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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define ll long long
using namespace std;

int x, y, N, ans[ 2003 ], ara[ 200000 + 3 ], hi;
vector <int> adj[ 2003 ];
int vis[ 2003 ], lev[ 2003 ];


void solve(int s)
{
  queue <int> q;
  q.push(s);
  for (int i=0; i<N+1; i++) {
    lev[ i ] = 0;
    vis[ i ] = 0;
  }
  vis[ s ] = 1;
  while ( !q.empty() ) {
    int u = q.front();
    for (int i=0; i<adj[ u ].size(); i++) {
      if ( !vis[ adj[ u ][ i ] ] ) {
       	q.push( adj[ u ][ i ] );
        vis[ adj[ u ][ i ] ] = 1;
        lev[ adj[ u ][ i ] ] = lev[ u ] + 1;
    	ans[ lev[ adj[ u ][ i ] ] ]++;// cout << lev[ adj[ u ][ i ] ] << " ";
      }
    }
    q.pop();
  }
}

int main()
{
  cin >> N >> x >> y;
  
  for (int i=0; i<N-1; i++){
    adj[ i ].push_back( i + 1 );
    adj[ i + 1 ].push_back( i );
  }
  
  adj[ x - 1 ].push_back( y - 1 );
  adj[ y - 1 ].push_back( x - 1 );
 
  for (int i=0; i<N; i++){
    solve(i);//cout << endl;
  }
  
  for (int i=1; i<N; i++)
  	cout << ans[ i ] / 2 << "\n";
  
  return 0;
}