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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int main()
{
	int n, x, y;
	scanf("%d %d %d", &n, &x, &y);
	int i, j;
	int p = y - x - 1;
	long long int s[2001] = {0};
	
	for(i = 1; i < n; i++)
	{
		for(j = i + 1; j <= n; j++)
		{
			if(j >= y && i <= x)
			{
				if(i > x && j >= y && j - i - p >= i - x + j - y + 1)
				{
					s[i - x + j - y + 1]++;
				}
				else if(j < y && i <= x && j - i - p >= x - i + y - j + 1)
				{
					s[x - i + y - j + 1]++;
				}
				else s[j - i - p]++;
			}
			else
			{
				s[j - i]++;
			}
		}
	}
	for(i = 1; i < n; i++)
	{
		printf("%lld\n", s[i]);
	}
}