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: Accepted)

#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
#define int int64_t
#define double long double
#define rep(i, N) for(int i=0; i<N; i++)
#define repr(i, N) for(int i=N-1; i>=0; --i)
#define all(v) v.begin(),v.end() 
using namespace std;
//using pr = pair<int, int>;
using tup  = tuple<int, int, int>;
using vec  = vector<int>;
using vvec = vector<vec>;

//プロトタイプ宣言

//グローバル変数
//const int INF = ;
//const int MAX_N = ;
//const int MAX_M = ;
//vvec dp(MAX_N, vec(MAX_M));

//関数：
/* メイン・プログラム */
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	int n, a, b; cin >> n >> a >> b;
	int d = abs(a-b);
	int ans;
	if(d%2 == 0) ans = d/2;
	else {
		ans = min(b-1, n-a);
		int tp = min(a-1, n-b);
		ans = tp + 1 + (ans-tp-1)/2;
	}
	
	cout << ans << endl;
	return 0;
}

//std::string str;
//str = std::to_string(int i);
//bitset<3> s(N);
//int ans = (char)grid - '0';
//int id = lower_bound(all(arr), x) - arr.begin();
//cout << fixed << setprecision(10) << tld << endl;
//vector<int> sub(itr,itr+d); //イテレーターで部分数列(i=0...d)を指定
		
/*struct data {
	char *name;
	int age;
	int height;
	double weight;
};*/

/*string ans = w;
string &ref = ans;
rep(i, N) {
	ref[i] = arr[i];
}*/

/*map<int,int> mp;
for(auto p : mp) {
	ct += p.second*(p.second-1); }

map<int,int>::iterator itr;
for(itr = mp.begin(); itr != mp.end(); itr++) {
	ans += itr->second * (itr->second-1);
}*/