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 int long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define rep(i,n) for(int i=0; i<(int)n; i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define COUT(x) cout << x << endl;
#define PB(x) push_back(x)
#define MP make_pair
int gcd(int a,int b){return b?gcd(b,a%b):a;}
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
const int INF = 1e9;
const int MOD = 1e9+7;

vector<vector<int>> g;
vector<int> dist;
int n,x,y; 

void bfs(int s, int limit){
    queue<int> que;
    rep(i,n) dist[i]=-1;
    dist[s]=0;
    que.push(s);
    while(!que.empty()){
        int v=que.front();
        que.pop();
        for(int nv:g[v]){
            if(dist[nv]==-1) {
                if(dist[nv]>limit) continue;
                dist[nv]=dist[v]+1;
                que.push(nv);
            }
        }
    }
}

signed main(){
    cin >> n >> x >> y;
    g.resize(n);
    dist.resize(n);
    rep(i,n-1){
        g[i].PB(i+1);
        g[i+1].PB(i);
    }
    g[x-1].PB(y-1);
    g[y-1].PB(x-1);

    for(int i=1; i<n; i++){//距離
        int res=0;
        rep(j,n){//始点
            bfs(j,i);
            rep(k,n){
                if(dist[k]==i) res++;
            }
        }
        cout<<res/2<<endl;
    }
}