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;
 
class Main{
  public static void main(String args[]){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int x = sc.nextInt();
    int y = sc.nextInt();
    int ans[] = new int[n];
    int c = y-x;
    int d;
    for(int i = 1; i < n; i++){
      for(int j = n; j > i; j--){
        d = j-i;
        if(i<=x&&y<=j){
          ans[x-i+j-y+1]++;
        } else if(i<=x&&j<y){
          if(d < x-i+1+y-j){
            ans[d]++;
          } else {
            ans[x-i+1+y-j]++;
          }
        } else if(x<i&&i<y&&y<=j){
          if(d < i-x+j-y+1){
            ans[d]++;
          } else {
            ans[i-x+j-y+1]++;
          }
        } else if(x<i&&i<y&&j<y){
          if(d < i-x+1+y-j){
            ans[d]++;
          } else {
            ans[i-x+1+y-j]++;
          }
        } else {
          ans[d]++;
        }
      }
    }
    for(int i = 1; i < n; i++){
      System.out.println(ans[i]);
    }
  }
}