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.io.IOException;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		int n =  Integer.parseInt(sc.next());
		int x =  Integer.parseInt(sc.next());
		int y =  Integer.parseInt(sc.next());
		int[][] to = new int[n][n];
		int[] ans = new int[n];


		for(int i = 0; i < n ; i++) {
			for(int j = 0; j < n ; j++){
				to[i][j] = Math.abs(j - i);
			}
		}
		to[x-1][y-1] = 1;
		to[y-1][x-1] = 1;


		for(int k = 0; k < n ; k++) {
			for(int i = 0; i < n ; i++) {
				for(int j = i+1 ; j < n ; j++) {
					to[i][j] = Math.min(to[i][j], to[i][k] + to[k][j]);
				}
			}
		}

		for(int i = 0; i < n ; i++) {
			for(int j = i+1; j < n ; j++){
				ans[to[i][j]]++;
			}
		}

		for(int i = 1 ; i < n ; i++) {
			System.out.println(ans[i]);
		}
	}
}