/*
A C++ (CPP) code to find how many subarrays ( contiguous part of an array) are present in an array whose product is less than a
 given number 'k'

 e.g.-
  Input array -  arr[5]= {1, 3, 4 ,5 ,2}
  Input No.   -   k = 5

  Subarrays of arr having product less than (k=5) are - {1} , {3} , {4} , {2} , {1 , 3}
*/

/*

Idea - Use Sliding Window Algorithm

Procedure -

Go in a loop ( j=0 to n) and create a sliding window each time starting from current element and increase its size (in second loop)
till both of these  conditions are true  ( product of all eleemnts in the window < k and iterator < last_index )

Proof -
 If at any instance , the product of all elements in the current window (subarray) exceeds k , then all further windows's product>k
 and hence we shift to the element to immediate right and do the same process .
*/

// Implementation

#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// required function

int count_SubArray_Product_Less_Than_K(int a[], int n, int k)
{
    // Actual answer
    int ans = 0;

    for (int j = 0; j < n; j++)
    {
        // traverse  each element
        int i = j;

        // initialize the product as 1 for current window .

        long long int product = 1;

        // making  a window that is expanding continuously till (product < k) AND (counter variable < size of array) .
        while (i < n && product < k)
        {
            product *= a[i];
            if (product < k)
            {
                ans++;
                i++;
            }
        }
    }

    return ans;
}

int main()
{

    // Taking INPUT from user
    int n;

    cout << "Enter the no. of elements in array : \n";
    cin >> n;

    int arr[n];

    cout << "Enter the elements : \n";

    for (int j = 0; j < n; j++)
    {
        cin >> arr[j];
    }

    int k;

    cout << "Enter the number k : \n";
    cin >> k;

    // Calling the required Function
    int ans = count_SubArray_Product_Less_Than_K(arr, n, k);

    // Giving Output
    cout << "The Number of subarrays in given array whose product is less than " << k << "are- " << endl;
    cout << ans << endl;

    return 0;
}