1. Abridged Problem Statement
Given a positive integer x, find the integer N such that N is exactly 1/x of the sum of all positive integers less than N. In other words, find N satisfying
  N = (1 + 2 + … + (N–1)) / x
Output that N.

2. Detailed Editorial

Problem restatement
We seek an integer N > 0 such that when you add up all integers from 1 to N–1, that total is x times N.

Key observation
The sum of the first N–1 positive integers is
  S = 1 + 2 + … + (N–1) = (N–1)·N/2

The requirement is
  S = x · N
Plug in the formula for S:
  (N–1)·N / 2 = x · N

Case analysis
— If N = 0, the sum is zero, but N must be positive, so we discard N = 0.
— Otherwise, divide both sides by N (since N ≠ 0):
  (N–1) / 2 = x
Multiply both sides by 2:
  N – 1 = 2x
Therefore
  N = 2x + 1

That single formula immediately gives the answer in O(1) time and O(1) memory, easily handling x up to 10^9.

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;
};

int64_t x;

void read() { cin >> x; }

void solve() {
    // We need an integer n whose positive predecessors 1..n-1 sum to x times
    // n, i.e. (n - 1) * n / 2 = x * n. Cancelling n gives n = 2 * x + 1.

    cout << x * 2 + 1 << '\n';
}

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

    read();
    solve();

    return 0;
}
```

4. Python Solution

```python
import sys

def main():
    # Read a single integer x from standard input
    data = sys.stdin.read().strip()
    x = int(data)

    # Derivation shows the answer is N = 2*x + 1
    result = 2 * x + 1

    # Print the result
    print(result)

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

Comments on the Python code:
- We read the entire input via sys.stdin.read(), strip whitespace, and convert to int.
- Using the derived formula N = 2·x + 1, we compute and print the result in O(1) time.

5. Compressed Editorial
Sum of 1..(N–1) = N(N–1)/2 must equal x·N ⇒ (N–1)/2 = x ⇒ N = 2x+1.
