生成一个具有 K 个正数的数组，使得 arr[i] 为 -1 或 1，并且数组的总和为正数

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，请打印其中任何一个。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)
{
    // Initialize 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;
}

Java

// Java program for the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to generate the resultant
// sequence of first N natural numbers
static void findSequence(int n, int k)
{
     
    // Initialize 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)
    {
        System.out.print(-1);
        return;
    }
 
    // Print the required sequence
    for(int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
 
// Driver code
public static void main(String args[])
{
    int N = 10, K = 6;
     
    findSequence(N, K);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python program for the above approach
# Function to generate the resultant
# sequence of first N natural numbers
def findSequence(n, k):
     
    # Initialize an array of size N
    arr = [0]*n
     
    # Stores the sum of positive and
    # negative elements
    sumPos = 0
    sumNeg = 0
     
    # Mark the first N - K elements
    # as negative
    for i in range(0, n - k):
       
        # 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 i in range(n - k, n):
         
        # 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 pr -1
    if (abs(sumNeg) >= sumPos):
        print(-1)
        return
     
    # Pr the required sequence
    for i in range(n):
        print( arr[i], end =" ")
 
# Driver Code
N = 10
K = 6
findSequence(N, K)
 
# This code is contributed by shivanisinghss2110

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)
    {
        // Initialize 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.

Javascript

// Javascript program for the above approach
 
// Function to generate the resultant
// sequence of first N natural numbers
function findSequence(n, k)
{
 
    // Initialize an array of size N
    let arr = new Array(n);
 
    // Stores the sum of positive and
    // negative elements
    let sumPos = 0, sumNeg = 0;
 
    // Mark the first N - K elements
    // as negative
    for (let 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 (let 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)
    {
        document.write(-1);
        return;
    }
 
    // Print the required sequence
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Driver Code
let N = 10, K = 6;
findSequence(N, K);
 
// This code is contributed by _saurabh_Jaiswal.

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)