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 <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef long double ld;

int dist[2005][2005];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, x, y;
	cin >> n >> x >> y;
	x--, y--;
	vector<vector<int>> G(n);
	REP(i,n-1){
		G[i].push_back(i+1);
		G[i+1].push_back(i);
	}
	G[x].push_back(y);
	G[y].push_back(x);

	REP(i,n) REP(j,n) dist[i][j] = -1;
	REP(i,n){
		queue<pair<int, int>> que;
		que.emplace(i, 1);

		while(!que.empty()){
			pair<int, int> cur = que.front();
			que.pop();

			for(auto nv: G[cur.first]){
				if(dist[i][nv]!=-1) continue;
				//cout << cur.first << " " << nv << " " << cur.second << '\n';
				dist[i][nv] = cur.second;
				que.emplace(nv, cur.second+1);
			}
		}
	}

	vector<int> ans(n-1, 0);
	REP(i,n) FOR(j,i+1,n) ans[dist[i][j]-1]++;

	REP(k,n-1) cout << ans[k] << '\n';	
	return 0;
}