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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function main(input) {
  const inp = input.split(' ');
	const n = Number(inp[0]);
	const x = Number(inp[1]);
	const y = Number(inp[2]);
	
  function dist(a, b) {
		return Math.min(b-a, Math.abs(x-a)+Math.abs(y-b)+1);
	}
  
  for(var i=1;i<n; i++) {
    var cnt = 0;
		
		function log(a,b) { 
			cnt++; 
		}
    
    for(var a=1; a<=n-i; a++) {
			const s = a+i;
			if(s <= n && dist(a,s) === i)
				log(a,s);
			
			const rmn = i - 1 - Math.abs(x-a);
			
			if(rmn > 0) {
				const prev = y - rmn;
				if(prev !== s && dist(a, prev) === i)
					log(a,prev); 
				
				const next = y + rmn;
				if(next !== s && next <= n && dist(a, next) === i)
					log(a, next);
			} else if(rmn === 0 && y !== s && dist(a, y) === i)
				log(a, y);
		}
		
    console.log(cnt);
  }
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));