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<bits/stdc++.h>
using namespace std;


int main(){
    int n, x, y;
    cin>>n>>x>>y;
    vector<vector<int>> d(n+1, vector<int>(n+1));
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            if(i<=x && j<=x){
                d[i][j] = j-i;
            }else if(i<=x && j<y && j>x){
                d[i][j] = x-i + min(j-x, y-j+1);
            }else if(i<=x && j>=y){
                d[i][j] = x-i + j-y + 1;
            }else if(i>x && i<y && j>=y){
                d[i][j] = j-y + min(i-x+1, y-i);
            }else{
                d[i][j] = min(j-i, i-x + y-j + 1);
            }
        }
    }
    map<int, int> umap;
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            // if(d[i][j]==2)cout<<i<<" "<<j<<endl;
            umap[d[i][j]]++;
        }
    }
    for(int i=1;i<n;i++){
        cout<<umap[i]<<endl;
    }
    return 0;
    
}