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>

#define repd(i, a, b) for (ll i = (a); i < (b); i++)
#define repb(i, n) for (ll i = (n)-1; i >= 0; i--)
#define rep(i, n) repd(i, 0, n)

using namespace std;

using ll = long long;
using ul = unsigned long long;
ll mod = 1000000007;

int main() {
source:
    ll n, x, y;
    cin >> n >> x >> y;
    ll ans[n + 1];
    memset(ans, 0, sizeof ans);

    repd(s, 1, n + 1) {
        repd(t, s, n + 1) {
            ll path0 = abs(s - t);
            ll path1 = abs(s - x) + 1 + abs(t - y);
            ll path2 = abs(s - y) + 1 + abs(t - x);
            ll temp = min(min(path0, path1), path2);
            ans[temp]++;
        }
    }

    repd(i, 1, n) { cout << ans[i] << endl; }

    return 0;
}