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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main()
{
	ll n,arr[2];
	cin>>n>>arr[0]>>arr[1];
	vector< pair<ll, pair<ll,ll> >  > v(n+n);
	
	for(int i=0;i<n;i++){ll x; cin>>x; v[i] = make_pair(x, make_pair(i, 0));}
	for(int i=0;i<n;i++){ll x; cin>>x; v[i] = make_pair(x, make_pair(i, 1));}
	
	sort(v.begin(), v.end());
	reverse(v.begin(), v.end());
	
	vector<bool> f(n); 
	for(int i=0;i<n;i++) f[i] = true;
	
	ll sum=0;
	for(int i=0; i<v.size(); i++)
	{
		if(f[v[i].second.first])
		{
			if(arr[v[i].second.second])
			{
				sum+= v[i].first;
				arr[v[i].second.second]--;
				f[v[i].second.first] = false;
			}
		}
	}
	
	cout<<sum<<endl;
}