允许负数的数组中成对乘积的最大和

给定n个元素的数组。查找成对乘法的最大和。总和可以更大，所以取10 ^ 9 + 7为mod。如果存在奇数个元素，则可以将任意一个元素(不成对)相加。
例子：

Input : arr[] = {-1, 4, 5, -7, -4, 9, 0}
Output : 77
So to get the maximum sum, the arrangement will 
be {-7, -4}, {-1, 0}, {9, 5} and {4}.
So the answer is (-7*(-4))+((-1)*0)+(9*5)+(4) ={77}.

Input : arr[] = {8, 7, 9}
Output : 79
Answer is (9*8) +(7) = 79.

1-对给定的数组进行排序。
2-首先，将负数从开始算起成对相乘，然后加到total_sum中。
3-第二，将正数从最后的数乘以total_sum。
4-检查负数和正数是否均为奇数，然后加上最后一对的乘积
即最后一个负数和正数。
5-或如果任何一个计数为奇数，则在左侧添加该元素。
6-返回总和。

C++

// C++ program for above implementation
#include 
#define Mod 1000000007
using namespace std;
 
// Function to find the maximum sum
long long int findSum(int arr[], int n)
{
    long long int sum = 0;
 
    // Sort the array first
    sort(arr, arr + n);
 
    // First multiply negative numbers pairwise
    // and sum up from starting as to get maximum
    // sum.
    int i = 0;
    while (i < n && arr[i] < 0) {
        if (i != n - 1 && arr[i + 1] <= 0) {
            sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
            i += 2;
        }
        else
            break;
    }
 
    // Second multiply positive numbers pairwise
    // and summed up from the last as to get maximum
    // sum.
    int j = n - 1;
    while (j >= 0 && arr[j] > 0) {
        if (j != 0 && arr[j - 1] > 0) {
            sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
            j -= 2;
        }
        else
            break;
    }
 
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if (j > i)
        sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
 
    // If one of them occurs odd times
    else if (i == j)
        sum = (sum + arr[i]) % Mod;
 
    return sum;
}
 
// Drivers code
int main()
{
    int arr[] = { -1, 9, 4, 5, -4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSum(arr, n);
    return 0;
}

Java

// Java program for above implementation
import java.io.*;
import java.util.*;
 
class GFG {
 
static int Mod = 1000000007;
 
// Function to find the maximum sum
static long findSum(int arr[], int n) {
    long sum = 0;
 
    // Sort the array first
    Arrays.sort(arr);
 
    // First multiply negative numbers
    // pairwise and sum up from starting
    // as to get maximum sum.
    int i = 0;
    while (i < n && arr[i] < 0) {
    if (i != n - 1 && arr[i + 1] <= 0) {
        sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
        i += 2;
    }
    else
        break;
    }
 
    // Second multiply positive numbers
    // pairwise and summed up from the
    // last as to get maximum sum.
    int j = n - 1;
    while (j >= 0 && arr[j] > 0) {
    if (j != 0 && arr[j - 1] > 0) {
        sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
        j -= 2;
    } else
        break;
    }
 
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if (j > i)
    sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
 
    // If one of them occurs odd times
    else if (i == j)
    sum = (sum + arr[i]) % Mod;
 
    return sum;
}
 
// Drivers code
public static void main(String args[]) {
    int arr[] = {-1, 9, 4, 5, -4, 7};
    int n = arr.length;
    System.out.println(findSum(arr, n));
}
}
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# Python3 code for above implementation
Mod= 1000000007
 
# Function to find the maximum sum
def findSum(arr, n):
    sum = 0
     
    # Sort the array first
    arr.sort()
     
    # First multiply negative numbers
    # pairwise and sum up from starting
    # as to get maximum sum.
    i = 0
    while i < n and arr[i] < 0:
        if i != n - 1 and arr[i + 1] <= 0:
            sum = (sum + (arr[i] * arr[i + 1])
                                 % Mod) % Mod
            i += 2
        else:
            break
         
    # Second multiply positive numbers
    # pairwise and summed up from the
    # last as to get maximum sum.
    j = n - 1
    while j >= 0 and arr[j] > 0:
        if j != 0 and arr[j - 1] > 0:
            sum = (sum + (arr[j] * arr[j - 1])
                                 % Mod) % Mod
            j -= 2
        else:
            break
         
    # To handle case if positive
    # and negative numbers both
    # are odd in counts.
    if j > i:
        sum = (sum + (arr[i] * arr[j]) % Mod)
                                       % Mod
         
    # If one of them occurs odd times
    elif i == j:
        sum = (sum + arr[i]) % Mod
     
    return sum
 
# Driver code
arr = [ -1, 9, 4, 5, -4, 7 ]
n = len(arr)
print(findSum(arr, n))
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program for above implementation
using System;
 
class GFG {
 
    static int Mod = 1000000007;
 
    // Function to find the maximum sum
    static long findSum(int[] arr, int n)
    {
        long sum = 0;
 
        // Sort the array first
        Array.Sort(arr);
 
        // First multiply negative numbers
        // pairwise and sum up from starting
        // as to get maximum sum.
        int i = 0;
        while (i < n && arr[i] < 0) {
            if (i != n - 1 && arr[i + 1] <= 0) {
                sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
                i += 2;
            }
            else
                break;
        }
 
        // Second multiply positive numbers
        // pairwise and summed up from the
        // last as to get maximum sum.
        int j = n - 1;
        while (j >= 0 && arr[j] > 0) {
            if (j != 0 && arr[j - 1] > 0) {
                sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
                j -= 2;
            }
            else
                break;
        }
 
        // To handle case if positive and negative
        // numbers both are odd in counts.
        if (j > i)
            sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
 
        // If one of them occurs odd times
        else if (i == j)
            sum = (sum + arr[i]) % Mod;
 
        return sum;
    }
 
    // Drivers code
    public static void Main()
    {
        int[] arr = { -1, 9, 4, 5, -4, 7 };
        int n = arr.Length;
        Console.WriteLine(findSum(arr, n));
    }
}
 
/*This code is contributed by vt_m.*/

PHP

= 0 && $arr[$j] > 0)
    {
        if ($j != 0 && $arr[$j - 1] > 0)
        {
            $sum = ($sum + ($arr[$j] *
                    $arr[$j - 1]) % $Mod) % $Mod;
            $j -= 2;
        }
        else
            break;
    }
 
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if ($j > $i)
        $sum = ($sum + ($arr[$i] *
                $arr[$j]) % $Mod) % $Mod;
 
    // If one of them occurs odd times
    else if ($i == $j)
        $sum = ($sum + $arr[$i]) % Mod;
 
    return $sum;
}
 
// Driver code
$arr = array (-1, 9, 4, 5, -4, 7 );
$n = sizeof($arr);
echo findSum($arr, $n);
 
// This code is contributed by ita_c
?>

Javascript

 
// JavaScript program for above implementation
 
 let Mod = 1000000007;
   
// Function to find the maximum sum
function findSum(arr, n) {
    let sum = 0;
   
    // Sort the array first
    arr.sort();
   
    // First multiply negative numbers
    // pairwise and sum up from starting
    // as to get maximum sum.
    let i = 0;
    while (i < n && arr[i] < 0) {
    if (i != n - 1 && arr[i + 1] <= 0) {
        sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;
        i += 2;
    }
    else
        break;
    }
   
    // Second multiply positive numbers
    // pairwise and summed up from the
    // last as to get maximum sum.
    let j = n - 1;
    while (j >= 0 && arr[j] > 0) {
    if (j != 0 && arr[j - 1] > 0) {
        sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;
        j -= 2;
    } else
        break;
    }
   
    // To handle case if positive and negative
    // numbers both are odd in counts.
    if (j > i)
    sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;
   
    // If one of them occurs odd times
    else if (i == j)
    sum = (sum + arr[i]) % Mod;
   
    return sum;
}
// Driver code
 
    let arr = [-1, 9, 4, 5, -4, 7];
    let n = arr.length;
    document.write(findSum(arr, n));
                             

输出：

87