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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <deque>

#define rep(i,n) for(int (i)=0;i < (n);(i)++)
#define ll long long
#define INF 20000000

using namespace std;

// input
int n,x,y;
int ad_list[2001][2001];

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
   cin >> n >> x >> y;

   rep(i,2001) rep(j,2001) ad_list[i][j] = INF;
   rep(i,n-1) {
	 ad_list[i][i+1] = 1;
	 ad_list[i+1][i] = 1;
   }
   ad_list[x-1][y-1] = 1;
   ad_list[y-1][x-1] = 1;

   rep(i,n) {
	 rep(j,n) {
	   rep(k,n) {
		 ad_list[j][k] = min(ad_list[j][k],ad_list[j][i] + ad_list[i][k]);
	   }
	 }
   }

   rep(i,n-1) {
	 int now = i+1;
	 int ans = 0;
	 rep(i,n) rep(j,n) {
	   if(i == j) continue;
		if(ad_list[i][j] == now) ans += 1;
	 }
	 cout << ans/2 <<endl;
   }
}