Problem:
Score : 300 points

Problem Statement

2N players are running a competitive table tennis training on N tables numbered from 1 to N.

The training consists of rounds.
In each round, the players form N pairs, one pair per table.
In each pair, competitors play a match against each other.
As a result, one of them wins and the other one loses.

The winner of the match on table X plays on table X-1 in the next round,
except for the winner of the match on table 1 who stays at table 1.

Similarly, the loser of the match on table X plays on table X+1 in the next round,
except for the loser of the match on table N who stays at table N.

Two friends are playing their first round matches on distinct tables A and B.
Let's assume that the friends are strong enough to win or lose any match at will.
What is the smallest number of rounds after which the friends can get to play a match against each other?

Constraints

2 \leq N \leq 10^{18}

1 \leq A < B \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the smallest number of rounds after which the friends can get to play a match against each other.

Sample Input 1

5 2 4

Sample Output 1

1

If the first friend loses their match and the second friend wins their match, they will both move to table 3 and play each other in the next round.

Sample Input 2

5 2 3

Sample Output 2

2

If both friends win two matches in a row, they will both move to table 1.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <unordered_map>
#include <vector>
#include <string.h>
#include <set>
#include <stack>

using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> void printvec(const vector<T>& v) {
  for (auto x : v) { cout << x << " "; }
  cout << endl;
}
template<class T> void printtree(const vector< vector<T> >& tree) {
  for (long long i = 0; i < tree.size(); ++i) {
    cout << i + 1 << ": "; printvec(tree[i]);
  }
}
template<class T, class U> void printmap(const map<T, U>& mp) {
  for (auto x : mp) { cout << x.first << "=>" << x.second << endl; }
}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

#define rep(i, n) for(ll i = 0; i < n; ++i)
#define all(s) s.begin(), s.end()
#define sz(x) (ll)(x).size()
#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define eb emplace_back

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> triple;
typedef double D;
typedef vector<ll> vl;
typedef vector<P> vp;

const ll INF = 1e9;
const ll MOD = 1000000007;  // 1e9 + 7

int main(int argc, char** argv) {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  //cout << setprecision(10) << fixed;

  ll n, a, b;
  cin >> n >> a >> b;
  ll diff = abs(a-b);
  if (a>b) { swap(a,b); }
  // Here, a<=b

  if (diff%2 == 1) {
    // Here, we battle in 1 or n
    // ll ans1 = max(abs(1-a),abs(1-b));
    // ll ans2 = max(abs(n-a),abs(n-b));

    // Here, we battle in middle
    ll ans3 = 0, ans4 = 0;
    {
      // go 1.
      ll _a = a, _b = b;
      ans3 += _a-1;
      _a -= ans3;
      _b -= ans3;
      assert(_a == 1);

      --_b;
      ++ans3;
      // Here, parity is same
      ans3 += (_b-1)/2;
    }
    {
      // go n.
      ll _a = a, _b = b;
      ans4 += n-_b;
      _a += ans4;
      _b += ans4;
      assert(_b == n);

      ++_a;
      ++ans4;
      // Here, parity is same
      ans4 += (n-_a)/2;
    }

    // cout << min(min(ans1, ans2),min(ans3,ans4)) << endl;
    cout << min(ans3,ans4) << endl;
  } else {
    cout << diff/2 << endl;
  }
}