1. Abridged Problem Statement  
-----------------------------  
You are given all pairwise XORs of an unknown set A of size n: B contains M = n·(n–1)/2 numbers, each equal to Ai⊕Aj for all 1≤i<j≤n, in arbitrary order. Moreover, A has no nontrivial subset (size≥2) whose XOR is 0 (equivalently, its elements are linearly independent over GF(2)). Recover any valid A (order doesn’t matter).

2. Detailed Editorial  
---------------------  
Key observations:  
- XOR is addition in the GF(2) vector space of 31-bit integers.  
- “No subset XORs to zero” means the elements of A form a linearly independent set (a basis) in that vector space.  
- The multiset B of all pairwise XORs then consists exactly of all Xi⊕Xj for basis elements Xi, Xj.  

Our goal is to find a basis {X1,…,Xn} such that the set of pairwise XORs matches B. Since a basis always can be “shifted” by XORing every element with a fixed vector without changing pairwise XORs, we may WLOG include 0 in our reconstructed set: if we recover {0, X2⊕X1, X3⊕X1,…}, then XORing the entire set with X1 yields a valid solution containing X1.  

Algorithm outline:  
1. Read m and array B of size m. Compute n from m = n(n–1)/2 via n = (1 + √(1 + 8m)) / 2.  
2. Build a hash set (or map) O marking all values in B for O(1) lookup.  
3. We will grow our answer vector ans starting from {0}. Maintain:  
   - pedo = all XORs of subsets of ans found so far (initially {0}).  
   - L = a hash set of those reachable XORs (initially {0}).  
4. Iterate through each b in B (in any order). If b is not yet in L, test whether adding b as a new basis element is valid: check that for every existing basis element y in ans, the xor b⊕y exists in O (i.e., that pairwise XORs are all present in B).  
5. If valid, incorporate b: For each existing subset‐XOR s in pedo, compute t = b⊕s and add t to L and pedo. Then append b to ans as a new basis vector.  
6. Continue until ans.size()==n. At that point ans = {0, X2, X3, …, Xn}. Print ans; it’s a valid set A′. If desired, XOR each by some a to shift off the 0, but that’s optional because 0 is allowed.  

Complexity:  
- n≤15 (because m≤100), B size m=O(n²).  
- For each candidate b we test O(n) existing basis elements, each lookup O(1).  
- Updating pedo costs O(2^k) where k is current basis size ≤n, so total ≈O(2^n·n) ≤ ~500k.  
Under m=100, this easily runs within time/memory limits.

3. The Provided C++ Solution With Detailed Comments  
----------------------------------------------------  
#include <bits/stdc++.h>  
using namespace std;

// Utility to update x = max(x, y)
template<class T, class T2>
inline void chkmax(T& x, const T2& y) {
    if (x < y) {
        x = y;
    }
}
// Utility to update x = min(x, y)
template<class T, class T2>
inline void chkmin(T& x, const T2& y) {
    if (x > y) {
        x = y;
    }
}

const int MAXN = (1 << 10);

int m, n;            // m = size of B; n = size of A to recover
int B[MAXN];         // array of pairwise XORs
int a0;              

// Read input, compute n from m = n(n-1)/2
void read() {
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> B[i];
    }
    // Solve n^2 - n - 2m = 0  =>  n = (1 + sqrt(1+8m)) / 2
    n = (1 + sqrt(1 + 8 * m)) / 2;
}

// L: keeps track of which XORs of chosen basis subsets we've generated  
// O: marks which values are present in B
map<int, bool> L, O;  

void solve() {
    // Mark all given pairwise XORs in O
    for (int i = 0; i < m; i++) {
        O[B[i]] = true;
    }

    vector<int> ans;   // will hold our basis, starting with 0
    vector<int> pedo;  // holds all subset XORs of ans
    ans.push_back(0);
    pedo.push_back(0);
    L[0] = true;

    // Try each b in B as a new basis element
    for (int i = 0; i < m; i++) {
        int b = B[i];
        if (L.count(b)) continue;  // Already representable by existing basis
        // Check validity: for every existing basis y, we need b⊕y in B
        bool ok = true;
        for (int y : ans) {
            if (!O.count(b ^ y)) {
                ok = false;
                break;
            }
        }
        if (!ok) continue;

        // Accept b as a new basis element:
        // Generate new subset XORs by XORing b with all previous subset XORs
        int oldSize = pedo.size();
        for (int idx = 0; idx < oldSize; idx++) {
            int s = pedo[idx];
            int t = b ^ s;
            if (!L.count(t)) {
                L[t] = true;
                pedo.push_back(t);
            }
        }
        ans.push_back(b);
        if ((int)ans.size() == n) break;  // done
    }

    // Output the reconstructed set A (contains 0 as one element)
    for (int x : ans) {
        cout << x << " ";
    }
    cout << "\n";
}

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

    read();
    solve();
    return 0;
}

4. Python Solution With Detailed Comments  
-----------------------------------------  
```python
import math
import sys
from collections import deque

def main():
    data = sys.stdin.read().split()
    m = int(data[0])
    Bs = list(map(int, data[1:]))

    # Recover n from m = n*(n-1)/2
    n = int((1 + math.isqrt(1 + 8*m)) // 2)

    # O: set of all given pairwise XORs for fast lookup
    O = set(Bs)

    # ans: our basis, start with 0
    ans = [0]
    # pedo: all XORs of subsets of ans
    pedo = [0]
    # L: set of reachable XORs
    L = {0}

    # Try each b in Bs as candidate basis element
    for b in Bs:
        if b in L:
            # Already spanned by current basis
            continue

        # Check that XOR with every existing basis element is in B
        valid = True
        for y in ans:
            if (b ^ y) not in O:
                valid = False
                break
        if not valid:
            continue

        # Accept b
        # Update pedo and L by XORing b with all existing subset XORs
        old_pedo = list(pedo)
        for s in old_pedo:
            t = b ^ s
            if t not in L:
                L.add(t)
                pedo.append(t)

        ans.append(b)
        if len(ans) == n:
            break

    # Print the reconstructed set (contains 0)
    print(" ".join(map(str, ans)))

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

5. Compressed Editorial  
------------------------  
- A’s elements are linearly independent vectors over GF(2). B lists all pairwise XORs Ai⊕Aj.  
- We can WLOG include 0 in our reconstruction (shifting by any A-element preserves B).  
- Algorithm: maintain a basis `ans` (starting with 0), a set `O` of all B-values, and track all subset XORs `pedo` plus a hash `L` of those XORs.  
- For each candidate b in B not yet in L, if for every existing basis y we have (b⊕y)∈O, accept b: update pedo/L by XORing b with all existing subset XORs, and append b to ans.  
- Stop once ans.size()==n. Output ans.