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 ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
const ll INF = 1e18;
#define pu push
#define pb push_back
#define mp make_pair
#define print(x) cout << (x) << endl;
#define rep(i,n) for(ll i = 0; i < (n); i++)
#define SORT(x) sort(x.begin(),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
#define fi first
#define sc second

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, X, Y;
    cin >> N >> X >> Y;
    X--;
    Y--;
    //start(0):S,end(N-1):Eと表記
    int SX = X-1;
    int XY = Y-X;
    int YE = N-1-Y;
    vector<int> dist(N-1);
    rep(i,N-1){
        for(int j = i+1; j < N; j++) {
            if(i<=X) {
                if(j<=(X+Y)/2) {
                    dist.at(j-i)++;
                }
                if((X+Y)/2<j&&j<=Y) {
                    dist.at(Y-j+1+X-i)++;
                }
                if(Y<j) {
                    dist.at(j-Y+1+X-i)++;
                }
            }
            else if(i<(X+Y)/2) {
                if(j<=Y) {
                    dist.at(min(j-i,Y-X+1-j+i))++;
                }
                if(Y<j) {
                    dist.at(j-Y+i-X+1)++;
                }
            }
            else if((X+Y)/2<=i) {
                dist.at(j-i)++;
            }
        }
    }
    rep(i,N-1){
        if(i==0) continue;
        cout << dist.at(i) << endl;
    }
    cout << 0 << endl;
}