1) Abridged problem statement
- You are given an odd integer L = 2N − 1 (5 ≤ L ≤ 2^31 − 1).
- A necklace has L beads colored black or white. It is called “beautiful” if there exist two black beads such that one of the two open arcs strictly between them contains exactly N beads.
- Find the minimal K such that every necklace with at least K black beads is necessarily beautiful.

2) Detailed editorial
Key observation:
- Let L = 2N − 1. For two distinct beads, the two open arcs strictly between them have sizes d − 1 and L − d − 1, where d is the number of edges along one direction (0 < d < L).
- Exactly N beads appear strictly between them iff d − 1 = N or L − d − 1 = N, i.e., iff d = N + 1 or d = L − 1 − N = N − 2.
- These two values are the same “undirected” separation: a pair at distance N + 1 in one direction is at distance N − 2 in the other. So it suffices to consider one direction, say d = N + 1 modulo L.

Graph reformulation:
- Build a graph G on L vertices (beads), with an edge between i and i + d (mod L), where d = N + 1.
- A necklace is not beautiful iff there is no pair of black beads joined by such an edge, i.e., the black beads form an independent set in G.
- Therefore, among all non-beautiful colorings, the maximum possible number of black beads equals the independence number α(G). The minimal K that forces beauty for every coloring is α(G) + 1.

Structure of G:
- The edges are “jump by d modulo L,” forming a 2-regular graph (each vertex connects to i ± d). This decomposes into g = gcd(L, d) disjoint cycles, each of length len = L / g.
- The maximum independent set in a cycle of length len is floor(len / 2), by alternating choices.
- Hence α(G) = g * floor((L / g) / 2). Therefore,
  K_min = α(G) + 1 = g * floor((L / g) / 2) + 1,
  where L = 2N − 1 and d = N + 1.

Useful simplification:
- g = gcd(L, d) = gcd(2N − 1, N + 1) = gcd(N + 1, 3). Thus g ∈ {1, 3}.
- If g = 1 (i.e., N + 1 not divisible by 3): K_min = floor(L/2) + 1 = N.
- If g = 3 (i.e., N ≡ 2 mod 3): K_min = 3 * floor((L/3)/2) + 1 = N − 1.
- So the closed form is:
  - If N ≡ 2 (mod 3): answer = N − 1
  - Else: answer = N

Complexity:
- O(log L) due to gcd; constant memory.

3) Provided C++ solution with detailed comments
#include <bits/stdc++.h>  // Pulls in standard headers; includes gcd in <numeric>
using namespace std;

// Stream output for pair<T1,T2> (not used here, part of a template)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Stream input for pair<T1,T2> (not used here)
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Stream input for vector<T> (not used here)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Stream output for vector<T> (not used here)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Global variables:
// two_n_minus_one stores L (the input), n stores N = (L+1)/2
int64_t n, two_n_minus_one;

// Read input L and compute N = (L+1)/2
void read() {
    cin >> two_n_minus_one;           // Read L = 2N-1 (problem guarantees odd)
    n = (two_n_minus_one + 1) / 2;    // N = (L+1)/2 (integer arithmetic)
}

void solve() {
    // Idea:
    // A coloring is non-beautiful iff no edge i -- (i + (n+1)) (mod L) has both endpoints black.
    // This graph is a union of g = gcd(L, n+1) cycles, each of length L/g.
    // Max black beads avoiding any "beautiful" pair equals
    //   max_avoid = g * floor( (L/g) / 2 ),
    // hence the minimal K that forces beauty is max_avoid + 1.

    int64_t L = two_n_minus_one;          // Rename for clarity
    int64_t dist = n + 1;                 // d = N + 1 (the step size that characterizes beauty)
    int64_t g = gcd(L, dist);             // Number of disjoint cycles
    int64_t max_avoid = g * ((L / g) / 2);// Sum of floor(len/2) over all cycles
    int64_t ans = max_avoid + 1;          // K_min = α(G) + 1

    cout << ans << endl;                  // Output the answer
}

int main() {
    ios_base::sync_with_stdio(false); // Fast IO
    cin.tie(nullptr);                 // Untie cin/cout

    int T = 1;                        // Single test case (problem specifies one)
    // cin >> T;                      // Left here for template completeness
    for(int test = 1; test <= T; test++) {
        read();                       // Read input
        // cout << "Case #" << test << ": ";
        solve();                      // Compute and print answer
    }

    return 0;                         // Successful termination
}

4) Python solution (commented)
import sys
import math

def main():
    # Read the first integer from input (robust to extra whitespace/newlines)
    data = sys.stdin.read().strip().split()
    if not data:
        return
    L = int(data[0])  # L = 2N - 1 (odd per problem statement)

    # Compute N = (L + 1) // 2 (since L = 2N - 1)
    N = (L + 1) // 2

    # The key "step" that characterizes beauty is d = N + 1.
    d = N + 1

    # The graph with edges i -- (i + d mod L) decomposes into g cycles.
    g = math.gcd(L, d)

    # Each cycle of length len = L // g has maximum independent set floor(len/2).
    # So the maximum number of black beads that still avoids beauty is:
    max_avoid = g * ((L // g) // 2)

    # Minimal K that forces beauty in every coloring:
    ans = max_avoid + 1

    print(ans)

if __name__ == "__main__":
    main()

5) Compressed editorial
- Beauty occurs iff there exist two black beads whose circular separation is d = N + 1 (mod L), where L = 2N − 1.
- Build graph G on L vertices with edges i -- (i + d). A non-beautiful coloring is precisely an independent set in G.
- G splits into g = gcd(L, d) cycles, each of length L/g. The maximum independent set per cycle is floor((L/g)/2). So α(G) = g * floor((L/g)/2).
- Minimal K that guarantees beauty is α(G) + 1 = g * floor((L/g)/2) + 1.
- Since gcd(2N − 1, N + 1) = gcd(N + 1, 3), the answer simplifies to:
  - If N ≡ 2 (mod 3): K = N − 1
  - Else: K = N
- Complexity: O(log L) time, O(1) memory.