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;
typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef vector<vector<vector<ll>>> vvvl;
struct edge{ll to,cost;};
const int inf = 1<<28;
const ll INF = 1LL<<53;
const int COMBMAX = 500005;
const ll MOD = 1e9+7;
#define _GLIBCXX_DEBUG
#define rep(i,n) for(int i=0;i<n;++i)
#define repf(i,n) for(int i=0;i<=n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
#define eachdo(v, e) for(const auto& e : v)
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (ll)distance(v.begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (ll)distance(v.begin(), upper_bound((v).begin(), (v).end(), e))
ll mpow(ll a, ll n,ll mod=MOD){ll res=1;while(n>0){if(n&1)res=res*a%mod;a=a*a%mod;n>>=1;}return res;}
template<class T1, class T2> ll bcount(T1 v, T2 a){return upper_index(v,a)-lower_index(v,a);} 
template<class T> bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}
template<class T> bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}
template<class T> void debug(T v){rep(i,v.size()) cout<<v[i]<<" ";cout<<endl;}
template<class T> void debug2(T v){rep(i,v.size()){rep(j,v[i].size())cout<<v[i][j]<<" ";cout<<endl;}}
template<class T> void rdv(T &v){rep(i,v.size())cin>>v[i];}
template<class T> void rdvv(T &v){rep(i,v.size()){rep(j,v[i].size())cin>>v[i][j];}}
// multimap
// mmap.emplace(key,value);
// auto ngw = list.equal_range("");
// for(auto e = ngw.first; it != ngw.second ; ++e)


int main(){
    ll N, X, Y; cin >> N >> X >> Y;
    --X;
    --Y;
    vector<ll> ans(N, 0);
    rep(j, N){
        rep(i, j){
            ll jdm = j - i;
            chmin(jdm,abs(X - i) + abs(Y - j) + 1);
            //cout << jdm << endl;
            ans[jdm]++; 
        }
    }
    rep(i, N-1) cout << ans[i + 1] << endl;
    return 0;
}