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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int N = input.nextInt();
		int X = input.nextInt();
		int Y = input.nextInt();
		ArrayList<Integer>[] graph = new ArrayList[N+1];
		for (int i = 1; i <= N; i++) {
			graph[i] = new ArrayList<Integer>();
		}
		for (int i = 1; i < N; i++) {
			graph[i].add(i+1);
			graph[i+1].add(i);
		}
		graph[X].add(Y);
		graph[Y].add(X);
		/*for (int i = 1; i <= N; i++) {
			System.out.print(i+"-->");
			for (int j = 0; j < graph[i].size(); j++) {
				System.out.print(graph[i].get(j)+"-->");
			}
			System.out.print("\n");
		}*/
		for (int k = 1; k < N; k++) {
			int track = 0;
			for (int a = 1; a < N; a++) {
				for (int b = a+1; b <= N; b++) {
					if (BFS(graph,a,b)==k) track++;
				}
			}
			System.out.println(track);
		}
	}
	public static int BFS(ArrayList<Integer>[] graph, int START, int END) {
		if (START==END) return 0;
		int cnt = 0;
		Queue<ArrayList<Integer>> cur = new LinkedList<ArrayList<Integer>>();
		boolean[] vis = new boolean[graph.length+1];
		cur.add(graph[START]);
		vis[START]=true;
		while (!cur.isEmpty()) {
			cnt++;
			int size = cur.size();
			for (int i = 0; i < size; i++) {
				ArrayList<Integer> x = cur.poll();
				for (int j = 0; j < x.size(); j++) {
					if (x.get(j)==END) return cnt;
					if (!vis[x.get(j)]) {
						vis[x.get(j)]=true;
						cur.add(graph[x.get(j)]);
					}
				}
			}
		}
		return -1;
	}
}