生成可以在 P 步中删除的最小和的数组

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，在一次操作中，您可以选择数组中的一个最小数，然后从所有数组元素中减去它。如果数组元素变为 0，那么您将删除它。
您必须打印数组和一个可能数组的最小可能总和，以便在精确应用 P 步后数组将消失。

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：这个问题可以通过遵循贪婪的方法来解决。首先，我们将放置前 P 个自然数，对于其余 (N – P) 个位置，我们将用 1 填充它，因为我们必须最小化总和。
所以总和将为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 */

Javascript

    // Javascript implementation of the approach
     
    // Function to find the required array
    function findArray(N, P)
    {
        // calculating minimum possible sum
        let ans = parseInt((P * (P + 1)) / 2, 10) + (N - P);
   
        // Array
        let arr = new Array(N + 1);
   
        // place firts P natural elements
        for (let i = 1; i <= P; i++)
        {
            arr[i] = i;
        }
   
        // Fill rest of the elements with 1
        for (let i = P + 1; i <= N; i++)
        {
            arr[i] = 1;
        }
   
        document.write("The Minimum Possible Sum is: " +
                                                ans + "</br>");
        document.write("The Array Elements are: " + "</br>");
   
        for (let i = 1; i <= N; i++)
        {
            document.write(arr[i] + " ");
        }
    }
     
    let N = 5, P = 3;
   
      findArray(N, P);
         

输出：

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

时间复杂度： O(N)

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