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 <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <queue>
#include <set>
using ll = long long;
using graph = std::vector<std::vector<ll>>;
using wGraph = std::vector<std::vector<std::pair<ll,ll>>>;
#define rep(i,n) for (int i=0; i < int(n); i++)

using namespace std;

struct edge {
    ll to;
    ll cost;
};

// <最短距離, 頂点の番号>
using P = pair<ll, ll>;

ll V;

#define MAX_V 100005
#define INF (1ll << 50)
vector<edge> G[MAX_V];
ll d[MAX_V];

void dijkstra(ll s) {
    priority_queue<P, vector<P>, greater<P> > que;
    fill(d, d+V, INF);
    d[s] = 0;
    que.push(P(0, s));

    while (!que.empty()) {
        P p = que.top();
        que.pop();
        ll v = p.second;
        if (d[v] < p.first) continue;

        for (ll i=0; i<G[v].size(); ++i) {
            edge e = G[v][i];
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P(d[e.to], e.to));
            }
        }
    }
}

int main() {

    cin >> V;
    ll X,Y;
    cin >> X >> Y;

    X--;Y--;
    for (ll i=0; i<V - 1; ++i) {
        auto e = edge{i + 1,1};
        G[i].push_back(e);
        e = edge{i,1};
        G[i + 1].push_back(e);
    }
    auto e = edge{Y,1};
    G[X].push_back(e);
    e = edge{X,1};
    G[Y].push_back(e);
    dijkstra(X);
    for (int i = 0; i < V; ++i) {
        //cout << d[i] << endl;
    }
    //cout << endl;
    ll count[V];
    for (int i = 0; i < V; ++i) {
        count[i] = 0;
    }
    for (int si = 0; si < V-1; ++si) {
        for (int gi = si + 1; gi < V; ++gi) {
            ll l = min((ll)(gi-si),d[gi] + d[si]);
            count[l - 1]++;
        }
    }
    for (int i = 0; i < V - 1; ++i) {
        cout << count[i] << endl;
    }
}