生成一个具有 K 个正数的数组，这样 arr[i] 要么是 -i 要么是 i 并且数组的总和是正数

给定两个正整数N和K ( 1 ≤ K ≤ N )，任务是构造一个数组arr[] (基于 1 的索引)，这样每个数组元素arr[i]要么是i要么是-i并且数组正好包含K 个正值，使得数组元素的总和为正。如果可以生成多个这样的序列，打印它们中的任何一个。否则，打印“-1” 。

例子：

Input: N = 10, K = 6
Output: -1 2 -3 4 -5 6 -7 8 9 10
Explanation:
Consider the sequence as {-1, 2, -3, 4, -5, 6, -7, 8, 9, 10}, the sum of the constructed sequence is (-1) + 2 + (-3) + 4 + (-5) + 6 + (-7) + 8 + 9 + 10 = 23, which is positive.

Input: N = 7, K = 2
Output: -1

方法：给定的问题可以通过使用贪心方法解决，方法是使序列中的前 (N – K) 个元素为负，其余K 个元素为正。请按照以下步骤解决问题：

• 初始化一个数组，比如arr[]来存储结果序列。
• 初始化两个变量，比如sumNeg和sumPos为0 ，分别存储第一个(N – K)和剩余元素的总和。
• 使用变量i迭代范围[0, N – K – 1]并将 arr[i]的值更新为-(i + 1)并将值arr[i] 添加到变量sumNeg 。
• 使用变量i迭代范围[N – K, N – 1]并将arr[i]的值更新为(i + 1)并将值 arr[i] 添加到变量sumPos 。
• 如果 sumNeg的绝对值大于sumPos ，则打印-1 。否则，打印数组arr[] 中的总和元素作为结果。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
  
// Function to generate the resultant
// sequence of first N natural numbers
void findSequence(int n, int k)
{
    // Initalize an array of size N
    int arr[n];
  
    // Stores the sum of positive and
    // negative elements
    int sumPos = 0, sumNeg = 0;
  
    // Mark the first N - K elements
    // as negative
    for (int i = 0; i < n - k; i++) {
  
        // Update the value of arr[i]
        arr[i] = -(i + 1);
  
        // Update the value of sumNeg
        sumNeg += arr[i];
    }
  
    // Mark the remaining K elements
    // as positive
    for (int i = n - k; i < n; i++) {
  
        // Update the value of arr[i]
        arr[i] = i + 1;
  
        // Update the value of sumPos
        sumPos += arr[i];
    }
  
    // If the sum of the sequence
    // is negative, then print -1
    if (abs(sumNeg) >= sumPos) {
        cout << -1;
        return;
    }
  
    // Print the required sequence
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
  
// Driver Code
int main()
{
    int N = 10, K = 6;
    findSequence(N, K);
  
    return 0;
}

C#

// C# program for the above approach
using System;
class GFG
{
    
    // Function to generate the resultant
    // sequence of first N natural numbers
    static void findSequence(int n, int k)
    {
        // Initalize an array of size N
        int[] arr = new int[n];
  
        // Stores the sum of positive and
        // negative elements
        int sumPos = 0, sumNeg = 0;
  
        // Mark the first N - K elements
        // as negative
        for (int i = 0; i < n - k; i++) {
  
            // Update the value of arr[i]
            arr[i] = -(i + 1);
  
            // Update the value of sumNeg
            sumNeg += arr[i];
        }
  
        // Mark the remaining K elements
        // as positive
        for (int i = n - k; i < n; i++) {
  
            // Update the value of arr[i]
            arr[i] = i + 1;
  
            // Update the value of sumPos
            sumPos += arr[i];
        }
  
        // If the sum of the sequence
        // is negative, then print -1
        if (Math.Abs(sumNeg) >= sumPos) {
            Console.Write(-1);
            return;
        }
  
        // Print the required sequence
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
  
    // Driver Code
    public static void Main()
    {
        int N = 10, K = 6;
        findSequence(N, K);
    }
}
  
// This code is contributed by ukasp.

输出：

-1 -2 -3 -4 5 6 7 8 9 10

时间复杂度： O(N)
辅助空间： O(N)