对于可被K整除的数组元素，可以通过将arr [i] / K附加到数组末尾K次来实现数组元素的总和

给定一个由N个整数和整数K组成的数组arr [] ，任务是通过遍历该数组并在数组末尾添加k次arr [i] / K来找到可能的数组元素之和，如果arr [i]可被K整除。否则，停止遍历。

例子：

Input: arr[] = {4, 6, 8, 2}, K = 2
Output: 44
Explanation:
Following operations are performed:

1. For arr[0](= 4): arr[0](= 4) is divisible by 2, therefore append 4/2 = 2, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2}.
2. For arr[1](= 6): arr[1](= 6) is divisible by 2, therefore append 6/2 = 3, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3}.
3. For arr[2](= 8): arr[2](= 8) is divisible by 2, therefore append 8/2 = 4, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4}.
4. For arr[3](= 2): arr[3](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1}.
5. For arr[4](= 2): arr[4](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1}.
6. For arr[5](= 2): arr[5](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1}.

After completing the above steps, the sum of the array element is = 4 + 6 + 8 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 1 + 1 + 1 + 1 + 1 + 1 = 44.

Input: arr[] = {4, 6, 8, 9}, K = 2
Output: 45

天真的方法：最简单的方法是解决给定的问题，即遍历给定的数组并在数组末尾添加K次(arr [i] / K)值。完成上述步骤后，打印数组元素的总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
    // Stores the sum of the array
    int sum = 0;
 
    vector v;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        v.push_back(arr[i]);
    }
 
    // Traverse the vector
    for (int i = 0;
         i < v.size(); i++) {
 
        // If v[i] is divisible by K
        if (v[i] % K == 0) {
 
            long long x = v[i] / K;
 
            // Iterate over the range
            // [0, K]
            for (int j = 0; j < K; j++) {
                // Update v
                v.push_back(x);
            }
        }
        // Otherwise
        else
            break;
    }
 
    // Traverse the vector v
    for (int i = 0; i < v.size(); i++)
        sum = sum + v[i];
 
    // Return the sum of the updated array
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sum(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{
     
    // Stores the sum of the array
    int sum = 0;
 
    ArrayList v = new ArrayList<>();
 
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
        v.add(arr[i]);
    }
 
    // Traverse the vector
    for(int i = 0; i < v.size(); i++)
    {
         
        // If v[i] is divisible by K
        if (v.get(i) % K == 0)
        {
            int x = v.get(i) / K;
 
            // Iterate over the range
            // [0, K]
            for(int j = 0; j < K; j++)
            {
                 
                // Update v
                v.add(x);
            }
        }
         
        // Otherwise
        else
            break;
    }
 
    // Traverse the vector v
    for(int i = 0; i < v.size(); i++)
        sum = sum + v.get(i);
 
    // Return the sum of the updated array
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = arr.length;
     
    System.out.println(sum(arr, N, K));
}
}
 
// This code is contributed by Kingash

输出：

44

时间复杂度： O(N * K * log N)
辅助空间： O(M)，M是数组的最大元素。

高效方法：还可以基于以下观察来优化上述方法：

• 如果arr [i]可被K整除，则将arr [i] / K相加， K次将总和增加arr [i] 。
• 因此，想法是在向量的末尾仅将arr [i] / K推送一次。

请按照以下步骤解决问题：

• 初始化一个变量，例如sum为0 ，该变量存储所有数组元素array A []的和。
• 初始化一个数组，比如说A [] ，并将所有数组元素arr []存储在A []中。
• 初始化一个变量，将其标记为0 ，以存储是否要在数组末尾添加元素。
• 遍历数组A []并执行以下步骤：
  • 如果值标志为0并且A [i]可被K整除，则在V的末尾按下A [i] 。
  • 否则，将flag的值更新为1 。
  • 将总和的值增加V [i％N] 。
• 完成上述步骤后，打印总和的值作为结果总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
    // Stores the sum of the array
    int sum = 0;
 
    // Stores the array elements
    vector v;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        v.push_back(arr[i]);
    }
 
    // Stores if the operation
    // should be formed or not
    bool flag = 0;
 
    // Traverse the vector V
    for (int i = 0; i < v.size(); i++) {
 
        // If flag is false and if
        // v[i] is divisible by K
        if (!flag && v[i] % K == 0)
            v.push_back(v[i] / K);
 
        // Otherwise, set flag as true
        else {
            flag = 1;
        }
 
        // Increment the sum by v[i % N]
        sum = sum + v[i % N];
    }
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sum(arr, N, K);
 
    return 0;
}

输出：

44

时间复杂度： O(N * log N)
辅助空间： O(N * log N)