getting the max 
                    // difference 
                    if (reduce > max) 
                        max = reduce; 
                } 
            } 
        } 
        System.out.println(sum - max); 
    } 
      
    // Driver Code 
    public static void main (String[] args) 
    { 
        int arr[] = { 1, 2, 3, 4, 5 }; 
        int n = arr.length; 
        findMin(arr, n); 
    }
}
  
// This code is contributed by AnkitRai01

Python3

# Function to return the minimum sum 
def findMin(arr, n):
    sum = 0
    for i in range(0, n): 
        sum = sum + arr[i]
  
    # sort the array to find the 
    # minimum element 
    arr.sort()
  
    min = arr[0]
    max = 0
  
    for i in range(n - 1, 0, -1): 
        num = arr[i]
        total = num + min
  
        # finding the number to 
        # divide 
        for j in range(2, num + 1): 
            if(num % j == 0):
                d = j
                now = (num // d) + (min * d)
  
                # Checking to what 
                # instance the sum 
                # has decreased 
                reduce = total - now
  
                # getting the max 
                # difference 
                if(reduce > max):
                    max = reduce
  
    print(sum - max)
  
# Driver Code 
arr = [1, 2, 3, 4, 5 ]
n = len(arr)
findMin(arr, n)
  
# This code is contributed by Sanjit_Prasad

C#

// C# implementation of the above approach
using System;
      
class GFG 
{
      
    // Function to return the minimum sum 
    static void findMin(int []arr, int n) 
    { 
        int sum = 0; 
        for (int i = 0; i < n; i++) 
            sum += arr[i]; 
      
        // sort the array to find the 
        // minimum element 
        Array.Sort(arr); 
      
        int min = arr[0]; 
        int max = 0; 
      
        for (int i = n - 1; i >= 1; i--) 
        { 
            int num = arr[i]; 
            int total = num + min; 
            int j; 
      
            // finding the number to 
            // divide 
            for (j = 2; j <= num; j++) 
            { 
                if (num % j == 0) 
                { 
                    int d = j; 
                    int now = (num / d) + 
                              (min * d); 
      
                    // Checking to what 
                    // instance the sum 
                    // has decreased 
                    int reduce = total - now; 
      
                    // getting the max 
                    // difference 
                    if (reduce > max) 
                        max = reduce; 
                } 
            } 
        } 
        Console.WriteLine(sum - max); 
    } 
      
    // Driver Code 
    public static void Main (String[] args) 
    { 
        int []arr = { 1, 2, 3, 4, 5 }; 
        int n = arr.Length; 
        findMin(arr, n); 
    }
}
  
// This code is contributed by PrinciRaj1992

输出：

14