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 (3.4.3) (judge verdict: Time Limit Exceeded)

N, X, Y = map(int, input().split())

g = [[]  for i in range(N)]
for i in range(N):
    u = i
    v = i + 1
    if v < N:
        g[u].append(v)
        g[v].append(u)

g[X - 1].append(Y - 1)
g[Y - 1].append(X - 1)

from collections import deque
dis = [[float('Inf')] * N for _ in range(N)]
count = [0] * (N - 1)
def bfs(i):
    d = [False] * N
    que = deque()
    #for j in g[i]:
    que.append(i)
    d[i] = True
    dis[i][i] = 0
    pre = i
    while que:
        p = que.popleft()

        for k in g[p]:
            if d[k]:
                continue
            else:
                d[k] = True
                que.append(k)
                dis[i][k] = dis[i][p] + 1
                dis[k][i] = dis[i][p] + 1
                count[dis[i][k] - 1] += 1 
                #count[dis[k][i] - 1] += 1
        pre = p

for i in range(N):
    bfs(i)

for i in range(N - 1):
    print(count[i] // 2)