最大化所选数组元素的总和，其值最多为 M

给定一个包含N个正数的数组arr[]和一个整数M 。任务是在arr[i] ≤ M时通过添加数组元素来最大化M的值。

注意：任何数组元素最多可以添加一次。

例子：

Input: arr[] = {3, 9, 19, 5, 21}, M = 10
Output: 67
Explanation: One way to getthe value is
M > 3; 3 is added to M and it becomes 10+3 = 13
M > 9; 9 is added to M and it becomes 13+9 = 22
M > 19; 19 is added to M and it becomes 22+19 = 41
M > 5; 5 is added to M and it becomes 41+5 = 46
M > 21; 21 is added to M and it becomes 46+21 = 67
Thus, M = 67 at the end.

Input: arr[] = {2, 13, 4, 19}, M = 6
Output: 12
Explanation: One way to get the value is
M > 4; 4 is added to M and it becomes 6+4 = 10
M > 2; 2 is added to M and it becomes 10+2 = 12
No other value in the array is smaller or equal to M.
Thus, M is 12 at the end.

编程需要懂一点英语

方法：解决方案是基于排序的概念。请按照以下步骤操作：

首先，按升序对数组进行排序。
对于从0 到 N-1的每个索引i ，请执行以下操作：
- 如果M ≥ arr[i] ，则将arr[i]与M相加。
- 如果M< arr[i] ，停止迭代。
返回 M 的最终值作为答案。

下面是上述方法的实现。

C++

// C++ code to implement the approach
#include 
using namespace std;
 
// Function to calculate
// the maximum value of M
// that can be obtained
int IsArrayHungry(int M, vector& arr)
{
    // Sort the array in increasing order.
    sort(arr.begin(), arr.end());
    long long sum = M;
    int N = arr.size();
 
    for (int i = 0; i < N; i++) {
        if (sum >= arr[i])
            sum += arr[i];
        else
            break;
    }
    return sum;
}
 
// Driver code
int main()
{
    vector arr{ 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    cout << res;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG
{
 
  // Function to calculate
  // the maximum value of M
  // that can be obtained
  static int IsArrayHungry(int M, int arr[ ])
  {
 
    // Sort the array in increasing order.
    Arrays.sort(arr);
    int sum = M;
    int N = arr.length;
 
    for (int i = 0; i < N; i++) {
      if (sum >= arr[i])
        sum += arr[i];
      else
        break;
    }
    return sum;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int arr[ ] = { 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    System.out.print(res);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python 3 code to implement the approach
 
# Function to calculate
# the maximum value of M
# that can be obtained
def IsArrayHungry(M,  arr):
 
    # Sort the array in increasing order.
    arr.sort()
    sum = M
    N = len(arr)
 
    for i in range(N):
        if (sum >= arr[i]):
            sum += arr[i]
 
        else:
            break
 
    return sum
 
# Driver code
if __name__ == "__main__":
 
    arr = [3, 9, 19, 5, 21]
    M = 10
    res = IsArrayHungry(M, arr)
    print(res)
 
    # This code is contributed by ukasp.

C#

// C# code to implement above approach
using System;
class GFG
{
 
  // Function to calculate
  // the maximum value of M
  // that can be obtained
  static int IsArrayHungry(int M, int []arr)
  {
 
    // Sort the array in increasing order.
    Array.Sort(arr);
    int sum = M;
    int N = arr.Length;
 
    for (int i = 0; i < N; i++) {
      if (sum >= arr[i])
        sum += arr[i];
      else
        break;
    }
    return sum;
  }
 
  // Driver Code:
  public static void Main()
  {
    int []arr = { 3, 9, 19, 5, 21 };
    int M = 10;
    int res = IsArrayHungry(M, arr);
    Console.WriteLine(res);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

时间复杂度： O(N * logN)
辅助空间： O(1)