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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
typedef long long int ll;
typedef pair<int, int> P;

int n, x, y, dist[2020], k[2020];

void bfs(int s)
{
    //初期化
    rep(i, n)
    {
        dist[i] = -1;
    }
    dist[s] = 0;
    queue<int> que;
    que.push(s);

    //本体
    while(que.size())
    {
        auto p = que.front();
        que.pop();
        if(p + 1 < n && dist[p + 1] == -1)
        {
            que.push(p + 1);
            dist[p + 1] = dist[p] + 1;
        }
        if(p - 1 >= 0 && dist[p - 1] == -1)
        {
            que.push(p - 1);
            dist[p - 1] = dist[p] + 1;
        }
        if(p == x && dist[y] == -1)
        {
            que.push(y);
            dist[y] = dist[p] + 1;
        }
    }

    for(int i = s; i < n; i++)
    {
        k[dist[i]]++;
    }

    return;
}

int main()
{
    cin >> n >> x >> y;
    x--; y--;
    rep(i, n)
    {
        bfs(i);
    }
    
    rep(i, n - 1)
    {
        cout << k[i + 1] << endl;
    }
    
    
    return 0;
}