给定一个由N个整数组成的数组arr [] ，任务是通过将每个数组元素arr [i]更改为(-1)* arr [i] – 1任意次，以使数组元素的乘积修改数组元素。数组最大化。
例子：
Input: arr[] = {-3, 0, 1}
Output: 2 -1 -2
Explanation:
Performing the following operations on the array elements maximizes the product:
Applying the operation on arr[0] once, arr[0] = (- 1) * (- 3) – 1 = 2.
Applying the operation on arr[1] once, arr[1] = 0 – 1 = -1.
Applying the operation on arr[2] once, arr[2] = -1 – 1 = -2.
The modified array is {2, -1, -2} with product 4, which is maximum possible.
Input: arr[] = {-9, 2, 0, 1, -5, 3, 16}
Output: -9 -3 -1 -2 -5 -4 16
方法：可以根据以下观察结果解决给定问题：
- 为了使乘积最大，当操作减小负数元素的绝对值时，元素的绝对值应更高。因此，给定的操作应仅应用于正值。
- 如果元素数为奇数，则对绝对值最高的元素执行运算，因为奇数个负数的乘积为负。因此，将产品的一个正数更改为正数和最大值。
- 对负数执行操作会降低其绝对值。因此，找到具有最大绝对值的元素并对其进行一次操作，因为这将使乘积最小。
请按照以下步骤解决问题：
- 遍历给定数组arr [] ，对于每个元素arr [i] ，检查arr [i]是否≥0 。如果发现为真，则更新arr [i] =(-1)* arr [i] – 1 。否则，请继续。
- 遍历数组后，检查N的值是否为奇数。如果发现为真，则找到具有最大绝对值的元素，然后对其进行一次运算。
- 完成上述步骤后，打印修改后的数组arr [] 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find array with maximum
// product by changing array
// elements to (-1)arr[i] - 1
void findArrayWithMaxProduct(int arr[],
int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
// Applying the operation on
// all the positive elements
if (arr[i] >= 0) {
arr[i] = -arr[i] - 1;
}
if (N % 2 == 1) {
// Stores maximum element in array
int max_element = -1;
// Stores index of maximum element
int index = -1;
for (int i = 0; i < N; i++)
// Check if current element
// is greater than the
// maximum element
if (abs(arr[i]) > max_element) {
max_element = abs(arr[i]);
// Find index of the
// maximum element
index = i;
}
// Perform the operation on the
// maximum element
arr[index] = -arr[index] - 1;
}
// Print the elements of the array
for (int i = 0; i < N; i++)
printf("%d ", arr[i]);
}
// Driver Code
int main()
{
int arr[] = { -3, 0, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
findArrayWithMaxProduct(arr, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to find array with maximum
// product by changing array
// elements to (-1)arr[i] - 1
static void findArrayWithMaxProduct(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
// Applying the operation on
// all the positive elements
if (arr[i] >= 0)
{
arr[i] = -arr[i] - 1;
}
if (N % 2 == 1)
{
// Stores maximum element in array
int max_element = -1;
// Stores index of maximum element
int index = -1;
for (int i = 0; i < N; i++)
// Check if current element
// is greater than the
// maximum element
if (Math.abs(arr[i]) > max_element) {
max_element = Math.abs(arr[i]);
// Find index of the
// maximum element
index = i;
}
// Perform the operation on the
// maximum element
arr[index] = -arr[index] - 1;
}
// Print the elements of the array
for (int i = 0; i < N; i++)
System.out.print(arr[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { -3, 0, 1 };
int N = arr.length;
// Function Call
findArrayWithMaxProduct(arr, N);
}
}
// This code is contributed by jithin.Python3
# Python3 program for the above approach
# Function to find array with maximum
# product by changing array
# elements to (-1)arr[i] - 1
def findArrayWithMaxProduct(arr, N):
# Traverse the array
for i in range(N):
# Applying the operation on
# all the positive elements
if (arr[i] >= 0):
arr[i] = -arr[i] - 1
if (N % 2 == 1):
# Stores maximum element in array
max_element = -1
# Stores index of maximum element
index = -1
for i in range(N):
# Check if current element
# is greater than the
# maximum element
if (abs(arr[i]) > max_element):
max_element = abs(arr[i])
# Find index of the
# maximum element
index = i
# Perform the operation on the
# maximum element
arr[index] = -arr[index] - 1
# Prthe elements of the array
for i in arr:
print(i, end = " ")
# Driver Code
if __name__ == '__main__':
arr = [-3, 0, 1]
N = len(arr)
# Function Call
findArrayWithMaxProduct(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
class GFG
{
// Function to find array with maximum
// product by changing array
// elements to (-1)arr[i] - 1
static void findArrayWithMaxProduct(int[] arr, int N)
{
// Traverse the array
for (int i = 0; i < N; i++)
// Applying the operation on
// all the positive elements
if (arr[i] >= 0)
{
arr[i] = -arr[i] - 1;
}
if (N % 2 == 1)
{
// Stores maximum element in array
int max_element = -1;
// Stores index of maximum element
int index = -1;
for (int i = 0; i < N; i++)
// Check if current element
// is greater than the
// maximum element
if (Math.Abs(arr[i]) > max_element) {
max_element = Math.Abs(arr[i]);
// Find index of the
// maximum element
index = i;
}
// Perform the operation on the
// maximum element
arr[index] = -arr[index] - 1;
}
// Print the elements of the array
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
static public void Main()
{
int[] arr = { -3, 0, 1 };
int N = arr.Length;
// Function Call
findArrayWithMaxProduct(arr, N);
}
}
// This code is contributed by sanjoy_62.输出：
2 -1 -2时间复杂度： O(N)
辅助空间： O(1)