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: Runtime Error)

#include <iostream>
#include <algorithm>
#include <vector>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N;
    int X;
    int Y;
    cin >> N >> X >> Y;

    queue<int> q1;

    // 1行目は、頂点1を始点とする場合
    //1,2は、1から2までいくときの最短距離、1.3は、1から3までいくときの最短距離
    vector<vector<int>> len(N + 1, vector<int>(N + 1, -1));

    vector<int> ans (N+1, 0);

    //i行目について考える（頂点iを始点とする場合について考える）
    for (int i = 1; i <= N - 1; i++)
    {
        len[i][i] = 0;

        q1.push(i);

        while (!q1.empty())
        {
            int temp = q1.front();
            q1.pop();

            if (temp + 1 <= N)
            {
                if (len[i][temp + 1] == -1)
                {
                    len[i][temp + 1] = len[i][temp] + 1;
                    q1.push(temp + 1);
                }
            }

            if( temp -1 >=1)
            {
                if(len[i][temp-1] == -1)
                {
                    len[i][temp-1] = len[i][temp] + 1;
                    q1.push(temp-1);
                }
            }

            //取り出した頂点が、Xかどうか確認
            if (temp == X)
            {
                if (len[i][Y] == -1)
                {
                    len[i][Y] = len[i][temp] + 1;
                    q1.push(Y);
                }


            }
        }
    }

    //ここまでで表はできている
    for(int i = 1; i<= N; i++)
    {
        for(int j = i; j<=N; j++)
        {
            int temp = len[i][j];
            ans[temp]++;
        }
    }

    for(int i = 1; i<=N; i++)
    {
        cout << ans[i] << endl;
    }

}