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
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, p;
vector<int> x;

void read() {
    cin >> n >> p;
    x.resize(n);
    cin >> x;
}

void solve() {
    // Each son contributes potential^p to the sum, and we may freely choose
    // which sons to include, so the optimum simply takes every son whose
    // contribution is positive: add max(0, v^p) over all sons.

    int64_t ans = 0;
    for(int v: x) {
        int64_t term = 1;
        for(int e = 0; e < p; e++) {
            term *= v;
        }

        ans += max<int64_t>(0, term);
    }

    cout << ans << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    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.