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 _GLIBCXX_DEBUG
typedef long long ll;
typedef pair<int, int> P;
typedef pair<pair<int, int>, int> PP;
const ll INFL = 1LL << 60;
const ll MOD = 1000000007;
const int INF = 1000000005;
const int MAX = 300000;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll a, ll b) { return (a * b) / GCD(a, b); }
//------------------------------------------------------------------------------------------------------------------------------

int main(void) {
    // code here
    int N, X, Y;
    cin >> N >> X >> Y;
    X--;
    Y--;
    vector<int> ans(N, 0);
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (i >= j) {
                continue;
            }
            int mi = min(j - i, abs(X - i) + abs(Y - j) + 1);
            ans[mi]++;
        }
    }
    for(int i = 1; i < N; i++) {
        cout << ans[i] << endl;
    }
}