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 (2.7.6) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>

using namespace std;

int main()
{
	int n,x,y;
	cin >> n >> x >> y;
	vector<vector<int> > d(n, vector<int>(n,-1));
	x--;y--;
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			d[i][j] = min(min(abs(i-j), abs(i-x) + abs(j-y)+1), abs(i-y)+abs(j-x)+1);
		}
	}

	vector<int> count(n,0);
	for(int i = 0; i < n;i++)
	{
		for(int j = 0; j < n; j++)
		{
				// cout << d[i][j] << " " ;
				count[d[i][j]]++;
		}
		// cout << endl;
	}

	for(int i = 1; i < n; i++)
		cout << count[i]/2 << endl;

}