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)

use std::cmp::min;

fn main() {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line);
    let input: Vec<i64> = line.split_whitespace().map(|a| a.parse().unwrap()).collect();
    let (n, x, y) = (
        input[0], input[1] - 1, input[2] - 1
    );
    let mut count: Vec<i64> = vec![0; (n as usize)];
    for j in 0..n {
        for i in 0..j {
            if j <= x || y <= i {
                count[(j - i) as usize] += 1
            } else if i <= x && y <= j {
                count[(x - i + j - y + 1) as usize] += 1
            } else if i <= x && x <= j && j <= y {
                count[(x - i + min(j - x, y - j + 1)) as usize] += 1
            } else if x <= i && x <= y && y <= j {
                count[(j - y + min(y - i, i - x + 1)) as usize] += 1
            } else if x <= i && j <= y {
                count[(min(j - i, i - x + y - j + 1)) as usize] += 1
            }
        }
    }
    for i in 1..(n as usize)  {
        println!("{}", count[i]);
    }
}