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;

int main() {
    int n , x , y;
    cin>>n>>x>>y;
    vector<vector<int>>g(n + 1);
    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);
    vector<int>ans(n + 1);
    for(int i = 1 ; i  <= n; ++i){
        vector<int>dist(n + 1 , -1);
        queue<int>q;
        q.push(i);
        dist[i] = 0;
        while(!q.empty()){
            int v = q.front();
            q.pop();
            for(int& u : g[v]){
                if(dist[u] == -1){
                    dist[u] = dist[v] + 1;
                    q.push(u);
                }
            }
        }
        for(int k = i + 1; k <= n; ++k){
            ans[dist[k]]++;
        }
    }
    for(int i = 1; i < n; ++i){
        cout<<ans[i]<<"\n";
    }
    return 0;
}