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: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <vector>

using namespace std;
using LLONG = long long;
const LLONG MOD = 1000000007;

int N, X, Y;
vector<int> pairNums(2001, 0);
vector<vector<int>> adList;

void BFS(const int v)
{
    vector<int> dists(N, 3000);
    deque<int> Q{ v };
    dists[v] = 0;
    int dist = 0;
    while (!Q.empty())
    {
        int u = Q.front(); Q.pop_front();
        int dist = dists[u] + 1;
        for (const int n : adList[u])
        {
            if (dist < dists[n])
            {
                Q.push_back(n);
                dists[n] = dist;
            }
        }
    }
    for (int i = v + 1; i < N; ++i)
    {
        ++pairNums[dists[i] - 1];
    }
}

int main()
{
    cin >> N >> X >> Y; --X; --Y;
    adList = vector<vector<int>>(N, vector<int>(N));
    for (int i = 0; i < N - 1; ++i)
    {
        adList[i].push_back(i + 1);
        adList[i + 1].push_back(i);
    }
    adList[X].push_back(Y);
    adList[Y].push_back(X);
    for (int v = 0; v < N; ++v)
    {
        BFS(v);
    }
    for (int i = 0; i < N - 1; ++i)
    {
        cout << pairNums[i] << endl;
    }
}