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 (1.42.0) (judge verdict: Accepted)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    dead_code
)]
//use permutohedron::LexicalPermutation as _;
use proconio::input;
//use proconio::marker::Chars;
use proconio::marker::Usize1;
//use std::collections::HashSet;
//use std::collections::HashMap;
//use std::collections::BTreeMap;
use std::cmp::*;
//use std::num;

fn main() {
    input! {
        n:usize,x:Usize1,y:Usize1

    }
    let mut d = vec![vec![0; n]; n];
    let mut e = vec![0; n];
    //println!("{},{}", x, y);
    for i in 0..n - 1 {
        for j in i + 1..n {
            // println!("{},{}", i, j);
            if j <= x || y <= i {
                //     println!("a");
                d[i][j] = j - i;
                e[d[i][j]] += 1;
                continue;
            }
            if i <= x && y <= j {
                //  println!("b");
                d[i][j] = x - i + j - y + 1;
                e[d[i][j]] += 1;
                continue;
            }
            if i <= x && j < y {
                //    println!("c");
                d[i][j] = x - i + min(j - x, y - j + 1);
                e[d[i][j]] += 1;
                continue;
            }
            if x < i && y <= j {
                //   println!("d");
                d[i][j] = j - y + min(i - x + 1, y - i);
                e[d[i][j]] += 1;
                continue;
            }
            if x < i && j < y {
                //  println!("e");

                d[i][j] = min(j - i, i - x + y - j + 1);
                e[d[i][j]] += 1;
            }
            // println!("f");
        }
    }
    for i in 0..n {
        //   println! {"{:?}", d[i]}
    }
    for i in e.iter().skip(1) {
        println!("{}", i);
    }
}