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)

from collections import deque
n,x,y=map(int,input().split())
edge=[[] for _ in range(n)]
edge[x-1].append(y-1)
edge[y-1].append(x-1)
for i in range(n-1):
  edge[i].append(i+1)
  edge[i+1].append(i)
visited=[[-1]*n for _ in range(n)]
def bfs(s):
  q=deque([s])
  visited[s][s]=0
  while q:
    x=q.popleft()
    for nx in edge[x]:
      if visited[s][nx]==-1:
        visited[s][nx]=visited[s][x]+1
        q.append(nx)
for i in range(n):
  bfs(i)
ans=[0]*n
for i in range(n):
  for j in range(n):
    if i<j:
      continue
    if visited[i][j]>=1:
      ans[visited[i][j]]+=1
for i in range(1,n):
  print(ans[i])