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)

import heapq
import sys
def main():
    input = sys.stdin.readline

    def dijkstra(s,edge):
        #始点sから各頂点への最短距離
        d = [10**20] * n
        used = [True] * n #True:未確定
        d[s] = 0
        used[s] = False
        edgelist = []
        for a,b in edge[s]:
            heapq.heappush(edgelist,a*(10**6)+b)
        while len(edgelist):
            minedge = heapq.heappop(edgelist)
            #まだ使われてない頂点の中から最小の距離のものを探す
            if not used[minedge%(10**6)]:
                continue
            v = minedge%(10**6)
            d[v] = minedge//(10**6)
            used[v] = False
            for e in edge[v]:
                if used[e[1]]:
                    heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
        return d

    n,x,y = map(int,input().split())

    edge = [[] for i in range(n)]
    #edge[i] : iから出る道の[重み,行先]の配列
    for i in range(n-1):
        edge[i].append((1,i+1))
        edge[i+1].append((1,i))
    edge[x-1].append((1,y-1))
    edge[y-1].append((1,x-1))

    for i in range(n):
        edge[i] = tuple(edge[i])
    edge = tuple(edge)

    ans = [0 for _ in range(n)]
    for i in range(n):
        #dijkstra(始点,edge)
        res = dijkstra(i,edge)
        for j in range(i+1,n):
            ans[res[j]] += 1

    for i in range(1,n):
        print(ans[i])

if __name__ == '__main__':
    main()