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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

require "bigdecimal"
require "bigdecimal/util"

def gi() gets.to_i end
def gs() gets.chomp end
def gim() gets.split.map(&:to_i) end
def gsm() gets.split.map(&:chomp) end
def YesNo1(bool, yes="Yes", no="No") puts (bool ? yes : no) end
def YESNO2(bool, yes="YES", no="NO") puts (bool ? yes : no) end

#####
n,x,y = gim
distDic = {}
resDic = {}
(1..(n-1)).each do |fir|
  ((fir+1)..n).each do |sec|
    dist = 0
    if fir <= x && y <= sec
      dist = x-fir + sec-y + 1
    else
      if fir > x && y > sec
        dist = [sec - fir, fir-x + y-sec +1].min
      else
        if y > sec
          dist = [sec - fir, x-fir + y-sec +1].min
        end
        if fir > x
          dist = [sec - fir, fir-x + sec-y +1].min
        end
        end
    end
    distDic[[fir,sec]] = dist
    resDic[dist] = 0 if resDic[dist] == nil
    resDic[dist] += 1 #ここ計算量食ってるかも
  end
end
# puts (1 <= x && y <= 3)
#
# print distDic
# puts
# print distDic.filter{|k,v| v == 1}
# puts
# print distDic[[4,6]]
# print resDic

(n-1).times do |i|
  if resDic[i+1] == nil
    puts 0
  else
    puts resDic[i+1]
  end
end