查找给定条件下的最大可能总和

给定大小为N的数组arr [] ，任务是通过遵循给定条件来找到数组的最大可能和：

• 在每一步骤中，只能使用一个元素来增加总和。
• 如果从数组中选择了某个元素K，则数组中的剩余数字将减少1。
• 数组中的元素不能减少到0以上。

例子：

Input: arr = {6, 6, 6}
Output: 15
Explanation:
Initially, the total sum is 0. Since all the elements are equal, any one element is chosen.
Sum after choosing the first six = 6. Remaining elements = {5, 5}.
Sum after choosing the five = 11. Remaining elements = {4}.
Finally, four is chosen making the maximum sum 15.

Input: arr = {0, 1, 0}
Output: 1
Explanation:
Initially, the total sum is 0. There is only one number that can be chosen in the array because rest of the elements are 0.
Therefore, the maximum sum = 1.

方法：由于所有其他元素的值都明显减少了1，因此，如果在每次迭代中选择最大元素，我们将获得最大和。因此，为了做到这一点，使用了分类。

• 想法是按降序对数组的元素进行排序。
• 现在，由于我们可以在每次迭代中选择最大值，因此我们可以在索引“ i”处将元素K的值计算为(K – i) 。
• 如果在任何索引处元素的值变为0，则该索引之后的所有元素将为0。

下面是上述方法的实现：

C++

// C++ program to find the maximum 
// possible Sum for the given conditions
#include
using namespace std;
  
// Function to find the maximum 
// possible sum for the 
// given conditions
int maxProfit(int arr[], int n)
{ 
      
    // Sorting the array
    sort(arr, arr + n, greater());
  
    // Variable to store the answer
    int ans = 0;
  
    // Iterating through the array
    for(int i = 0; i < n; i++)
    {
         
       // If the value is greater than 0
       if ((arr[i] - (1 * i)) > 0)
           ans += (arr[i] - (1 * i));
         
       // If the value becomes 0 
       // then break the loop because 
       // all the weights after this 
       // index will be 0
       if ((arr[i] - (1 * i)) == 0)
           break;
    }
      
    // Print profit
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = {6, 6, 6};
    int n = sizeof(arr) / sizeof(arr[0]);
      
    cout << maxProfit(arr, n);
    return 0;
}
  
// This code is contributed by ankitkumar34

Java

// Java program to find the maximum 
// possible Sum for the given conditions 
import java.util.Arrays; 
import java.util.Collections; 
  
public class GFG{
  
    // Function to find the maximum 
    // possible sum for the 
    // given conditions 
    static int maxProfit(Integer [] arr)
    { 
          
        // Sorting the array 
        Arrays.sort(arr, Collections.reverseOrder());
      
        // Variable to store the answer 
        int ans = 0;
      
        // Iterating through the array 
        for(int i = 0; i < arr.length; i++)
        {
      
           // If the value is greater than 0 
           if ((arr[i] - (1 * i)) > 0)
               ans += (arr[i] - (1 * i));
      
           // If the value becomes 0 
           // then break the loop because 
           // all the weights after this 
           // index will be 0 
           if ((arr[i] - (1 * i)) == 0) 
               break;
        }
          
        // Print profit 
        return ans;
    }
      
// Driver code 
public static void main(String[] args)
{
    Integer arr[] = { 6, 6, 6 };
    System.out.println(maxProfit(arr));
}
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 program to find the maximum 
# possible Sum for the given conditions
  
# Function to find the maximum 
# possible sum for the 
# given conditions
def maxProfit(arr):
      
    # Sorting the array
    arr.sort(reverse = True)
  
    # Variable to store the answer
    ans = 0
  
    # Iterating through the array
    for i in range(len(arr)):
  
        # If the value is greater than 0
        if (arr[i]-(1 * i))>0:
            ans+=(arr[i]-(1 * i))
  
        # If the value becomes 0 
        # then break the loop because 
        # all the weights after this 
        # index will be 0
        if (arr[i]-(1 * i))== 0:
            break
  
    # print profit
    return ans    
  
# Driver code
if __name__ == "__main__":
   
    arr = [6, 6, 6]
  
    print(maxProfit(arr))

C#

// C# program to find the maximum 
// possible Sum for the given conditions 
using System;
  
class GFG{
  
// Function to find the maximum 
// possible sum for the 
// given conditions 
static int maxProfit(int[] arr)
{ 
          
    // Sorting the array 
    Array.Sort(arr);
    Array.Reverse(arr);
      
    // Variable to store the answer 
    int ans = 0;
      
    // Iterating through the array 
    for(int i = 0; i < arr.Length; i++)
    {
         
       // If the value is greater than 0 
       if ((arr[i] - (1 * i)) > 0)
           ans += (arr[i] - (1 * i));
         
       // If the value becomes 0 
       // then break the loop because 
       // all the weights after this 
       // index will be 0 
       if ((arr[i] - (1 * i)) == 0) 
           break;
    }
          
    // Print profit 
    return ans;
}
      
// Driver code 
static public void Main ()
{
    int[] arr = { 6, 6, 6 };
      
    Console.Write(maxProfit(arr));
}
}
  
// This code is contributed by Shubhamsingh10

输出：

15