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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define E "\n"

using namespace std;
const long long MOD = (long long)1e9 + 7;

int n, x, y, peri, ans;

int fun(int a, int b)
{
	int ret = 0;
	if (b - a <= peri / 2) {
		if (b <= a) {
			ret += a - b + 1;
		}
		int t1 = max(1, b - a);
		int t2 = min(peri / 2, b - 1);
		ret += (t2 - t1 + 1) * 2;
		if (peri % 2 == 0 && t2 == peri / 2) ret--;
	}
	return ret;
}

int main()
{
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> x >> y;
	peri = y - x + 1;
	for (int i = 1; i < n; i++) {
		ans = 0;
		if (i <= peri / 2) ans += peri;
		int tmp = x - 1;
		ans += fun(tmp, i);
		tmp = n - y;
		ans += fun(tmp, i);
		tmp = x - 1 + n - y + 1;
		if (1 < x && y < n && i <= tmp && i > 1) {
			int t1 = max(i, x + 1);
			int t2 = min(tmp, x - 2 + i);
			ans += t2 - t1 + 1;
		}
		cout << ans << E;
	}
	//system("pause");
	return 0;
}