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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

var N, X, Y int
var arr = make([]bool, 2001)
var ans = map[int]int{}

func main() {
	fmt.Scan(&N, &X, &Y)
	for i := 1; i<=N; i++ {
		queue := []int{i}
		visited := map[int]bool{ i: true }
		for k := 1; k<=N-1; k++ {
			nextQueue := []int{}
			for _, v := range queue {
				if v != 1 && !visited[v-1] {
					visited[v-1] = true
					nextQueue = append(nextQueue, v-1)
				}
				if v != N && !visited[v+1] {
					visited[v+1] = true
					nextQueue = append(nextQueue, v+1)
				}
				if v == X && !visited[Y] {
					visited[Y] = true
					nextQueue = append(nextQueue, Y)
				}
				if v == Y && !visited[X] {
					visited[X] = true
					nextQueue = append(nextQueue, X)
				}
			}
			ans[k] += len(nextQueue)
			queue = nextQueue 
		}
	}
	for i := 1; i <= N-1; i++ {
		fmt.Println(ans[i]/2)
	}
}