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 <iostream>
#include<algorithm>
#include<string>
#include <cmath>
#include <vector>
#include <map>
#include <cstdio>
#include <iomanip>
#include<set>
#include <numeric>
#include <queue>

#pragma region Macros
#define int long long
#define double long double
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
const  int INF = 1e12;
const int dx[8] = { 1, 0, -1, 0, 1, -1, -1, 1 };
const int dy[8] = { 0, 1, 0, -1, 1, 1, -1, -1 };
const int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
#define krep(i,k,n) for(int i=(k);i<n+k;i++)
#define Krep(i,k,n) for(int i=(k);i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define Rrep(i,n) for(int i=n;i>0;i--)
#define REP(i,n) for(int i=1;i<=n;i++)
#define LAST(x) x[x.size()-1]
#define ALL(x) (x).begin(),(x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x)
#define swap(a,b) (a += b,b = a - b,a -= b)
#define DIVCEIL(a,b) ((a+b-1)/b)
#define SUM1n(n) ((n)*(n+1)/2)
#define SUM1n2(n) (n*(2*n+1)*(n+1))/6
#define SUMkn(k,n) (SUM1n(n)-SUM1n(k-1))
#define PB push_back
#define Fi first
#define Se second 

int intpow(int a, int n) {
	// a^nのint ver
	if (n == 0)
		return 1;
	else {
		rep(i, n - 1)
			a *= a;
		return a;
	}
}
int MODPOW(int a, int n, int mod) {
	// a^n mod
	int ans = 1;
	while (n > 0) {
		if (n & 1)
			ans = ans * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return ans;
}
int factorial(int a) {
	if (a == 0)
		return 1;
	else
		return a * factorial(a - 1);
}
int MODFACT(int a, int mod) {
	int ans = 1;
	REP(i, a) {
		ans *= i;
		ans %= MOD;
	}
	return ans;
}
int nPr(int n, int r) {
	int s = n - r + 1;
	int sum = 1;
	for (int i = s; i <= n; i++)
		sum *= i;
	return sum;
}
int MODnPr(int n, int r, int mod) {
	int s = n - r + 1;
	int sum = 1;
	for (int i = s; i <= n; i++) {
		sum *= i;
		sum = sum % MOD;
	}
	return sum;
}
//int nCr(int n, int r)
int nCr2(int n, int r) {
	return factorial(n) / (factorial(r) * factorial(n - r));
}
int GCD(int a, int b)
{
	if (a < b)
		swap(a, b);
	if (b == 0)
		return a;
	if (a % b == 0)
		return b;
	return GCD(b, a % b);
}
int LCM(int a, int b)
{
	return  a * b / GCD(a, b);
}
int divisor_count(int n) {
	//約数の数（だっけ？）
	int ans = 0;
	REP(i, n) {
		if (n % i == 0)
			ans++;
	}
	return ans;
}
int CEIL1(int n) {
	//1桁目切り上げ
	return (n + 9) / 10 * 10;
}
int getdigit(int n) {
	return log10(n) + 1;
}
int digit(int n, int k) {
	//nのk桁目
	rep(i, k - 1)
		n /= 10;
	return n % 10;
}
int digit_sum(int n) {
	int sum = 0, dig;
	while (n) {
		dig = n % 10;
		sum += dig;
		n /= 10;
	}
	return sum;
}
int DIVTIME(int n, int k) {
	//nをkで何回割れるか的な
	int div = 0;
	while (n % k == 0) {
		div++;
		n /= k;
	}
	return div;
}
int binary(int n) {
	//10進数→2進数
	int ans = 0;
	for (int i = 0; n > 0; i++) {
		ans += n % 2 * intpow(10, i);
		n /= 2;
	}
	return ans;
}
int intabs(int n) {
	if (n < 0)
		return -1 * n;
	else
		return n;
}
int Kaibun(int n) {
	int ans = 0;
	int d = getdigit(n);
	REP(i, d)
		ans += digit(n, i) * pow(10, d-i);
	return ans;
}
inline bool is_uru(int y) {
	if (y % 400 == 0)
		return 1;
	if (y % 100 == 0)
		return 0;
	if (y % 4 == 0)
		return 1;
	return 0;
}
void next_date(int &y, int &m, int &d) {
	int day = days[m];
	if (m == 2 && is_uru(y))
		day++;
	d++;
	if (day < d) {
		m++;
		d = 1;
	}
	if (m == 13) {
		y++;
		m = 1;
	}
}
double LOG(int a, int b) {
	return log(b) / log(a);
}
double DISTANCE(int x1, int y1, int x2, int y2) {
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
double clock_angle(int h, int m) {
	h %= 12;
	double mm = 6.0 * m;
	double nn = 30.0 * h + 0.5 * m;
	return std::min(fabs(mm - nn), 360.0 - fabs(nn - mm));
}
inline bool BETWEEN(int x, int min, int max) {
	if (min <= x && x <= max)
		return true;
	else
		return false;
}
inline bool between(int x, int min, int max) {
	if (min < x && x < max)
		return true;
	else
		return false;
}
inline bool is_prime(int x) {
	if (x == 1)
		return false;
	if (x == 2)
		return true;
	if (x % 2 == 0)
		return false;
	double sqrtx = sqrt(x);
	for (int i = 3; i <= sqrtx; i += 2) {
		if (x % i == 0)
			return false;
	}
	return true;
}
inline bool is_sqrt(int n) {
	if (sqrt(n) == (int)sqrt(n))
		return true;
	else
		return false;
}
template<class T>
inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T>
inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
// 順位付け
using namespace std;
#pragma endregion


typedef vector<int> vint;
typedef vector<string> vstring;
typedef map<int, int> mint;
typedef pair<int, int> pint;
using Graph = vector<vint>;

mint m;


vector<pint> prime_factorize(int N) {
	vector<pint>  res;
	Krep(i,2,sqrt(N)) {
		if (N % i != 0) 
			continue;
		int ex = 0;
		while (N % i == 0) {
			++ex;
			N /= i;
		}
		res.push_back({ i, ex });
	}
	if (N != 1) 
		res.push_back({ N, 1 });
	return res;
}

/*int NumberofDivsors(int N) {
	vector<pint> a = Prime_factorize(N);
	int ans = 1;
	for (pint p : a)
		ans *= p.second() + 1;
	return ans;
}
*/

double median(vint a) {
	int N = a.size();
	if (N % 2 == 1)
		return (double)a[N / 2];
	else
		return (double)(a[N / 2-1] + a[N / 2]) / 2;
}

int collatz(int n,int cou) {
	if (m.count(n))
		return m[n]+cou;
	if (n == 1)
		return cou;
	if (n % 2 == 0)
		return collatz(n / 2, cou + 1);
	return collatz(3 * n + 1, cou + 1);
}

signed main() {
	int N, A, B;
	cin >> N >> A >> B;
	int ans;
	if ((B - A) % 2 == 0)
		ans = (B - A) / 2;
	else
		ans = min(A-1,N-B)+1+(B-A-1)/2;
	cout << ans;
}