范围乘积为正整数的数组中索引对的计数

给定非零整数数组A ，任务是找到对数(l，r) ，其中(l <= r)使得A [l] * A [l + 1] * A [l + 2 ]….A [r]为正。

例子：

Input: A = {5, -3, 3, -1, 1}
Output: 7
Explanation:
First pair, (1, 1) = 5 is positive
Second pair, (3, 3) = 3 is positive
Third pair, (1, 4) = 5 * -3 * 3 * -1 = 45 is positive
Forth pair, (2, 4) = -3 * 3 * -1 = 9 is positive
Fifth pair, (1, 5) = 5 * -3 * 3 * -1 * 1 = 45 is positive
Sixth pair, (2, 5) = -3 * 3 * -1 * 1 = 9 is positive
Seventh pair, (5, 5) = 1 is positive
So, there are seven pairs with positive product.

Input: A = {4, 2, -4, 3, 1, 2, -4, 3, 2, 3}
Output: 27

方法：
这个想法是检查每个数组元素可能的数字对。

• 遍历数组，对数组中的每个元素执行以下步骤。
• 跟踪之前的负数个元素为偶数的元素数(作为even_count)以及在其前的负数个元素为奇数的元素数(作为odd_count) 。
• 到目前为止存储否定元素的总数(作为total_count) 。
• 如果total_count是偶数，则将even_count添加到答案中。否则，添加odd_count 。

下面是上述方法的实现：

C++

// C++ Program to find the
// count of index pairs
// in the array positive
// range product
  
#include 
using namespace std;
  
void positiveProduct(int arr[], int n)
{
    int even_count = 0;
    int odd_count = 0;
    int total_count = 0;
    int ans = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Condition if number of
        // negative elements is even
        // then increase even_count
        if (total_count % 2 == 0)
            even_count++;
  
        // Otherwise increase odd_count
        else
            odd_count++;
  
        // Condition if current element
        // is negative
        if (arr[i] < 0)
            total_count++;
  
        // Condition if number of
        // negative elements is even
        // then add even_count
        // in answer
        if (total_count % 2 == 0)
            ans += even_count;
  
        // Otherwise add odd_count
        // in answer
        else
            ans += odd_count;
    }
  
    cout << ans << "\n";
}
  
// Driver Code
int main()
{
    int A[] = { 5, -3, 3, -1, 1 };
  
    int size = sizeof(A) / sizeof(A[0]);
  
    positiveProduct(A, size);
  
    return 0;
}

Java

// Java program to find the count of
// index pairs in the array positive 
// range product 
class GFG{
      
public static void positiveProduct(int arr[],
                                   int n) 
{ 
    int even_count = 0; 
    int odd_count = 0; 
    int total_count = 0; 
    int ans = 0; 
      
    for(int i = 0; i < n; i++)
    {
          
       // Condition if number of 
       // negative elements is even 
       // then increase even_count 
       if (total_count % 2 == 0)
       {
           even_count++; 
       } 
         
       // Otherwise increase odd_count 
       else
       {
           odd_count++;
       }
         
       // Condition if current element 
       // is negative 
       if (arr[i] < 0)
       {
           total_count++;
       }
         
       // Condition if number of 
       // negative elements is even 
       // then add even_count 
       // in answer 
       if (total_count % 2 == 0) 
           ans += even_count;
             
       // Otherwise add odd_count 
       // in answer 
       else
           ans += odd_count; 
    } 
    System.out.println(ans);
      
} 
  
// Driver Code    
public static void main(String[] args) 
{
    int A[] = { 5, -3, 3, -1, 1 }; 
    int size = A.length; 
      
    positiveProduct(A, size); 
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find the count 
# of index pairs in the array 
# positive range product
def positiveProduct(arr, n):
  
    even_count = 0
    odd_count = 0
    total_count = 0
    ans = 0
  
    for i in range(n):
  
        # Condition if number of
        # negative elements is even
        # then increase even_count
        if(total_count % 2 == 0):
            even_count += 1
  
        # Otherwise increase odd_count
        else:
            odd_count += 1
  
        # Condition if current element
        # is negative
        if(arr[i] < 0):
            total_count += 1
  
        # Condition if number of
        # negative elements is even
        # then add even_count
        # in answer
        if(total_count % 2 == 0):
            ans += even_count
  
        # Otherwise add odd_count
        # in answer
        else:
            ans += odd_count
  
    print(ans)
  
# Driver Code
if __name__ == '__main__':
      
    A = [ 5, -3, 3, -1, 1 ]
    size = len(A)
  
    positiveProduct(A, size)
  
# This code is contributed by Shivam Singh

C#

// C# program to find the count of
// index pairs in the array positive 
// range product 
using System;
  
class GFG{
      
public static void positiveProduct(int []arr,
                                   int n) 
{ 
    int even_count = 0; 
    int odd_count = 0; 
    int total_count = 0; 
    int ans = 0; 
      
    for(int i = 0; i < n; i++)
    {
          
       // Condition if number of 
       // negative elements is even 
       // then increase even_count 
       if (total_count % 2 == 0)
       {
           even_count++; 
       } 
         
       // Otherwise increase odd_count 
       else
       {
           odd_count++;
       }
         
       // Condition if current element 
       // is negative 
       if (arr[i] < 0)
       {
           total_count++;
       }
         
       // Condition if number of 
       // negative elements is even 
       // then add even_count 
       // in answer 
       if (total_count % 2 == 0) 
           ans += even_count;
         
       // Otherwise add odd_count 
       // in answer 
       else
           ans += odd_count; 
    } 
    Console.WriteLine(ans);
} 
  
// Driver Code 
public static void Main(String[] args) 
{
    int []A = { 5, -3, 3, -1, 1 }; 
    int size = A.Length; 
      
    positiveProduct(A, size); 
}
}
  
// This code is contributed by 29AjayKumar

输出：

7

时间复杂度： O(N)
空间复杂度： O(1)