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)

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<algorithm>

using namespace std;

int a[2010];
int n , x, y;
int main()
{
	cin >> n >> x >> y;
	for(int i = 1; i <= n-1; i++)
	{
		a[i] = n - i;
		printf("%d ", a[i]);
	}
	cout << endl;
	int l1 = y-1;
	int l2 = n - x;
	int maxl = max(l1, l2);
	int minl = min(l1, l2);
	
	int cnt = 1;
	int d = y - x - 1;
	printf("\n%d %d %d\n", l1, l2, d);
	for(int l = y-x; l <= n-1; l++)
	{
		if(l < minl)
		{
			
			a[l] -= cnt;
			a[l-d] += cnt;
			printf("%d ", cnt);
			for(int i = 1; i <= n-1; i++)
			{
				cout << a[i] << " ";
			}
			cout << endl;
			cnt++;
		}
		else if(l >= minl && l <= maxl)
		{
			a[l] -= cnt;
			a[l-d] += cnt;
			printf("%d ", cnt);
			for(int i = 1; i <= n-1; i++)
			{
				cout << a[i] << " ";
			}
			cout << endl;
		}
		else
		{
			cnt--;
			a[l] -= cnt;
			a[l-d] += cnt;
			printf("%d ", cnt);
			for(int i = 1; i <= n-1; i++)
			{
				cout << a[i] << " ";
			}
			cout << endl;
		}
	}
	for(int i = 1; i <= n-1; i++)
	{
		printf("%d\n", a[i]);
	}
	return 0;
}