1. Problem Understanding

In this problem, we are given an array of integers and a "mystical divisor" $K$. Our task is to count the number of contiguous subarrays whose sums are divisible by $K$.
A subarray is any continuous segment of the array. For example, given an array $[a_1, a_2, \dots, a_N]$, a subarray could be defined by indices $i$ and $j$ such that the subarray is $[a_i, a_{i+1}, \dots, a_j]$. The sum of this subarray is $S_{i,j} = \sum_{k=i}^j a_k$.
The condition for a subarray to be "mystical" is that $S_{i,j} \mod K = 0$. There might be multiple test cases in a single input.

2. Solution Analysis

A brute force approach to solve the problem would involve iterating over all possible subarrays and computing their sum, then checking for divisibility by $K$. However, since the number of subarrays in an array of size $N$ is on the order of $\mathcal{O}(N^2)$, and computing each sum in a naive manner may also take linear time, this brute force approach becomes inefficient for large arrays.

A more efficient approach leverages prefix sums and the properties of the modulo operation. Here’s the outline of the algorithm:

• Compute a prefix sum array, where $P[i] = \sum_{j=0}^{i} a_j$.
• For any subarray defined by indices $i$ and $j$ (with $i \leq j$), the subarray sum can be expressed as $P[j] - P[i-1]$.
• The key observation is that the subarray sum $S_{i,j}$ is divisible by $K$ if and only if: $$(P[j] - P[i-1]) \mod K = 0 \quad \Longleftrightarrow \quad P[j] \mod K = P[i-1] \mod K$$
• We can count how many times each remainder (after taking modulo $K$) appears while iterating through the prefix sum array.
• If a particular remainder $r$ appears $n_r$ times, then there are $\frac{n_r \times (n_r - 1)}{2}$ pairs of indices (i.e., subarrays) that give a sum divisible by $K$.
• Additionally, whenever the prefix sum itself is divisible by $K$ (i.e., remainder $0$), it means that the subarray from the beginning of the array to that point is mystical.

This method avoids explicitly checking every subarray, reducing the time complexity significantly.

3. Time and Space Complexity Analysis

• Time Complexity:

  • Calculating the prefix sum by iterating through the array takes $\mathcal{O}(N)$ time.
  • Iterating over the prefix sums to count modulo frequencies also takes $\mathcal{O}(N)$ time.
  • Thus, the overall time complexity per test case is $\mathcal{O}(N)$.

• Space Complexity:

  • Storing the prefix sums requires $\mathcal{O}(N)$ space.
  • The frequency dictionary (or array) used to count remainders takes $\mathcal{O}(K)$ space.
  • In total, the space complexity is $\mathcal{O}(N + K)$ per test case.

4. Implementation Details

• Read the input, taking into account multiple test cases.
• For each test case:
  • Parse the values of $N$ and $K$, and then the array of integers.
  • Initialize a variable for the running prefix sum.
  • Use a dictionary (or an array of size $K$) to maintain the count of each prefix sum modulo $K$. Initially, set the count for remainder $0$ to be $1$, as having a prefix sum exactly divisible by $K$ on its own (from index $0$ to some $i$) is valid.
  • As you iterate through the array:
    • Update the running prefix sum.
    • Compute the current remainder, ensuring it is non-negative (using modulo $K$ properly).
    • Add the count of the current remainder (from the dictionary) to the answer because each matching earlier prefix signifies a subarray with a sum divisible by $K$.
    • Increment the count for the current remainder.
• Finally, output the count for mystical subarrays for each test case.

This solution is efficient enough to handle large inputs within the problem constraints without resorting to a brute force technique.