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 Rust (1.15.1) (judge verdict: Accepted)

use std::collections::VecDeque;
use std::io::stdin;

const INF: usize = 100000000;

fn main() {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    let mut itr = s.split_whitespace().map(|e| e.parse::<usize>().unwrap());
    let n = itr.next().unwrap();
    let x = itr.next().unwrap() - 1;
    let y = itr.next().unwrap() - 1;

    let mut ans = vec![0; n];
    for i in 0..n {
        let mut q = VecDeque::new();
        let mut dist = vec![INF; n];
        q.push_back(i);
        dist[i] = 0;
        while !q.is_empty() {
            let v = q.pop_front().unwrap();
            let d = dist[v];
            if v > 0 && dist[v - 1] == INF {
                dist[v - 1] = d + 1;
                q.push_back(v - 1);
            }
            if v + 1 < n && dist[v + 1] == INF {
                dist[v + 1] = d + 1;
                q.push_back(v + 1);
            }

            if v == x && dist[y] == INF {
                dist[y] = d + 1;
                q.push_back(y);
            }

            if v == y && dist[x] == INF {
                dist[x] = d + 1;
                q.push_back(x);
            }
        }

        for i in 0..n {
            ans[dist[i]] += 1;
        }
    }

    for i in 1..n {
        println!("{}", ans[i] / 2);
    }
}