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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

template<typename T> class Dijkstra {
public:
  struct edge {
    int to; T cost;
  };
  int V;
  vector<vector<edge>> G;
  vector<T> dist;
  using pti = pair<T,int>;
  Dijkstra(int n) : V(n), G(V), dist(V, numeric_limits<T>::max()) {}
  void add(int u, int v, T cost) {
    G.at(u).push_back((edge) {v, cost});
    G.at(v).push_back((edge) {u, cost});
  }
  void solve(int s) {
    priority_queue<pti, vector<pti>, greater<pti>> Q;
    dist.at(s) = 0;
    Q.push(pti(0, s));
    while (!Q.empty()) {
      pti p = Q.top();
      Q.pop();
      int v = p.second;
      if (dist.at(v) < p.first) continue;
      for (auto &w : G.at(v)) {
        if (dist.at(w.to) > dist.at(v) + w.cost) {
          dist.at(w.to) = dist.at(v) + w.cost;
          Q.push(pti(dist.at(w.to), w.to));
        }
      }
    }
  }
};

int main() {
  int N, X, Y;
  cin >> N >> X >> Y;
  X--, Y--;
  Dijkstra<int> D(N);
  for (int i = 0; i + 1 < N; i++) {
    D.add(i, i + 1, 1);
  }
  D.add(X, Y, 1);
  vector<int> ans(N);
  for (int i = 0; i < N; i++) {
    Dijkstra<int> tmp = D;
    tmp.solve(i);
    for (auto d : tmp.dist) ans.at(d)++;
  }
  for (int i = 1; i < N; i++) {
    cout << ans.at(i) / 2 << "\n";
  }
}