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: Accepted)

#include<bits/stdc++.h>
using namespace std;
using Graph=vector<vector<int>>;
using P=pair<int,int>;

vector<int> dist;
vector<int> answer;

int main(){
    int n,x,y,i;cin>>n>>x>>y;
    x--;y--;

    Graph G(n);
    for(i=0;i<n-1;i++){//Graphの作成
        G[i].push_back({i+1});
    }
    for(i=1;i<n;i++){
        G[i].push_back({i-1});
    }
    G[x].push_back({y});
    G[y].push_back({x});

    dist.assign(n,-1);
    answer.assign(n,0);

    for(int num=0;num<n;num++){//全頂点をスタート位置としてそこからの最短距離を調べる
        queue<int> que;
        que.emplace(num);//numからstart
        dist[num]=0;//start地点は距離0
        while(que.size()!=0){
            int state=que.front();
            que.pop();
            for(auto next:G[state]){
                if(dist[next]==-1){//まだ訪れていないとき
                    dist[next]=dist[state]+1;
                    que.emplace(next);
                    answer[dist[next]]++;//最短距離ごとに加算
                }
            }
        }
        dist.assign(n,-1);//初期化
    }
    
    for(i=1;i<n;i++){
        cout<<answer[i]/2<<" ";//重複して2回数えているので
    }
    cout<<endl;
    return 0;
}