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;
#define ll long long int
#define inf 99999
int main()
{
    ios_base::sync_with_stdio(false);
cin.tie(NULL);
    int n,x,y,i,j,k;
    cin>>n>>x>>y;
    vector<int>key(n);
    int G[n][n];
    for(i=0;i<n;++i)
        {for(j=0;j<n;++j)
        {
            if(j-i==1)
            {G[i][j]=1;}
            else if(i-j==1)
            G[i][j]=1;
            else if(i==j)
            G[i][j]=0;
            else G[i][j]=inf;
        }
            key[i]=0;
        }
    
    G[x-1][y-1]=1;
    G[y-1][x-1]=1;
    
    for (k = 0; k < n; k++)  
    {  
        for (i = 0; i < n; i++)  
        {    
            for (j = 0; j < n; j++)  
            {  
                if (G[i][k] + G[k][j] < G[i][j])  
                    G[i][j] = G[i][k] + G[k][j]; 
            }  
        }  
    }
    
    for(i=0;i<n;++i)
        {for(j=0;j<n;++j)
            key[G[i][j]]++;  
        }
    for(i=1;i<=n-1;++i)
        cout<<key[i]/2<<endl;
}