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 (1.3.71) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val x = sc.nextInt()
    val y = sc.nextInt()
    println(problem160d(n, x, y))
}

fun problem160d(n: Int, x: Int, y: Int): String {
    val routes = Array(n) { mutableListOf<Int>() }
    for (i in 0 until n - 1) {
        routes[i].add(i + 1)
        if (i in x-1..y-1) {
            routes[i + 1].add(i)
        }
    }
    routes[x - 1].add(y - 1)
    routes[y - 1].add(x - 1)

    val distances= Array(n - 1) { IntArray(n) { -1 } }

    for (i in 0 until n - 1) {
        val queue = ArrayDeque<Int>()
        queue.offer(i)
        distances[i][i] = 0

        while(!queue.isEmpty()) {
            val current = queue.poll()
            for (j in 0 until routes[current].size) {
                val next = routes[current][j]
                if (distances[i][next] != -1) continue
                distances[i][next] = distances[i][current] + 1
                queue.offer(next)
            }
        }
    }

//    debugLog(distances.mapIndexed { i, it -> it.drop(i) })

    val ans = Array(n - 1) { 0 }
    for (i in 0 until n - 1) {
        for (j in distances.mapIndexed { i, it -> it.drop(i).groupBy { it } }) {
            ans[i] += j[i + 1]?.count() ?: 0
        }
    }

    return ans.joinToString("\n")
}