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;

fn bfs(graph: &Vec<Vec<usize>>, n: usize, start: usize) -> Vec<Option<usize>> {
    let mut distance = vec![None; n];
    let mut q = VecDeque::new();
    q.push_back((start, 0));
    while !q.is_empty() {
        let (current, d) = q.pop_front().unwrap();
        if distance[current].is_some() {
            continue;
        }
        distance[current] = Some(d);
        for &next in graph[current].iter() {
            q.push_back((next, d + 1));
        }
    }
    distance
}

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(&mut std::io::stdin(), &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
        ($t:ty) => {
            get().parse::<$t>().unwrap()
        };
        () => {
            get!(i64)
        };
    }

    let (n, mut x, mut y) = (get!(usize), get!(usize), get!(usize));
    x -= 1;
    y -= 1;
    let mut graph = vec![vec![]; n];
    for i in 0..n - 1 {
        graph[i].push(i + 1);
        graph[i + 1].push(i);
    }
    graph[x].push(y);
    graph[y].push(x);

    let mut ans = vec![0; n - 1];
    for i in 0..n {
        let distance = bfs(&graph, n, i);
        for &d in distance.iter() {
            if d.unwrap() == 0 {
                continue;
            }
            ans[d.unwrap() - 1] += 1;
        }
    }

    for &x in ans.iter() {
        println!("{}", x / 2);
    }
}