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 ll;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
const int MAX = 1e5 + 10;
const double exps = 1e-8;

inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a > b ? b : a; }
int line[2010][2010];
int vis[2010];
int n, x, y;
void Floyd() {
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i <= n; ++i) {
            if (line[i][k] != INF) {
                for (int j = 1; j <= n; ++j) {
                    if (line[i][j] > line[i][k] + line[k][j]) {
                        line[i][j] = line[i][k] + line[k][j];
                    }
                }
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            if (line[i][j] != INF) vis[line[i][j]]++;
        }
    }
}
int main() {
    scanf("%d%d%d", &n, &x, &y);
    memset(line, 0x3f, sizeof(line));
    for (int i = 1; i <= n - 1; ++i) line[i][i + 1] = 1, line[i + 1][i] = 1;
    line[x][y] = 1;
    line[y][x] = 1;
    Floyd();
    for (int k = 1; k <= n - 1; ++k) {
        printf("%d\n", vis[k]);
    }
}