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;
int N,X,Y;
vector<vector<int>> G;
vector<vector<int>> dist;
map<int, map<int, int>> D;

void bfs(int s) {
    dist[s][s] = 0;
    queue<int> Q;
    Q.push(s);
    while(!Q.empty()) {
        int v = Q.front(); Q.pop();
        for(int to : G[v]) {
            if(dist[s][to] == -1) {
                dist[s][to] = dist[s][v] + 1;
                Q.push(to);
            }
        }
    }

    for(int i=0; i<N; ++i) {
        if(i == s) continue;
        D[s][dist[s][i]] += 1;
    }
}

int main() {
    cin >> N >> X >> Y;
    G.resize(N);
    dist.resize(N, vector<int>(N, -1));
    for(int i=0; i+1<N; ++i) {
        G[i].push_back(i + 1);
        G[i + 1].push_back(i);
    }
    G[X - 1].push_back(Y - 1);
    G[Y - 1].push_back(X - 1);

    for(int i=0; i<N; ++i) bfs(i);

    for(int k=1; k<N; ++k) {
        int res = 0;
        for(int i=0; i<N; ++i) res += D[i][k];
        cout << res / 2 << '\n';
    }
}