1. Abridged Problem Statement  
Given an integer N (≤100), an exponent p (1 ≤ p ≤ 3), and a list x of N integers (–3 ≤ x_i ≤ 3), choose any subset of these integers to maximize the sum of their p-th powers. You may also choose the empty subset (sum = 0). Output the maximum possible sum.

2. Detailed Editorial  
We want to maximize  
    S = ∑_{i in chosen subset} (x_i)^p  
over all subsets of {1…N}, allowing the empty set (sum = 0). Observe:

- Since N ≤ 100 and |x_i| ≤ 3, brute-forcing subsets (2^100) is impossible. But p ≤ 3 and x_i is tiny, so we look for a greedy rule.
- Compute a_i = (x_i)^p for each i. There are only 7 possible x_i values (–3, –2, –1, 0, 1, 2, 3) and p ≤3, so the mapping x_i → a_i is small.
- For each a_i:
  - If a_i > 0, adding it to the sum helps.
  - If a_i ≤ 0, adding it cannot increase the sum (it can only lower or leave it unchanged), so we skip it.
- Therefore the answer is simply  
      ans = ∑_{i=1..N} max(0, x_i^p).  
- Edge cases:
  - All a_i ≤ 0 → answer stays at 0 (empty subset).
  - p = 2: every a_i = (x_i)^2 ≥ 0 → ans = ∑(x_i)^2.
  - p = 1 or 3: negative x_i yield negative a_i → we skip those.

Time complexity: O(N).  
Memory: O(N) for storing inputs.

3. C++ Solution with Detailed Comments
```cpp
#include <iostream>
#include <vector>
#include <cmath>    // for std::pow, but we can also multiply directly
using namespace std;

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

    int N;
    cin >> N;               // Read number of sons (N ≤ 100)

    int p;
    cin >> p;               // Read exponent (1 ≤ p ≤ 3)

    vector<int> x(N);
    for (int i = 0; i < N; i++) {
        cin >> x[i];        // Read each son's mental potential (–3 ≤ x[i] ≤ 3)
    }

    long long ans = 0;      // Use long long to be safe, though max sum is small
    for (int i = 0; i < N; i++) {
        int v = x[i];
        // Compute v^p. For small p and small |v| we can do it directly:
        int vp;
        if (p == 1) {
            vp = v;
        } else if (p == 2) {
            vp = v * v;
        } else { // p == 3
            vp = v * v * v;
        }
        // Only add positive contributions
        if (vp > 0) {
            ans += vp;
        }
    }

    cout << ans << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments
```python
def main():
    # Read the number of sons (not really needed except to know how many potentials follow)
    n = int(input().strip())
    # Read the exponent p (1, 2, or 3)
    p = int(input().strip())
    # Read the list of mental potentials
    potentials = list(map(int, input().split()))

    total = 0  # This will accumulate the maximum sum
    for v in potentials:
        # Compute v**p efficiently for small p
        if p == 1:
            vp = v
        elif p == 2:
            vp = v * v
        else:  # p == 3
            vp = v * v * v
        # Only add if the power is positive
        if vp > 0:
            total += vp

    # Print the final maximum sum (zero if all vp ≤ 0)
    print(total)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
Compute each value x_i^p and sum only the positive ones; if none are positive, answer 0.