阳性乘积的子阵列数

给定N个整数的数组arr [] ，任务是查找具有正积的子数组的数量。

例子：

Input: arr[] = {-1, 2, -2}
Output: 2
Subarrays with positive product are {2} and {-1, 2, -2}.

Input: arr[] = {5, -4, -3, 2, -5}
Output: 7

方法：本文讨论了寻找负积子阵列的方法。如果cntNeg是阴性乘积子阵列的计数，而total是给定阵列的所有可能的子阵列的计数，则阳性乘积子阵列的计数将是cntPos = total – cntNeg 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of
// subarrays with negative product
int negProdSubArr(int arr[], int n)
{
    int positive = 1, negative = 0;
    for (int i = 0; i < n; i++) {
  
        // Replace current element with 1
        // if it is positive else replace
        // it with -1 instead
        if (arr[i] > 0)
            arr[i] = 1;
        else
            arr[i] = -1;
  
        // Take product with previous element
        // to form the prefix product
        if (i > 0)
            arr[i] *= arr[i - 1];
  
        // Count positive and negative elements
        // in the prefix product array
        if (arr[i] == 1)
            positive++;
        else
            negative++;
    }
  
    // Return the required count of subarrays
    return (positive * negative);
}
  
// Function to return the count of
// subarrays with positive product
int posProdSubArr(int arr[], int n)
{
  
    // Total subarrays possible
    int total = (n * (n + 1)) / 2;
  
    // Count to subarrays with negative product
    int cntNeg = negProdSubArr(arr, n);
  
    // Return the count of subarrays
    // with positive product
    return (total - cntNeg);
}
  
// Driver code
int main()
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << posProdSubArr(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
      
// Function to return the count of 
// subarrays with negative product 
static int negProdSubArr(int arr[], int n) 
{ 
    int positive = 1, negative = 0; 
    for (int i = 0; i < n; i++)
    { 
  
        // Replace current element with 1 
        // if it is positive else replace 
        // it with -1 instead 
        if (arr[i] > 0) 
            arr[i] = 1; 
        else
            arr[i] = -1; 
  
        // Take product with previous element 
        // to form the prefix product 
        if (i > 0) 
            arr[i] *= arr[i - 1]; 
  
        // Count positive and negative elements 
        // in the prefix product array 
        if (arr[i] == 1) 
            positive++; 
        else
            negative++; 
    } 
  
    // Return the required count of subarrays 
    return (positive * negative); 
} 
  
// Function to return the count of 
// subarrays with positive product 
static int posProdSubArr(int arr[], int n) 
{ 
  
    // Total subarrays possible 
    int total = (n * (n + 1)) / 2; 
  
    // Count to subarrays with negative product 
    int cntNeg = negProdSubArr(arr, n); 
  
    // Return the count of subarrays 
    // with positive product 
    return (total - cntNeg); 
} 
  
// Driver code 
public static void main (String[] args)
{ 
    int arr[] = { 5, -4, -3, 2, -5 }; 
    int n = arr.length; 
  
    System.out.println(posProdSubArr(arr, n)); 
}
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach 
  
# Function to return the count of 
# subarrays with negative product 
def negProdSubArr(arr, n): 
  
    positive = 1
  
    negative = 0
  
    for i in range(n): 
  
        # Replace current element with 1 
        # if it is positive else replace 
        # it with -1 instead 
        if (arr[i] > 0): 
  
            arr[i] = 1
  
        else: 
  
            arr[i] = -1
  
        # Take product with previous element 
        # to form the prefix product 
        if (i > 0): 
  
            arr[i] *= arr[i - 1] 
  
        # Count positive and negative elements 
        # in the prefix product array 
        if (arr[i] == 1): 
  
            positive += 1
  
        else: 
  
            negative += 1
  
    # Return the required count of subarrays 
    return (positive * negative) 
  
# Function to return the count of
# subarrays with positive product
def posProdSubArr(arr, n):
  
    total = (n * (n + 1)) / 2;
  
    # Count to subarrays with negative product
    cntNeg = negProdSubArr(arr, n);
  
    # Return the count of subarrays
    # with positive product
    return (total - cntNeg);
  
# Driver code 
arr = [5, -4, -3, 2, -5] 
n = len(arr) 
print(posProdSubArr(arr, n)) 
  
# This code is contributed by Mehul Bhutalia

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the count of 
// subarrays with negative product 
static int negProdSubArr(int []arr, int n) 
{ 
    int positive = 1, negative = 0; 
    for (int i = 0; i < n; i++)
    { 
  
        // Replace current element with 1 
        // if it is positive else replace 
        // it with -1 instead 
        if (arr[i] > 0) 
            arr[i] = 1; 
        else
            arr[i] = -1; 
  
        // Take product with previous element 
        // to form the prefix product 
        if (i > 0) 
            arr[i] *= arr[i - 1]; 
  
        // Count positive and negative elements 
        // in the prefix product array 
        if (arr[i] == 1) 
            positive++; 
        else
            negative++; 
    } 
  
    // Return the required count of subarrays 
    return (positive * negative); 
} 
  
// Function to return the count of 
// subarrays with positive product 
static int posProdSubArr(int []arr, int n) 
{ 
  
    // Total subarrays possible 
    int total = (n * (n + 1)) / 2; 
  
    // Count to subarrays with negative product 
    int cntNeg = negProdSubArr(arr, n); 
  
    // Return the count of subarrays 
    // with positive product 
    return (total - cntNeg); 
} 
  
// Driver code 
public static void Main (String[] args)
{ 
    int []arr = { 5, -4, -3, 2, -5 }; 
    int n = arr.Length; 
  
    Console.WriteLine(posProdSubArr(arr, n)); 
}
}
  
// This code is contributed by 29AjayKumar

输出：

7