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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args : Array<String>) {
    val (n, x, y) = readLine()!!.split(' ').map(String::toInt)
    val q = ArrayDeque<Int>()
    val r = IntArray(n + 1) { -1 }
    val s = IntArray(n)
    for (i in 1..n) {
        q.add(i)
        r[i] = 0
        while (q.isNotEmpty()) {
            val v = q.remove()
            val d = r[v]
            if (v != 1 && r[v - 1] == -1) {
                q.add(v - 1)
                r[v - 1] = d + 1
            }
            if (v != n && r[v + 1] == -1) {
                q.add(v + 1)
                r[v + 1] = d + 1
            }
            if (v == x && r[y] == -1) {
                q.add(y)
                r[y] = d + 1
            }
            if (v == y && r[x] == -1) {
                q.add(x)
                r[x] = d + 1
            }
        }
        for (j in 1..n) {
            s[r[j]]++
            r[j] = -1
        }
    }
    for (i in 1 until n) {
        println(s[i] shr 1)
    }
}