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 Python (3.8.2) (judge verdict: Time Limit Exceeded)

from collections import deque

N, X, Y = [int(x) for x in input().split()]
result = {x: 0 for x in range(N)}

# 面倒なので0オリジンに修正
X -= 1
Y -= 1

# 全位置から移動できるパターンを幅優先探索で網羅する
for i in range(N):

    queue = deque()
    visited = set()

    # 開始位置と距離を初期値として入れる
    queue.appendleft((i, 0))

    while queue:
        now_pos, distance = queue.pop()

        # 訪れていない場合、その位置に移動するまでの距離に結果を加算
        # 訪問済フラグを立てる
        if not now_pos in visited:
            result[distance] += 1
            visited.add(now_pos)

        # 前、後ろ、X <-> Y の移動を実行
        # すでに訪問済の場合は追加しない
        if now_pos - 1 >= 0:
            if not now_pos - 1 in visited:
                queue.appendleft((now_pos - 1, distance + 1))

        if now_pos + 1 < N:
            if not now_pos + 1 in visited:
                queue.appendleft((now_pos + 1, distance + 1))

        if now_pos == X:
            if not Y in visited:
                queue.appendleft((Y, distance+1))

        if now_pos == Y:
            if not X in visited:
                queue.appendleft((X, distance+1))

for i in range(1, N):
    print(result[i] // 2)