We can use sliding window technique to solve the problem.
Here, i and j represent starting and ending points of the sliding window.
Initially i = j = 0.
Now, we will traverse the whole array and try to add elements.

• If sum < k, j++. So, number of sub-arrays produced = j-i. Also add arr[j] to the sum.
  
• If sum >= k, we will subtract arr[i] from sum so that again sum<k. So, i++.

#include <iostream>
using namespace std;
int main()
{
    int n, k;
    cin>>n>>k;
    int arr[n];
    for (int i=0;i<n;i++) cin>>arr[i];
    int i=0, j=0, count=0, sum=arr[0];
	while (i<n && j<n){
		if (sum<=k){
			j++;
			if (j>=i) count+=j-i;
			if (j<n) sum += arr[j];
		}
		else{
			sum -= arr[i];
			i++;
		}
	}
	cout<<count;
}

Convert the statement to it's equivalent in terms of prefix sums i.e $$$pre_r - pre_l \lt = k$$$.

Now iterating over $$$r$$$, we need to find number of $$$l$$$ such that $$$pre_l \gt = pre_r - k$$$, which can be computed using ordered set of all $$$pre_l$$$.