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>
#define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define reps(i, n) for(int i=1, i##_len=(n); i<=i##_len; ++i)
#define rrep(i, n) for(int i=((int)(n)-1); i>=0; --i)
#define rreps(i, n) for(int i=((int)(n)); i>0; --i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define reps2(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define inf 2e9
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

#define NMAX 2002

int M[NMAX][NMAX];
int B[NMAX];

int main(){
    int n, x, y;
    cin >> n >> x >> y;
    reps(i, n - 1){
        B[i] = 0;
    }
    reps(i, n){
        reps(j, n){
            if(i > j){
                B[min(abs(i - j), abs(i - y) + abs(j - x) + 1)]++;
            }
        }
    }
    reps(i, n - 1){
        printf("%d\n", B[i]);
    }
    return 0;
}