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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>

using namespace std;

int min_distance_counts[2000];

int dist(int pos1, int pos2);

int main(void)
{
    int nodes_count;
    int x, y;

    cin >> nodes_count >> x >> y;
    x -= 1;
    y -= 1;

    for(int i = 0; i < 2000; i++){
        min_distance_counts[i] = 0;
    }

    for(int i = 0; i < nodes_count - 1; i++){
        for(int j = i + 1; j < nodes_count; j++){
            int pure_dist = dist(i, j);
            int xtoy_dist = dist(i, x) + 1 + dist(y, j);
            /*
            int ytox_dist = dist(i, y) + 1 + dist(x, j);

            min_distance_counts[min(min(pure_dist, xtoy_dist), ytox_dist)] += 1;
            */
            min_distance_counts[min(pure_dist, xtoy_dist)] += 1;
        }
    }

    for(int i = 1; i < nodes_count; i++){
        cout << min_distance_counts[i] << endl;
    }

    return 0;
}

int dist(int pos1, int pos2)
{
    if(pos1 >= pos2){
        return pos1 - pos2;
    }else{
        return pos2 - pos1;
    }
}