Problem:
Score : 400 points

Problem Statement

We have an undirected graph G with N vertices numbered 1 to N and N edges as follows:

For each i=1,2,...,N-1, there is an edge between Vertex i and Vertex i+1.

There is an edge between Vertex X and Vertex Y.

For each k=1,2,...,N-1, solve the problem below:

Find the number of pairs of integers (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j in G is k.

Constraints

3 \leq N \leq 2 \times 10^3

1 \leq X,Y \leq N

X+1 < Y

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X Y

Output

For each k=1, 2, ..., N-1 in this order, print a line containing the answer to the problem.

Sample Input 1

5 2 4

Sample Output 1

5
4
1
0

The graph in this input is as follows:

There are five pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 1: (1,2)\,,(2,3)\,,(2,4)\,,(3,4)\,,(4,5).

There are four pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 2: (1,3)\,,(1,4)\,,(2,5)\,,(3,5).

There is one pair (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 3: (1,5).

There are no pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 4.

Sample Input 2

3 1 3

Sample Output 2

3
0

The graph in this input is as follows:

Sample Input 3

7 3 7

Sample Output 3

7
8
4
2
0
0

Sample Input 4

10 4 8

Sample Output 4

10
12
10
8
4
1
0
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <stdio.h>
#include <time.h>
#define int long long
#define endl '\n'
#define INF 1000000000000000000
#define EPS 1e-10
#define PI 3.141592653589793238
#define all(a) a.begin(),a.end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define pb push_back
#define double long double
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<P, ll> PP;
typedef complex<double> com;
template<class T, class S> bool chmax(T &a, const S &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T, class S> bool chmin(T &a, const S &b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll n, ll m) { return (m ? gcd(m, n%m) : n); }
ll lcm(ll n, ll m) { return n / gcd(n, m)*m; }


signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);
	//modcalc();

	int n, x, y;
	cin >> n >> x >> y;
	x--; y--;
	vector<vector<int>> graph(n, vector<int>());
	rep(i, n - 1) {
		graph[i].pb(i + 1);
		graph[i + 1].pb(i);
	}
	graph[x].pb(y);
	graph[y].pb(x);
	vector<int> ans(n - 1);
	rep(i, n) {
		queue<P> que;
		vector<bool> vis(n);
		que.push(P(i, 0)); vis[i] = true;
		while (!que.empty()) {
			P p = que.front(); que.pop();
			int px = p.first; int py = p.second;
			for (int j : graph[px]) {
				if (!vis[j]) {
					vis[j] = true;
					ans[py]++;
					que.push(P(j, py + 1));
				}
			}
		}
	}
	rep(i, n - 1) cout << ans[i] / 2 << endl;
}