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 <iostream>
#include <vector>
#include <queue>
using namespace std;

int N, X, Y, ans[1 << 18], dist[1 << 18];
vector<int> G[1 << 12];

int main() {
	cin >> N >> X >> Y;
	G[X].push_back(Y);
	G[Y].push_back(X);
	for (int i = 1; i <= N - 1; i++) {
		G[i].push_back(i + 1);
		G[i + 1].push_back(i);
	}
	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= N; j++) dist[j] = (1 << 30);
		dist[i] = 0; queue<int>Q; Q.push(i);
		while (!Q.empty()) {
			int pos = Q.front(); Q.pop();
			for (int j : G[pos]) {
				if (dist[j] != (1 << 30)) continue;
				dist[j] = dist[pos] + 1;
				Q.push(j);
			}
		}
		for (int j = i + 1; j <= N; j++) ans[dist[j]]++;
	}
	for (int i = 1; i <= N - 1; i++) cout << ans[i] << endl;
	return 0;
}