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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
#define INF 11000000;

int main(){
  int N, X, Y;
  cin >> N >> X >> Y;
  X--;
  Y--;
  vector<int> ans(N - 1, 0);
  for (int i = 0; i < N; i++) {
      for (int j = i + 1; j < N; j++) {
          if (i <= X && j >= Y)
              ans.at(j - i - (Y - X))++;
          else if (i <= X && j < Y && Y - j + 1 < j - X)
              ans.at(j - i - (Y - X) + (Y - j) + 1)++;
          else if (i > X && j >= Y && i - X + 1 < Y - i)
              ans.at(j - i - (Y - X) + (i - X) + 1)++;
          else if (i > X && j < Y && (Y - j) + (i - X) + 1 < j - i)
              ans.at(j - i - (Y - X) + (Y - j) + (i - X) + 2)++;
          else
              ans.at(j - i - 1)++;
      }
  }
  for (int i = 0; i < N - 1; i++) cout << ans.at(i) << endl;
  return 0;
}