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.Scanner;

public class Main {

    // クソコードを生み出してしまった。。。
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int X = sc.nextInt();
        int Y = sc.nextInt();
        sc.close();

        for (int k = 1; k <= N - 1; k++) {
            int ans = 0;
            for (int i = 1; i <= N - 1; i++) {
                if (i <= X) {
                    if (i + k <= X) {
                        ans++;
                    } else {
                        for (int j = i + k; j <= N; j++) {
                            int d = j - i;
                            if (j < Y) {
                                d = Math.min((X - i) + 1 + (Y - j), d);
                            } else {
                                d = Math.min((X - i) + 1 + (j - Y), d);
                            }

                            if (d == k) {
                                ans++;
                            }
                        }
                    }

                } else if (i >= Y) {
                    if (i + k <= N) {
                        ans++;
                    }
                } else { // X < i < Y
                    for (int j = i + k; j <= N; j++) {
                        int d = j - i;
                        if (j < Y) {
                            d = Math.min((i - X) + 1 + (Y - j), d);
                        } else {
                            d = Math.min((i - X) + 1 + (j - Y), d);
                        }

                        if (d == k) {
                            ans++;
                        }
                    }
                }

            }
            System.out.println(ans);
        }

    }

}