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>
#include <ext/pb_ds/assoc_container.hpp>
 
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,tune=native")
 
using namespace __gnu_pbds;
using namespace std;
     
#define mp make_pair
#define sz(x) (int)((x).size())
#define all(x) (x).begin(),(x).end()
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define eb emplace_back
 
typedef pair< int, int > pii;
typedef pair< long long, long long > pll;
typedef long long ll;
typedef unsigned long long ull;
typedef vector< int > vi;
typedef vector< vi > vvi;
typedef vector< ll > vll;
typedef vector< vll > vvll;
typedef vector< pii > vpii;
typedef vector< vpii > vvpii;
typedef vector< pll > vpll;
typedef long double ld;
typedef vector< ld > vld;
typedef unsigned int uint;
 
const ll MOD2 = 1e9 + 7;
const int MOD = 1e9 + 7;
 
const ull infull = numeric_limits<unsigned long long>::max();
 
void fix(int &x, ll MOD) {
    x = (x % MOD);
    if(x < 0) x += MOD;
    return;
}
 
ll lgput(ll a, ll b, ll MOD) {
    ll ret = 1;
    a %= MOD;
    while(b) {
        if(b&1) ret = ret*a % MOD;
        a = a*a % MOD;
        b >>= 1;
    }
 
    return ret;
}
 
ll inv(ll a, ll MOD) {
    return lgput(a, MOD-2, MOD);
}
 
template <uint MD> struct ModInt {
    using M = ModInt;
    const static M G;
    uint v;
    ModInt(ll _v = 0) { set_v(uint(_v % MD + MD)); }
    M& set_v(uint _v) {
        v = (_v < MD) ? _v : _v - MD;
        return *this;
    }
    explicit operator bool() const { return v != 0; }
    M operator-() const { return M() - *this; }
    M operator+(const M& r) const { return M().set_v(v + r.v); }
    M operator-(const M& r) const { return M().set_v(v + MD - r.v); }
    M operator*(const M& r) const { return M().set_v(uint(ull(v) * r.v % MD)); }
    M operator/(const M& r) const { return *this * r.inv(); }
    M& operator+=(const M& r) { return *this = *this + r; }
    M& operator-=(const M& r) { return *this = *this - r; }
    M& operator*=(const M& r) { return *this = *this * r; }
    M& operator/=(const M& r) { return *this = *this / r; }
    bool operator==(const M& r) const { return v == r.v; }
    const bool operator<(const M& o) const {return v < o.v; }
    M pow(ll n) const {
        M x = *this, r = 1;
        while (n) {
            if (n & 1) r *= x;
            x *= x;
            n >>= 1;
        }
        return r;
    }
    M inv() const { return pow(MD - 2); }
    friend ostream& operator<<(ostream& os, const M& r) { return os << r.v; }
};
 
using Mint = ModInt<998244353>;

int main() { 
    #ifdef BLAT
        freopen("stdin", "r", stdin);
        freopen("stderr", "w", stderr);
    #endif
 
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(12);
    srand(time(NULL));

    int n, x, y;
    cin >> n >> x >> y;

    vector< vector< int > > gr(n + 1);

    for(int i = 1; i < n; ++i) {
        gr[i].emplace_back(i+1);
        gr[i+1].emplace_back(i);
    }

    gr[x].eb(y);
    gr[y].eb(x);

    vector< int > ans(n);
    vi dist(n + 1);

    for(int i = 1; i <= n; ++i) {
        fill(all(dist), -1);
        dist[i] = 0;
        queue< int > q;
        q.push(i);
        while(q.size()) {
            int node = q.front();
            if(node > i) {
                ans[dist[node]] ++;
            }
            q.pop();
            for(auto &x : gr[node]) {
                if(dist[x] == -1) {
                    dist[x] = dist[node] + 1;
                    q.push(x);
                }
            }
        }
    }

    for(int i = 1; i < n; ++i) cout << ans[i] << '\n';
    return 0; 
}