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;
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9 + 5;

int main()
{
  int n, x, y;
  cin >> n >> x >> y;
  x--, y--;
  vector<int> ans(n);
  for (int i = 0; i < n; i++) {
    vector<int> dist(n, INF);
    queue<int> q;
    q.push(i);
    dist[i] = 0;
    auto push = [&](int v, int d) {
      if (dist[v] != INF) return;
      dist[v] = d;
      q.push(v);
    };
    while(!q.empty()) {
      int v = q.front(); q.pop();
      int d = dist[v];
      if (v - 1 >= 0) push(v - 1, d + 1);
      if (v + 1 < n) push(v + 1, d + 1);
      if (v == x) push(y, d + 1);
      if (v == y) push(x, d + 1);
    }
    for (int i = 0; i < n; i++) {
      ans[dist[i]]++;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << ans[i] / 2 << endl;
  }
  return 0;
}