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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

#define ll long long
#define ld long double
typedef vector <ll> vi;                 // Vector of long long
typedef vector <vi> vvi;                // Vector of vi
typedef vector <pair<ll,ll>> ii;        // Vector of pairs
#define ff first                        // For pairs
#define ss second                       // For pairs
#define sz(a) int((a).size())           // Function to determine size of any container
#define pb push_back                    // Pushback to vector
#define mp make_pair                    // Makes pairs to be stored as pair
#define all(c) (c).begin(), (c).end()   // Mainly used by me in sorting
// ordered_set adds two new functions to set - (set).find_by_order([kth element based on zero indexing]) and order_of_key()
// order_of_key returns number of elements less that parameter. If element exists, that order is its index
#define ordered_set tree < ll ,  null_type ,  less<ll> ,  rb_tree_tag ,  tree_order_statistics_node_update >

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

    ll n, a, b;
    cin >> n >> a >> b;

    if (a > b)
        swap(a, b);
    
    if ((b - a) % 2 == 0)
        cout << (b - a) / 2 << "\n";
    else
        cout << min(b - 1, n - a) << "\n";
}