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 ll long long
#define INF 0x3f3f3f3f
using namespace std;
const ll maxn = 2 * 1e6 + 5;
const ll mod = 1e9 + 7;
int ans[maxn],n,x,y,vis[maxn];
int book[maxn];
vector<int> G[maxn];
void bfs(int x)
{
    book[x] = 1;
    vis[x] = 1;
    queue<int> qu;
    qu.push(x);
    while(qu.size())
    {
        ll t = qu.front();
        qu.pop();
        for(int i = 0; i < G[t].size(); i++)
        {
            ll endd = G[t][i];
            if(vis[endd] == 0)
            {
                vis[endd] = vis[t] + 1;
                if(!book[endd])
                    ans[vis[endd] - 1]++;
                qu.push(endd);
            }
        }
    }
}
int main()
{
    cin >> n >> x >> y;
    for(int i = 1; i <= n - 1; 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 i = 1; i <= n; i++)
    {
        bfs(i);
        for(int i = 1; i <= n; i++)
            vis[i] = 0;
    }
    for(int i = 1; i <= n - 1; i++)
        cout << ans[i] << endl;
    return 0;
}