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())


graph = {}

for i in range(n):
    if(i == 0):
        graph[i] = [1]
    elif(i == n-1):
        graph[i] = [n-2]
    else:
        graph[i] = [i-1,i+1]


#print(graph)

xx = graph[x-1]
xx.append(y-1)
graph[x-1] = xx

yy = graph[y-1]
yy.append(x-1)
graph[y-1] = yy

#print(graph)

def bfs(graph,st):
    q = deque()
    q.append(st)
    dist = [-1]*n
    dist[st] = 0
    
    while len(q) > 0:
    # キューの先頭（左端）からnodeを取り出す
        v = q.popleft()
        
        for i in graph[v]:
            if (dist[i] != -1):
                continue
            dist[i] = dist[v] + 1;
            q.append(i)
    
    return dist

res = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
    res[i] = bfs(graph,i)

#print(res)

for k in range(1,n):
    ress = 0
    for i in range(n):
        ress += res[i].count(k)
    
    
    print(ress//2)