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)]

use std::cmp;
use std::cmp::{max, min, Ordering};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::io::stdin;
use std::ops::*;
use std::process::exit;
use std::str::FromStr;

pub const MOD: usize = 1000000007;

fn main() {
    let (n, x, y): (usize, i64, i64) = input_three();
    let mut ans = vec![0; n - 1];

    for i in 0..n - 1 {
        for j in i + 1..n {
            let d = min(
                j - i,
                (x - 1 - i as i64).abs() as usize + (y - 1 - j as i64).abs() as usize + 1,
            );
            // dbg!(i, j, d);
            // println!();
            ans[d - 1] += 1;
        }
    }

    for i in ans {
        println!("{}", i);
    }
}

pub fn input<T: FromStr>() -> T {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().ok().unwrap()
}

pub fn input_vec<T: FromStr>() -> Vec<T> {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|x| x.parse().ok().unwrap())
        .collect::<Vec<_>>()
}

pub fn input_lines<T: FromStr>(l: usize) -> Vec<T> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input());
    }
    res
}

pub fn input_lines_two<T: FromStr, U: FromStr>(l: usize) -> Vec<(T, U)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_two());
    }
    res
}

pub fn input_lines_three<T: FromStr, U: FromStr, V: FromStr>(l: usize) -> Vec<(T, U, V)> {
    let mut res = Vec::new();
    for _i in 0..l {
        res.push(input_three());
    }
    res
}

pub fn input_two<T: FromStr, U: FromStr>() -> (T, U) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
    )
}

pub fn input_three<T: FromStr, U: FromStr, V: FromStr>() -> (T, U, V) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse::<T>().ok().unwrap(),
        s[1].parse::<U>().ok().unwrap(),
        s[2].parse::<V>().ok().unwrap(),
    )
}

pub fn input_four<T: FromStr, U: FromStr, V: FromStr, W: FromStr>() -> (T, U, V, W) {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    let s: Vec<&str> = buf.split_whitespace().collect();
    (
        s[0].parse().ok().unwrap(),
        s[1].parse().ok().unwrap(),
        s[2].parse().ok().unwrap(),
        s[3].parse().ok().unwrap(),
    )
}