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 (GCC 5.4.1) (judge verdict: Accepted)

#include <assert.h>
#include <stdio.h>

int abs(int value) {
  if (value < 0) {
    return -value;
  }
  return value;
}

int main() {
  int n, x, y, i, j, target, dist, count;
  assert(scanf("%d %d %d", &n, &x, &y) == 3);
  x -= 1;
  y -= 1;
  for (i = 1; i < n; i++) {
    count = 0;
    for (j = 0; j < n - i; j++) {
      target = j + i;
      dist = abs(x - j);
      if (dist >= i) {
        count++;
      } else {
        if (target < y && dist + 1 + y - target >= i) {
          count++;
        }

        if (i - dist == 1) {
          count++;
        } else {
          target = y - (i - dist - 1);

          if (target - j > i) {
            count++;
          }

          target = y + (i - dist - 1);

          if (target < n) {
            count++;
          }
        }
      }
    }
    printf("%d\n", count);
  }
}