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: Wrong Answer)

//https://atcoder.jp/contests/abc160/submissions/11483851

fun main(A: Array<String>) {
    val (n, x, y) = readLine()!!.split(" ").map { it.toInt() }
    for (i in 1..n - 1) {
        var count = 0
        for (j in 1..n - i) {
            count++
            val distanceXtoJ = Math.abs(x - j)
            if (i > distanceXtoJ) {
                val destinationBig = y + i - distanceXtoJ - 1
                val destinationSmall = y - i + distanceXtoJ + 1

                if (destinationBig == destinationSmall) {
                    if (destinationBig - j > i) {
                        count++
                    }
                } else if (destinationBig <= n) {
                    if (i + j <= y) {
                        if (destinationBig - j > i) {
                            count++
                        }
                        if (destinationSmall - j > i) {
                            count++
                        } else if (destinationSmall - j < i) {
                            count--
                        }
                    } else {
                        if (destinationBig - j < i) {
                            count++
                        } else if (destinationBig - j > i) {
                            count--
                        }
                        if (destinationSmall - j > i) {
                            count++
                        }
                    }
                } else {
                    if (destinationSmall - j > i) {
                        count++
                    } else if (destinationSmall - j < i) {
                        count--
                    }
                }
            }
        }
        println(count)
    }
}