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: Wrong Answer)

/**
 * 最短経路の練習：「Warshall-Floyd法」
 * ABC160 D
 * https://atcoder.jp/contests/abc160/tasks/abc160_d
 * 
 * 期待する入力
 * N X Y
 * 3 <= N <= 2000
 * 1 <= X, Y <= N
 * X+1 < Y
 */

 package main

 import (
	 "fmt"
	 "bufio"
	 "os"
	 "strconv"
 )
 
 func main() {
	 sc := bufio.NewScanner(os.Stdin)
	 sc.Split(bufio.ScanWords)
 
	 N, X, Y := nextInt(sc), nextInt(sc), nextInt(sc)
	 
	 // 距離計算「Warshall-Floyd法」
	 // 今回、エッジの距離は全て１だが、可変でも行けるはず...
	 // 参考：https://qiita.com/ta-ka/items/a023a11efe17ab097433#warshall-floyd%E6%B3%95
	 distance := make([][]int, N)
	 counter := make([]int, N-1)
	 ks := []int{0, X-1}  // 0は隣接行列?を初期化するため
	 for index, k := range ks {
		 for i := 0; i < N; i++ {
			 if k == 0 {
				 distance[i] = make([]int, N)
				 if i == (X-1) {
					 distance[i][Y-1] = 1
				 } else if i == (Y-1) {
					 distance[i][X-1] = 1
				 }
			 }
			 for j := 0; j < i; j++ {
				 if k == 0 {
					 isSetedDistance := false
					 isSetedDistance = isSetedDistance || i == j
					 isSetedDistance = isSetedDistance || i == (X - 1) && j == (Y - 1)
					 isSetedDistance = isSetedDistance || j == (X - 1) && i == (Y - 1)
					 if !isSetedDistance {
						 distance[i][j] = i - j  // 迂回路が無い場合の距離を計算
					 }
				 }
				 distance[i][j] = min(distance[i][j], distance[i][k] + distance[k][j])
				 distance[j][i] = distance[i][j]  // 無向グラフのため、対称行列
				 if index == 1 {
					 counter[distance[i][j]-1] += 1
				 }
			 }
		 }
	 }
 
	 // 表示
	 for _, num := range counter {
		 fmt.Println(num)
	 }
 
 }
 
 func min(a, b int) int {
	 if a < b {
		 return a
	 }
	 return b
 }
 
 func nextInt(sc *bufio.Scanner) int {
	 sc.Scan()
	 t, _ := strconv.Atoi(sc.Text())
	 return t
 }