1. Abridged Problem Statement  
Given an integer N (0 ≤ N ≤ 65535), compute the maximum number of regions into which an infinite sheet of paper can be divided by drawing N straight lines, assuming no two lines are parallel and no three pass through the same point.

2. Detailed Editorial  

  Problem restatement  
  We want the maximum number of regions (zones) determined by N lines in the plane, with generic position (every pair of lines intersects and no three lines meet at one point).

  Key observation and inductive argument  
  - With 0 lines, there is exactly 1 region (the whole plane).  
  - When you add the i-th line, it intersects each of the previous (i–1) lines in a distinct point (because no two are parallel and no three concurrent). These intersection points split the new line into i segments. Each segment divides an existing region into two, thereby increasing the total region count by 1 per segment—i.e., by i.  
  - Hence if R(N−1) is the maximum regions with N−1 lines, then  
      R(N) = R(N−1) + (number of new segments)  
            = R(N−1) + (N).  

  Summation formula  
  Starting from R(0) = 1, we get  
      R(1) = 1 + 1 = 2  
      R(2) = R(1) + 2 = 4  
      …  
      R(N) = 1 + (1 + 2 + … + N).  
  The sum of the first N positive integers is N·(N+1)/2, so  
      R(N) = 1 + N·(N+1)/2.  

  Implementation details  
  - Read integer N.  
  - Compute result as 1 + (N × (N+1)/2).  
  - Print the result.  
  - Data types: N can be up to ~65 k, so N·(N+1)/2 ~ 2.1 × 10⁹ fits in 32-bit signed, but we use 64-bit to be safe.

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;

void read() { cin >> n; }

void solve() {
    /*
     * Each new line crosses all previous lines in distinct points, adding one
     * more region than the (k-1) it crosses, i.e. the k-th line adds k regions.
     * Summing from the initial single region gives the maximum
     * 1 + 1 + 2 + ... + n = 1 + n(n+1)/2.
     */

    int64_t answer = 1 + (int64_t)n * (n + 1) / 2;
    cout << answer << '\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
import sys

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])  
    # The maximum number of regions formed by n lines in general position
    # is 1 plus the sum of integers from 1 to n:
    #   regions = 1 + (1 + 2 + ... + n) = 1 + n*(n+1)//2
    result = 1 + n * (n + 1) // 2
    print(result)

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

5. Compressed Editorial  
Maximum regions = 1 + sum_{i=1..N} i = 1 + N(N+1)/2. Each new line intersects all previous lines in distinct points, creating i new segments that each add one region.