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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<usize>();
    let (n, x, y) = (v[0], v[1] - 1, v[2] - 1);
    let mut edges = vec![vec![]; n];
    for i in 0..n - 1 {
        edges[i].push(Edge { to: i + 1, cost: 1 });
        edges[i + 1].push(Edge { to: i, cost: 1 });
    }
    edges[x].push(Edge { to: y, cost: 1 });
    edges[y].push(Edge { to: x, cost: 1 });

    let mut answers = vec![0; n];
    for i in 0..n {
        let costs = solve(&edges, i);
        for cost in costs {
            answers[cost as usize] += 1;
        }
    }
    for &ans in answers.iter().skip(1) {
        println!("{}", ans / 2);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

use std::cmp::Ordering;
use std::collections::BinaryHeap;
const INF: i64 = 100000_00000_00000;

#[derive(PartialEq, Debug)]
struct MinInt {
    value: i64,
}

impl Eq for MinInt {}

impl PartialOrd for MinInt {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        other.value.partial_cmp(&self.value)
    }
}

impl Ord for MinInt {
    fn cmp(&self, other: &MinInt) -> Ordering {
        other.value.cmp(&self.value)
    }
}

fn make_pair(x: i64, y: usize) -> (MinInt, usize) {
    (MinInt { value: x }, y)
}

#[derive(Debug, Clone)]
struct Edge {
    to: usize,
    cost: i64,
}

fn solve(edges: &Vec<Vec<Edge>>, start_idx: usize) -> Vec<i64> {
    let num_apexes = edges.len();
    let mut d = vec![INF; num_apexes];
    d[start_idx] = 0;
    let mut que = BinaryHeap::new();
    que.push(make_pair(0, start_idx));

    while let Some((u, v)) = que.pop() {
        if d[v] < u.value {
            continue;
        }
        for e in &edges[v] {
            if d[v] != INF && d[e.to] > d[v] + e.cost {
                d[e.to] = d[v] + e.cost;
                que.push(make_pair(d[e.to], e.to));
            }
        }
    }
    d
}