具有奇数乘积的子数组的计数

给定一个大小为N的整数数组arr[] ，任务是计算具有奇数乘积的子数组的数量。
例子：

Input : arr[] = {5, 1, 2, 3, 4} 
Output : 4 
Explanation: The sub-arrays with odd product are- 
{5}, {1}, {3}, {5, 1}. Hence the count is 4.
Input : arr[] = {12, 15, 7, 3, 25, 6, 2, 1, 1, 7} 
Output : 16 
 

朴素的方法：一个简单的解决方案是计算每个子数组的乘积并检查它是否为奇数并相应地计算计数。
时间复杂度： O(N ² )
有效方法：奇数乘积只有奇数乘积才有可能。因此，对于数组中的每K 个连续奇数，具有奇数乘积的子数组的数量增加K*(K+1)/2 。计算连续奇数的一种方法是计算每两个连续偶数的索引之间的差值并减去1 。对于计算， -1和N被视为偶数的索引。
下面是上述方法的实现：

C++

// C++ program to find the count of
// sub-arrays with odd product
#include 
using namespace std;
 
// Function that returns the count of
// sub-arrays with odd product
int countSubArrayWithOddProduct(int* A, int N)
{
    // Initialize the count variable
    int count = 0;
 
    // Initialize variable to store the
    // last index with even number
    int last = -1;
 
    // Initialize variable to store
    // count of continuous odd numbers
    int K = 0;
 
    // Loop through the array
    for (int i = 0; i < N; i++) {
        // Check if the number
        // is even or not
        if (A[i] % 2 == 0) {
            // Calculate count of continuous
            // odd numbers
            K = (i - last - 1);
 
            // Increase the count of sub-arrays
            // with odd product
            count += (K * (K + 1) / 2);
 
            // Store the index of last
            // even number
            last = i;
        }
    }
 
    // N considered as index of
    // even number
    K = (N - last - 1);
 
    count += (K * (K + 1) / 2);
 
    return count;
}
 
// Driver Code
int main()
{
    int arr[] = { 12, 15, 7, 3, 25,
                  6, 2, 1, 1, 7 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << countSubArrayWithOddProduct(arr, n);
 
    return 0;
}

Java

// Java program to find the count of
// sub-arrays with odd product
class GFG {
 
// Function that returns the count of
// sub-arrays with odd product
static int countSubArrayWithOddProduct(int A[],
                                       int N)
{
     
    // Initialize the count variable
    int count = 0;
 
    // Initialize variable to store the
    // last index with even number
    int last = -1;
 
    // Initialize variable to store
    // count of continuous odd numbers
    int K = 0;
 
    // Loop through the array
    for(int i = 0; i < N; i++)
    {
 
       // Check if the number
       // is even or not
       if (A[i] % 2 == 0)
       {
 
           // Calculate count of continuous
           // odd numbers
           K = (i - last - 1);
            
           // Increase the count of sub-arrays
           // with odd product
           count += (K * (K + 1) / 2);
            
           // Store the index of last
           // even number
           last = i;
       }
    }
 
    // N considered as index of
    // even number
    K = (N - last - 1);
    count += (K * (K + 1) / 2);
     
    return count;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 12, 15, 7, 3, 25, 6, 2, 1, 1, 7 };
    int n = arr.length;
 
    // Function call
    System.out.println(countSubArrayWithOddProduct(arr, n));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program to find the count of
# sub-arrays with odd product
 
# Function that returns the count of
# sub-arrays with odd product
def countSubArrayWithOddProduct(A, N):
     
    # Initialize the count variable
    count = 0
 
    # Initialize variable to store the
    # last index with even number
    last = -1
 
    # Initialize variable to store
    # count of continuous odd numbers
    K = 0
 
    # Loop through the array
    for i in range(N):
         
        # Check if the number
        # is even or not
        if (A[i] % 2 == 0):
             
            # Calculate count of continuous
            # odd numbers
            K = (i - last - 1)
 
            # Increase the count of sub-arrays
            # with odd product
            count += (K * (K + 1) / 2)
 
            # Store the index of last
            # even number
            last = i
 
    # N considered as index of
    # even number
    K = (N - last - 1)
 
    count += (K * (K + 1) / 2)
    return count
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 12, 15, 7, 3, 25, 6, 2, 1, 1, 7 ]
    n = len(arr)
 
    # Function call
    print(int(countSubArrayWithOddProduct(arr, n)))
 
# This code is contributed by Bhupendra_Singh

C#

// C# program to find the count of
// sub-arrays with odd product
using System;
class GFG{
 
// Function that returns the count of
// sub-arrays with odd product    
static int countSubArrayWithOddProduct(int[] A,
                                       int N)
{
         
    // Initialize the count variable
    int count = 0;
     
    // Initialize variable to store the
    // last index with even number
    int last = -1;
     
    // Initialize variable to store
    // count of continuous odd numbers
    int K = 0;
     
    // Loop through the array
    for(int i = 0; i < N; i++)
    {
        
       // Check if the number
       // is even or not
       if (A[i] % 2 == 0)
       {
            
           // Calculate count of continuous
           // odd numbers
           K = (i - last - 1);
            
           // Increase the count of sub-arrays
           // with odd product
           count += (K * (K + 1) / 2);
            
           // Store the index of last
           // even number
           last = i;
       }
    }
     
    // N considered as index of
    // even number
    K = (N - last - 1);
    count += (K * (K + 1) / 2);
     
    return count;
}
 
// Driver code
static void Main()
{
    int[] arr = { 12, 15, 7, 3, 25,
                  6, 2, 1, 1, 7 };
    int n = arr.Length;
     
    // Function call
    Console.WriteLine(countSubArrayWithOddProduct(arr, n));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

// Javascript program to find the count of
// sub-arrays with odd product
 
// Function that returns the count of
// sub-arrays with odd product
function countSubArrayWithOddProduct(A, N)
{
 
    // Initialize the count variable
    var count = 0;
 
    // Initialize variable to store the
    // last index with even number
    var last = -1;
 
    // Initialize variable to store
    // count of continuous odd numbers
    var K = 0;
 
    // Loop through the array
    for (var i = 0; i < N; i++)
    {
     
        // Check if the number
        // is even or not
        if (A[i] % 2 == 0)
        {
         
            // Calculate count of continuous
            // odd numbers
            K = (i - last - 1);
 
            // Increase the count of sub-arrays
            // with odd product
            count += (K * (K + 1) / 2);
 
            // Store the index of last
            // even number
            last = i;
        }
    }
 
    // N considered as index of
    // even number
    K = (N - last - 1);
    count += (K * (K + 1) / 2);
    return count;
}
 
// Driver Code
var arr = [ 12, 15, 7, 3, 25,
              6, 2, 1, 1, 7 ];
var n = arr.length;
 
// Function call
document.write( countSubArrayWithOddProduct(arr, n));
 
// This code is contributed by rrrtnx.

输出：

16

时间复杂度： O(N)
辅助空间： O(1)

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