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<bits/stdc++.h>
using namespace std;
typedef long long unsigned int ll;
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i, start, end) for(int i = start; i < (int)(end); ++i)
int cnt[2020][2020];
bool edge[2020][2020];
int main(){
    int n, x, y;
    cin >> n >> x >> y;
    rep(i, 0, n+1) rep(j, 0, n+1) cnt[i][j] = INF;
    rep(i, 0, n+1) cnt[i][i+1] = 1, cnt[i+1][i] = 1;
    cnt[x-1][y-1] = 1; cnt[y-1][x-1] = 1;
    rep(i,0,n){
        rep(j,0,n){
            rep(k, 0, n){
                rep(l,0,n){
                    cnt[j][k] = min(cnt[j][k], cnt[j][l] + cnt[l][k]);
                }
                cnt[i][j] = min(cnt[i][j], cnt[i][k] + cnt[k][j]);
            }
        }
    }
    rep(i,1,n){
        long long ans = 0;
        rep(j,0,n-1){
            rep(k,j+1,n){
                if(cnt[j][k] == i) ans++;
            }
        }
        cout << ans << endl;
    }
    return 0;
}