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 <stdio.h>
#include <algorithm>
using namespace std;
int X, Y;
int n, m, u, v, w, in[2001][2001], cnt[2001];
int main()
{
	scanf("%d%d%d", &n, &X, &Y); X--, Y--;
	for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (i - j)in[i][j] = 1e9;
	for (int i = 0; i < n - 1; i++) in[i][i + 1] = in[i + 1][i] = 1;
	for (int i = 0; i < n-1; i++) for (int j = i+1; j < n; j++) in[i][j] = in[j][i] = j - i;
	for (int i = 0; i <= X; i++)for (int j = Y; j < n; j++) {
		in[i][j]-=Y-X-1;
	}
	in[X][Y] = in[Y][X] = 1;
	for (int i = 0; i <= X; i++)for (int j = X + 1; j < Y; j++) {
		in[i][j] = min(in[i][j], in[i][X] + in[X][Y] + in[j][Y]);
	}
	for (int i = X + 1; i < Y; i++)for (int j = Y; j < n; j++) {
		in[i][j] = min(in[i][j], in[X][i] + in[X][Y] + in[Y][j]);
	}

	for (int i = X + 1; i < Y; i++)for (int j = X + 1; j < Y; j++) {
		in[i][j] = min(in[i][j], in[X][i] + in[X][Y] + in[Y][j]);
	}


	for (int i = 0; i < n - 1; i++) for (int j = i + 1; j < n; j++)
		cnt[in[i][j]]++;
	for (int k = 0; k < n - 1; k++)
		printf("%d\n", cnt[k + 1]);
	return 0;
}