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 int long long
#define INF (1LL<<30)

vector<int> adjlist[2005];
vector<int> d[2005];

signed main() {
  int N,X,Y;
  cin >> N >> X >> Y;
  X--; Y--;
  adjlist[0].push_back(1);
  adjlist[N-1].push_back(N-2);
  for(int i=1; i<N-1; i++) {
    adjlist[i].push_back(i-1);
    adjlist[i].push_back(i+1);
  }
  adjlist[X].push_back(Y);
  adjlist[Y].push_back(X);

  for(int s = 0; s<N; s++) {
    auto &ds = d[s];
    ds.resize(N);
    fill(ds.begin(), ds.end(), INF);
    queue<int> q;
    q.push(s);
    ds[s] = 0;
    while(q.size()) {
      int t = q.front(); q.pop();
      for(auto tt : adjlist[t]) {
        if(ds[tt] != INF) continue;
        ds[tt] = ds[t] + 1;
        q.push(tt);
      }
    }
  }

  map<int, int> ans;

//  cerr << "=========" << endl;
  for(int s=0; s<N; s++) {
    for(int t=0; t<N; t++) {
//      cerr << s << "\t" << t << "\t" << d[s][t] << endl;
      ans[d[s][t]]++;
    }
  }
//  cerr << "=========" << endl;

  for(int k=1; k<=N-1; k++) {
    cout << ans[k]/2 << endl;
  }  
}