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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
using ll = long long int;
constexpr ll M = 1e9 + 7;
inline ll sum_1(const ll n) { return ((n * (n + 1)) / 2) % M; }
inline ll sum_2(const ll n) { return ((n * (n + 1) * (2 * n + 1)) / 6) % M; }
inline ll sum_3(const ll n) { return (((n * (n + 1)) / 2) % M * ((n * (n + 1)) / 2) % M) % M; }
inline ll sum_1(const ll i, const ll j) { return (sum_1(j) - sum_1(i - 1) + M) % M; }
inline ll sum_2(const ll i, const ll j) { return (sum_2(j) - sum_2(i - 1) + M) % M; }
inline ll sum_3(const ll i, const ll j) { return (sum_3(j) - sum_3(i - 1) + M) % M; }

struct MyHash {
    size_t operator()(const pair<int,int> &x) const {
        return x.first + x.second;
    }
};

void solve() {
    int n, x, y;
    cin >> n >> x >> y;
    unordered_map<pair<int,int>, int, MyHash> m;

    for (int i = 1; i <= n; ++i) {
        for (int j = i+1; j <= n; ++j) {
            m[{i,j}] = min({
                abs(i-j),
                abs(x-i) + 1 + abs(j-y),
                abs(x-j) + 1 + abs(i-y),
            });
        }
    }
    for (int k = 1; k < n; ++k) {
        int ans = 0;
        for (auto &i: m) {
            if (i.second==k) {
                ++ans;
            }
        }
        cout << ans << endl;
    }

}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifdef D_LOCAL
        freopen("in.in", "r", stdin);
        freopen("out.out", "w", stdout);
    #endif

    solve();

    return 0;
}