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;
using ll = long long;
using P = pair<int,int>;
#define rep(i,n) for(ll i = 0; i < (ll)(n); i++)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = 1 << 30;
vector<int> to[2020];
int n,x,y;
int ans[2020];

void bfs(int i){
    vector<int> dist(n,INF);
    dist[i] = 0;
    queue<int> q;
    q.push(i);
    while(!q.empty()){
        int v = q.front(); q.pop();
        for(auto tv : to[v]){
            if(dist[tv]!=INF) continue;
            dist[tv] = dist[v]+1;
            ans[dist[tv]]++;
            q.push(tv);
        }
    }
    
}

int main(){
    cin >> n >> x >> y;
    x--; y--;
    rep(i,n-1) {
        to[i].push_back(i+1);
        to[i+1].push_back(i);
    }
    to[x].push_back(y);
    to[y].push_back(x);
    rep(i,n){
        bfs(i);
    }
    for(int i=1; i<=n-1; i++){
        cout << ans[i]/2 << endl;
    }
    return 0;
}