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 ll long long
#define vec vector<ll>
#define sortv(a); sort(a.begin(), a.end());
#define rev(a); reverse(a.begin(), a.end());
#define FOR(i,a,b) for (ll i = (ll)(a); i < (b); i++)
#define ROF(i,a,b) for (ll i = (ll)(b)-1; i >= (a); i--)

ll MOD = 1000000007;
ll INF = 1LL << 60;

/*****************************************************************************/
ll n,x,y;

void solve() {
	map<int,int> mp;
	FOR(i,0,n-1) {
		FOR(j,i+1,n) {
			ll d = j-i;
			d=min(d,abs(x-i)+abs(y-j)+1);
			d=min(d,abs(y-i)+abs(x-j)+1);
			mp[d]++;
		}
	}
	FOR(k,1,n) { // 1 to n-1
		cout<<mp[k]<<'\n';
	}
}

int main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin>>n>>x>>y;
	x--;y--;
	solve();
}