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;
#define rep(i, n) for(int i=0; i<n; i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
#define repr(i, n) for(int i=n-1; i>=0; i--)
#define repr1(i, n) for(int i=n; i>=1; i--)
#define all(v) v.begin(),v.end()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = 1e9+7;
const double EPS = 1e-10;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int n, x, y;
ll dist[2010][2010];
ll cnt[2010];

int main() {
    cin >> n >> x >> y;
    x--; y--;
    vector<vector<int>> g(n);
    rep(i, n) {
        if (i > 0) g[i-1].push_back(i);
        if (i < n-1) g[i+1].push_back(i);
    }
    g[x].push_back(y);
    g[y].push_back(x);

    rep(i, n) rep(j, n) dist[i][j] = LLINF;
    rep(i, n) dist[i][i] = 0;
    rep(i, n) {
        priority_queue<pli, vector<pli>, greater<pli>> q;
        q.emplace(0, i);
        while (!q.empty()) {
            pli top = q.top();
            q.pop();
            int u = top.second;
            for (auto v: g[u]) {
                if (chmin(dist[i][v], dist[i][u]+1)) {
                    q.emplace(dist[i][v], v);
                }
            }
        }
    }

    rep(i, n) rep(j, n) {
        if (i >= j) continue;
        cnt[dist[i][j]]++;
    }
    for (int k = 1; k <= n-1; k++) {
        cout << cnt[k] << endl;
    }

}