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)

fn main() {
    proconio::input! {
        n: i64,
        x: i64,
        y: i64,
    }
    let mut hm = std::collections::HashMap::new();
    for i in 1..n {
        for j in i + 1..=n {
            let not_use_xy = j - i;
            let use_x_y = (x - i).abs() as i64 + 1 + (j - y).abs() as i64;
            let use_y_x = (y - i).abs() as i64 + 1 + (j - x).abs() as i64;
            let min = [not_use_xy, use_x_y, use_y_x]
                .iter()
                .cloned()
                .min()
                .unwrap();
            *hm.entry(min).or_insert(0) += 1;
        }
    }
    for k in 1..n {
        println!("{}", hm.get(&k).or(Some(&0)).unwrap());
    }
}