生成最小和的数组，可以通过 P 步
删除
原文:https://www . geesforgeks . org/generate-array-with-minimum-sum-哪些可以在 p-steps 中删除/
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，在一个操作中，您可以选择数组中的一个最小数量，并从所有数组元素中减去它。如果数组元素变为 0，那么您将移除它。 你必须打印该数组和一个可能数组的最小可能和，这样在精确应用 P 步后，该数组将消失。
示例:
输入: N = 4，P = 2 输出: 最小可能和为:5 数组元素为:1 2 1 1 说明: 数组可以是【1，2，1，1】第一步后变成【0，1，0，0】变成【1】，第二步后消失。因此总和是 5，它是最小可能值。 输入 : N = 3，P = 1 输出 : 最小可能和为:3 数组元素为:1 1 1
方法:遵循贪婪的方法可以解决问题。首先，我们将首先放置 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。 所以总和将是P *(P+1)/2+(N–P)。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
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 first 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 语言(一种计算机语言，尤用于创建网站)
// 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 first 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
Python 3
# 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 first 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 first 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 */
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to find the required array
function findArray($N, $P)
{
// calculating minimum possible sum
$ans = ($P * ($P + 1)) / 2 + ($N - $P);
// Array
$arr[$N + 1] = array();
// place first P natural elements
for ($i = 1; $i <= $P; $i++)
$arr[$i] = $i;
// Fill rest of the elements with 1
for ($i = $P + 1; $i <= $N; $i++)
$arr[$i] = 1;
echo "The Minimum Possible Sum is: ",
$ans, "\n";
echo "The Array Elements are: \n";
for ($i = 1; $i <= $N; $i++)
echo $arr[$i], ' ';
}
// Driver Code
$N = 5;
$P = 3;
findArray($N, $P);
// This code is contributed by ajit.
?>
java 描述语言
<script>
// 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 first 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);
</script>
Output:
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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