1. Abridged Problem Statement  
Given N segment lengths (each up to 10 500 digits), determine any three that can form a non-degenerate triangle (i.e. the sum of lengths of any two exceeds the third). If such a triple exists, output their lengths; otherwise output `0 0 0`.

2. Detailed Editorial  

Problem restatement  
• You have N segment lengths, each an integer possibly much larger than built-in types.  
• A triangle condition for lengths a, b, c is that all three inequalities hold:  
  a + b > c,  
  a + c > b,  
  b + c > a.  
• Because lengths are positive, after sorting them in non-decreasing order, only the “adjacent triple” check a + b > c (for a ≤ b ≤ c) is needed.

Key observations and strategy  
1. Sorting reduces the number of comparisons. Once the list is sorted as L[0] ≤ L[1] ≤ … ≤ L[N−1], if for some i we have L[i] + L[i+1] > L[i+2], then automatically:  
   L[i] + L[i+2] > L[i+1] and L[i+1] + L[i+2] > L[i], because L[i+2] ≥ L[i+1] ≥ L[i] > 0.  
2. You only need to check i = 0, 1, …, N−3, testing whether L[i] + L[i+1] > L[i+2].  
3. If you find such i, print that triple and stop; otherwise print “0 0 0”.

Handling big integers  
• C++: use a bigint library capable of addition, comparison, and sorting.  
• Python: its built-in `int` type supports arbitrary precision.

Complexity  
• Sorting N numbers costs O(N log N) comparisons; each comparison for M-digit numbers costs O(M).  
• The linear scan is O(N) big-integer additions/comparisons.  
• With N ≤ 1000 and M ≤ 10 500, this is efficient.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// A minimal big-integer wrapper is assumed available via the original code.
// Here we just show the main logic with comments explaining each part.

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

    int n;
    cin >> n;  // Read number of segments
    vector<bigint> segs(n);

    // Read each segment length into a bigint
    for(int i = 0; i < n; i++) {
        cin >> segs[i];
    }

    // Sort the list in non-decreasing order
    sort(segs.begin(), segs.end());

    // Check each consecutive triple
    for(int i = 0; i + 2 < n; i++) {
        // If segs[i] + segs[i+1] > segs[i+2], we found a triangle
        if(segs[i] + segs[i+1] > segs[i+2]) {
            // Output the three segment lengths
            cout << segs[i] << " " << segs[i+1] << " " << segs[i+2] << "\n";
            return 0;  // Terminate after finding one valid triangle
        }
    }

    // If no valid triple was found, print three zeros
    cout << "0 0 0\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
# Python handles arbitrarily large integers with the built-in int type.

import sys

def main():
    data = sys.stdin.read().strip().split()
    n = int(data[0])
    # Read segment lengths as Python ints
    segs = list(map(int, data[1:]))

    # Sort the segments in non-decreasing order
    segs.sort()

    # Scan through all adjacent triples
    for i in range(n - 2):
        a, b, c = segs[i], segs[i+1], segs[i+2]
        # Check triangle condition for sorted a <= b <= c
        if a + b > c:
            # Print the first found triple and exit
            print(a, b, c)
            return

    # If none form a triangle, print three zeros
    print(0, 0, 0)

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

5. Compressed Editorial  
Sort the segment lengths, then for each consecutive triple (a, b, c) check if a + b > c. If so, output that triple; otherwise output “0 0 0”. Sorting ensures only adjacent triples need checking.