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: Time Limit Exceeded)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::io::Read;
use std::io::{stdout, BufWriter, Write};
use std::cmp::min;

use std::collections::HashMap;
 
fn main() {
    let mut buf = String::new();
    
    let out = stdout();
    let mut out = BufWriter::new(out.lock());
 
    std::io::stdin().read_to_string(&mut buf).unwrap();
 
    let mut iter = buf.split_whitespace();
 
    let n: usize = iter.next().unwrap().parse().unwrap();
    let x: usize = iter.next().unwrap().parse().unwrap();
    let y: usize = iter.next().unwrap().parse().unwrap();
    
    let mut d = vec![vec![2*n; n]; n];
    
    for i in 0..n-1 {
        d[i][i+1] = 1;
        d[i+1][i] = 1;
    }
    
    d[x-1][y-1] = 1;
    d[y-1][x-1] = 1;
    
    //println!("{:?}",d);
    
    for k in 0..n {
        for i in 0..n {
            for j in 0..n {
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
            }
        }
    }
    
    let mut ans = vec![0; n-1];
    for i in 0..n-1 {
        for j in i+1..n {
            ans[d[i][j]-1] += 1;
        }
    }
    
    //println!("{:?}",tmp);
    //println!("{:?}",d);
    for item in &ans {
        println!("{:?}",item);
    }
    
}