1. Abridged Problem Statement  
-----------------------------  
Given integers N, M, K and a list of N positive integers v₁…vₙ, count how many of them satisfy (vᵢ)^M ≡ 0 (mod K).

2. Detailed Editorial  
----------------------  
**Understanding the Task**  
We have a sequence of N positive integers. For each element v, we want to raise it to the power M and check whether the result is divisible by K. That is, check if v^M mod K == 0. Finally, we output the count of such v.

**Constraints and Implications**  
- N, M, K ≤ 10 000  
- Each v ≤ 10 001  
- Time limit is tight (0.25 s), so an O(N·M) algorithm is too slow in the worst case (10^8 multiplications).  

**Key Observation: Fast Modular Exponentiation**  
To compute v^M mod K efficiently, we use the binary exponentiation (a.k.a. exponentiation by squaring) algorithm. This reduces the exponentiation from O(M) multiplications to O(log M) multiplications under modulus K.

**Algorithm Steps**  
1. Read N, M, K.  
2. Initialize a counter `answer = 0`.  
3. For each of the N values v:  
   a. Compute `r = v^M mod K` using fast modular exponentiation in O(log M).  
   b. If r == 0, increment `answer`.  
4. Print `answer`.

**Complexity**  
- Each modular exponentiation takes O(log M) steps.  
- Total is O(N·log M), which for N, M ≤ 10^4 is O(10^4·14) ≈ 1.4·10^5 modular multiplications, easily within the time limit.

3. C++ Solution with Detailed Comments  
---------------------------------------  
```cpp
#include <bits/stdc++.h>      // Includes all standard headers
using namespace std;

// Fast exponentiation: compute (base^exp) % mod in O(log exp)
int modPow(int base, int exp, int mod) {
    int result = 1 % mod;           // Initialize result = 1 mod mod
    base %= mod;                    // Reduce base modulo mod upfront
    while (exp > 0) {
        if (exp & 1) {              // If current least-significant bit is 1
            result = int((long long)result * base % mod);
        }
        base = int((long long)base * base % mod);  // Square base
        exp >>= 1;                  // Shift exp right by 1 bit
    }
    return result;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M, K;
    cin >> N >> M >> K;            // Read N, M, K

    int answer = 0;
    for (int i = 0; i < N; i++) {
        int v;
        cin >> v;                   // Read the next value
        // Compute v^M mod K
        if (modPow(v, M, K) == 0) {
            ++answer;               // If divisible by K, count it
        }
    }

    cout << answer << "\n";        // Output the final count
    return 0;
}
```

4. Python Solution with Detailed Comments  
------------------------------------------  
```python
import sys
data = sys.stdin.read().split()
# Parse input
# data[0]=N, data[1]=M, data[2]=K, then N numbers follow
N, M, K = map(int, data[:3])
values = map(int, data[3:])

def mod_pow(base, exp, mod):
    """
    Compute (base^exp) % mod using binary exponentiation.
    Time complexity: O(log exp).
    """
    base %= mod
    result = 1 % mod
    while exp > 0:
        if exp & 1:                  # If current bit of exp is 1
            result = (result * base) % mod
        base = (base * base) % mod   # Square the base
        exp >>= 1                    # Shift exp right
    return result

answer = 0
for v in values:
    # If v^M mod K == 0, increment our count
    if mod_pow(v, M, K) == 0:
        answer += 1

print(answer)
```

5. Compressed Editorial  
-----------------------  
Use O(N·log M) time: for each v compute (v^M mod K) via binary exponentiation and count how many results equal zero.