生成具有最小总和的数组，可以在P个步骤中将其删除

给定两个数字N和P。任务是生成一个包含所有正元素的数组，并且在一个操作中，您可以在数组中选择一个最小值，然后从所有数组元素中减去它。如果数组元素变为0，则将其删除。

您必须打印该数组和一个可能的数组的最小和，以便在精确执行P个步骤后，该数组将消失。

例子：

Input : N = 4, P = 2
Output :
The Minimum Possible Sum is: 5
The Array Elements are: 1 2 1 1
Explanation:
The array can be [1, 2, 1, 1] after 1st step it becomes [0, 1, 0, 0] and it becomes [1] and after step 2 it will be vanished.Thus the sum is 5 and it is minimum possible value.

Input : N = 3 , P = 1
Output :
The Minimum Possible Sum is: 3
The Array Elements are: 1 1 1

方法：可以通过遵循贪婪的方法来解决问题。首先，我们将放置第一个自然数，对于其余(N – P)个数，我们将其填充为1，因为我们必须使总和最小化。

因此，总和为P *(P + 1)/ 2 +(N – P) 。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// Function to find the required array
void findArray(int N, int P)
{
    // calculating minimum possible sum
    int ans = (P * (P + 1)) / 2 + (N - P);
  
    // Array
    int arr[N + 1];
  
    // place firts P natural elements
    for (int i = 1; i <= P; i++)
        arr[i] = i;
  
    // Fill rest of the elements with 1
    for (int i = P + 1; i <= N; i++)
        arr[i] = 1;
  
    cout << "The Minimum Possible Sum is: " << ans << "\n";
    cout << "The Array Elements are: \n";
  
    for (int i = 1; i <= N; i++)
        cout << arr[i] << ' ';
}
  
// Driver Code
int main()
{
    int N = 5, P = 3;
  
    findArray(N, P);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
      
    // Function to find the required array
    static void findArray(int N, int P) 
    {
        // calculating minimum possible sum
        int ans = (P * (P + 1)) / 2 + (N - P);
  
        // Array
        int arr[] = new int[N + 1];
  
        // place firts P natural elements
        for (int i = 1; i <= P; i++) 
        {
            arr[i] = i;
        }
  
        // Fill rest of the elements with 1
        for (int i = P + 1; i <= N; i++) 
        {
            arr[i] = 1;
        }
  
        System.out.print("The Minimum Possible Sum is: " +
                                                ans + "\n");
        System.out.print("The Array Elements are: \n");
  
        for (int i = 1; i <= N; i++) 
        {
            System.out.print(arr[i] + " ");
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int N = 5, P = 3;
  
        findArray(N, P);
    }
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of above approach
  
# Function to find the required array
def findArray(N, P):
      
    # calculating minimum possible sum
    ans = (P * (P + 1)) // 2 + (N - P);
  
    # Array
    arr = [0] * (N + 1);
  
    # place firts P natural elements
    for i in range(1, P + 1):
        arr[i] = i;
  
    # Fill rest of the elements with 1
    for i in range(P + 1, N + 1):
        arr[i] = 1;
  
    print("The Minimum Possible Sum is: ", ans);
    print("The Array Elements are: ");
  
    for i in range(1, N + 1):
        print(arr[i], end = " ");
  
# Driver Code
N = 5;
P = 3;
findArray(N, P);
  
# This code is contributed by mits

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to find the required array
    static void findArray(int N, int P) 
    {
        // calculating minimum possible sum
        int ans = (P * (P + 1)) / 2 + (N - P);
  
        // Array
        int []arr = new int[N + 1];
  
        // place firts P natural elements
        for (int i = 1; i <= P; i++) 
        {
            arr[i] = i;
        }
  
        // Fill rest of the elements with 1
        for (int i = P + 1; i <= N; i++) 
        {
            arr[i] = 1;
        }
  
        Console.Write("The Minimum Possible Sum is: " +
                                                ans + "\n");
        Console.Write("The Array Elements are: \n");
  
        for (int i = 1; i <= N; i++) 
        {
            Console.Write(arr[i] + " ");
        }
    }
  
    // Driver Code
    public static void Main()
    {
        int N = 5, P = 3;
  
        findArray(N, P);
    }
}
  
/* This code contributed by PrinciRaj1992 */

输出：

The Minimum Possible Sum is: 8
The Array Elements are: 
1 2 3 1 1

时间复杂度： O(N)