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 (
	"bufio"
	"fmt"
	"os"
)

func main() {
	r := bufio.NewReader(os.Stdin)
	var n, x, y int
	fmt.Fscan(r, &n)
	fmt.Fscan(r, &x)
	fmt.Fscan(r, &y)
	x--
	y--

	ans := make([]int, n)
	for i := 0; i < n; i++ {
		// 各ノードから他のノードへの最短距離を求める

		cost := make([]int, n)
		for j := 0; j < n; j++ {
			cost[j] = -1
		}
		cost[i] = 0
		var q Queue
		q.push(i)

		for !q.empty() {
			current, _ := q.pop()

			// 1つ前のノード
			if current-1 >= 0 && cost[current-1] == -1 {
				q.push(current - 1)
				cost[current-1] = cost[current] + 1
			}
			// 1つ後のノード
			if current+1 < n && cost[current+1] == -1 {
				q.push(current + 1)
				cost[current+1] = cost[current] + 1
			}
			// ショートカット
			if current == x && cost[y] == -1 {
				q.push(y)
				cost[y] = cost[current] + 1
			}
			if current == y && cost[x] == -1 {
				q.push(x)
				cost[x] = cost[current] + 1
			}
		}

		for j := 0; j < n; j++ {
			ans[cost[j]]++
		}
	}

	w := bufio.NewWriter(os.Stdout)
	defer w.Flush()
	for i := 1; i < n; i++ {
		fmt.Fprintln(w, ans[i]/2)
	}
}

// Union-Find
type unionFind struct {
	d []int
}

func newUnionFind(n int) *unionFind {
	uf := new(unionFind)
	uf.d = make([]int, n)
	for i := 0; i < n; i++ {
		uf.d[i] = -1
	}
	return uf
}

func (uf *unionFind) find(x int) int {
	if uf.d[x] < 0 {
		return x
	}
	uf.d[x] = uf.find(uf.d[x])
	return uf.d[x]
}

func (uf *unionFind) unite(x, y int) bool {
	rootX := uf.find(x)
	rootY := uf.find(y)
	if rootX == rootY {
		return false
	}

	if uf.d[rootX] < uf.d[rootY] {
		uf.d[rootX] += uf.d[rootY]
		uf.d[rootY] = rootX
	} else {
		uf.d[rootY] += uf.d[rootX]
		uf.d[rootX] = rootY
	}

	return true
}

func (uf *unionFind) same(x, y int) bool {
	return uf.find(x) == uf.find(y)
}

func (uf *unionFind) size(x int) int {
	return -uf.d[uf.find(x)]
}

// mod combination
func modpow(a, n, mod int) int {
	res := 1
	for n > 0 {
		if n%2 != 0 {
			res = res * a % mod
		}
		a = a * a % mod
		n = n / 2
	}
	return res
}

func modcomb(n, a, mod int) int {
	x := 1
	y := 1
	for i := 0; i < a; i++ {
		x = x * (n - i)
		x %= mod
		y = y * (i + 1)
		y %= mod
	}
	return x * modpow(y, mod-2, mod) % mod
}

func modfactorial(n, mod int) int {
	result := 1
	for i := 1; i <= n; i++ {
		result = (result * i) % mod
	}
	return result
}

// Utility

func min(x, y int) int {
	if x < y {
		return x
	}
	return y
}

func max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

func primeFactor(x int) map[int]int {
	res := make(map[int]int)
	for i := 2; i*i <= x; i++ {
		for x%i == 0 {
			res[i]++
			x = x / i
		}
	}
	if x != 1 {
		res[x] = 1
	}
	return res
}

func divisor(x int) []int {
	res := make([]int, 0)
	for i := 1; i*i <= x; i++ {
		if x%i == 0 {
			res = append(res, i)
			if i != x/i {
				res = append(res, x/i)
			}
		}
	}
	return res
}

func gcd(x, y int) int {
	if y == 0 {
		return x
	}
	return gcd(y, x%y)
}

func lcm(x, y int) int {
	return x / gcd(x, y) * y
}

func reverseString(s string) string {
	r := []rune(s)
	for i, j := 0, len(r)-1; i < len(r)/2; i, j = i+1, j-1 {
		r[i], r[j] = r[j], r[i]
	}
	return string(r)
}

// Queue ...
type Queue []int

// pop ...
func (q *Queue) empty() bool {
	return len(*q) == 0
}

// push ...
func (q *Queue) push(i int) {
	*q = append(*q, i)
}

// pop ...
func (q *Queue) pop() (int, bool) {
	if q.empty() {
		return 0, false
	} else {
		res := (*q)[0]
		*q = (*q)[1:]
		return res, true
	}
}

// Stack ...
type Stack []int

// pop ...
func (s *Stack) empty() bool {
	return len(*s) == 0
}

// push ...
func (s *Stack) push(i int) {
	*s = append(*s, i)
}

// pop ...
func (s *Stack) pop() (int, bool) {
	if s.empty() {
		return 0, false
	} else {
		index := len(*s) - 1
		res := (*s)[index]
		*s = (*s)[:index]
		return res, true
	}
}