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 <stdio.h>
#include <vector>
#include <iostream>
#include <queue>
#include <bits/stdc++.h>
 
#define rep(i,n) for (int i=0;i<(n);++i)
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<double> vd;
typedef pair<int, int> pi;

const int INF = 1001001001;

int main() {
  int n, x, y;
  cin >> n >> x >> y;
  vector<vi> g(n, vi(n, INF));
  rep(i, n){
    for(int j=i+1;j<n;j++){
      g[i][j] = j - i;
      g[j][i] = j - i;
    }
  }  
  g[x-1][y-1] = 1;
  g[y-1][x-1] = 1;

  vi kv;
  kv.push_back(x-1);
  kv.push_back(y-1);
  rep(_k, 2){
    int k = kv[_k];
    rep(i, n){
      rep(j, n){
        g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
      }
    }
  }
  vi hist(n);
  rep(i, n){
    for(int j=i+1;j<n;j++){
      hist[g[i][j]]++;
    }
  }
  for(int i=1;i<n;i++){
    cout << hist[i] << endl;
  }
  return 0;
}