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 < n; i++)
#define rep1(i, n) for(int i = 1; i < n+1; i++)
#define sort(A) sort(A.begin(),A.end())
#define reverse(A) reverse(A.begin(),A.end());

typedef long long ll;

int main(){
  int n,x,y;
  cin >> n >> x >> y;
  
  vector<vector<int>> to(n+1,vector<int>(0));
  
  rep1(i,n-1){
    to[i].push_back(i+1);
    to[i+1].push_back(i);
  }
  to[x].push_back(y);
  to[y].push_back(x);
  
  vector<int> ans(n,0);
  
  rep1(i,n){
      
    //こっからBFS
    int INF = 1000000000;
    vector<int> d(n+1,INF);
    vector<bool> seen(n+1,false);
    queue<int> q;
    q.push(i);
    d[i] = 0;
    while(!q.empty()){
      int now = q.front();
      q.pop();
      if(seen[now]) continue;
      seen[now] = true;
      for(int next : to[now]){
        d[next] = min(d[now]+1,d[next]);
        q.push(next);
      }
    }//while
    
    for(int j=i+1;j<=n;j++) ans[d[j]] ++;
  }//i
  
  rep1(i,n-1) cout << ans[i] << endl;
}