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>
#define maxn 2005
using namespace std;
vector<int> G[maxn];
int vis[maxn],dis[maxn];
int ans[maxn];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,x,y;
	cin >> n >> x >> y;
	for(int i=1;i<n;i++) {
		G[i].push_back(i + 1);
		G[i + 1].push_back(i);
	}
	G[x].push_back(y);
	G[y].push_back(x);
	for(int st=1;st<=n;st++) {
		dis[st] = 0;
		vis[st] = st;
		queue<int> q;
		q.push(st);
		while(!q.empty()) {
			int u = q.front();
			q.pop();
			for(int v : G[u])
				if(vis[v] != st) {
					vis[v] = st;
					dis[v] = dis[u] + 1;
					q.push(v);
				}
		}
		for(int ed=1;ed<=n;ed++)
			ans[dis[ed]]++;
	}
	for(int i=1;i<n;i++)
		cout << ans[i] / 2 << '\n';
 	return 0;
}