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>

#define fst(t) std::get<0>(t)
#define snd(t) std::get<1>(t)
#define thd(t) std::get<2>(t)
#define unless(p) if(!(p))
#define until(p) while(!(p))

using ll = std::int64_t;
using P = std::tuple<int,int>;

int N, X, Y;
int cnt[2100];

int dist(int v, int w){
    if(v > w){std::swap(v, w);}

    if(v <= X){
        return std::min(w - v, (X - v) + 1 + std::abs(w - Y));
    }

    if(Y <= w){
        return std::min(w - v, (w - Y) + 1 + std::abs(v - X));
    }

    return std::min(w - v, (Y - X) - (w - v) + 1);
}

int main(){
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);

  std::cin >> N >> X >> Y;

  for(int i=1;i<=N;++i){
      for(int j=i+1;j<=N;++j){
          cnt[dist(i, j)] += 1;
      }
  }

  for(int k=1;k<N;++k){
      std::cout << cnt[k] << std::endl;
  }
}