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. Provided C++ Solution with Line-by-Line Comments  

```cpp
#include <bits/stdc++.h>        // includes all standard headers for convenience
#define endl '\n'               // shorthand for newline when using cout

using namespace std;           // avoid prefixing std:: everywhere

// Utility to update a variable if the new value is larger
template<class T, class T2>
inline void chkmax(T& x, const T2& y) {
    if (x < y) {
        x = y;
    }
}

// Utility to update a variable if the new value is smaller
template<class T, class T2>
inline void chkmin(T& x, const T2& y) {
    if (x > y) {
        x = y;
    }
}

const int MAXN = (1 << 20);   // not used in this problem but left from template

int64_t x;                    // to store the input x

// Reads the input from stdin into variable x
void read() {
    cin >> x;
}

// Computes and prints the answer N = 2*x + 1
void solve() {
    cout << x * 2 + 1 << endl;
}

int main() {
    ios_base::sync_with_stdio(false); // speed up I/O
    cin.tie(NULL);                    // untie cin from cout to avoid flushes

    read();       // read input
    solve();      // compute and output result
    return 0;     // exit
}
```

4. Python Solution with Detailed Comments  

```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.