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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
class ProgramD
    {
        static void Main(string[] args)
        {
            
            //入力
            string[] a = Console.ReadLine().Split(' ');
            long n = long.Parse(a[0]);
            long x = long.Parse(a[1]);
            long y = long.Parse(a[2]);
            long[] ans = new long[n];

            for(long i = 1;i <= n;i++)
                for(long j = i+1;j <= n;j++)
                {
                    if(j <= x)
                    {
                        ans[j - i]++;
                        continue;
                    }

                    if(y <= i)
                    {
                        ans[j - i]++;
                        continue;
                    }

                    
                    ans[Math.Min(j - i, Math.Abs(y - j)+ Math.Abs(x - i) + 1)]++;

                }

            for(long i = 1;i <= n - 1;i++)
                Console.WriteLine(ans[i]);

            
        }
 
 
    }