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: Accepted)



import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        int n = Integer.parseInt(scan.next());
        int x = Integer.parseInt(scan.next());
        int y = Integer.parseInt(scan.next());

        scan.close();

        int[] k = new int[n];

        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                int direct = j - i;

                // 順に進んでワープ
                int warp;
                if (x < i || j < x) {
                    // ワープせずにたどり着くので考えない
                    warp = n;
                } else if (y <= j) {
                    // jがワープ先よりも後
                    warp = (x - i) + 1 + (j - y);
                } else {
                    // jがワープ先よりも前
                    warp = (x - i) + 1 + (y - j);
                }

                // 戻ってワープ
                int reverseWarp;
                if (i < x) {
                    // ワープできない
                    reverseWarp = n;
                } else if (y <= j) {
                    // jがワープ先よりも後
                    reverseWarp = (i - x) + 1 + (j - y);
                } else {
                    // jがワープ先よりも前
                    reverseWarp = (i - x) + 1 + (y - j);
                }

                int min = Math.min(Math.min(direct, warp), reverseWarp);
                k[min]++;
            }
        }

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

}