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 lint = long long;
using lint128 = __int128_t;
const lint mod = 998244353;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountll((lint)(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
inline void YES(bool condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
inline void Yes(bool condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
inline void assert_NO(bool condition){ if(!condition){ cout << "NO" << endl; exit(0); } }
inline void assert_No(bool condition){ if(!condition){ cout << "No" << endl; exit(0); } }
inline void assert_minus_1(bool condition){ if(!condition){ cout << -1 << endl; exit(0); } }
lint power(lint base, lint exponent, lint module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ lint root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr first, itr last){ for(auto i = first; i != last; i++){ cin >> (*i); } }
template<class T> T gcd_calc(T a, T b){ if(b){ return gcd_calc(b, a % b); }else{ return a; }}
template<class T> T gcd(T a, T b){ if(a < b) swap(a, b); return gcd_calc(a, b); }
template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<lint> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } lint C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };
template<class itr> bool next_sequence(itr first, itr last, int max_bound){ itr now = last; while(now != first){ now--; (*now)++; if((*now) == max_bound){ (*now) = 0; }else{ return true; } } return false; }
template<class T> bool chmax(T &a, const T &b){ if(a < b){ a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b){ if(b < a){ a = b; return 1; } return 0; }
inline int at(lint x, int k){ return (x >> k) & 1; }
random_device rnd;
#define rep(i, n) for(int i = 0; i < n; i++)

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