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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>


int main(void) {
  long long int n, x, y;
  scanf("%lld %lld %lld", &n, &x, &y);
  x--;
  y--;
  long long int center_xy = (x + y) / 2;

  long long int dis[n];
  for (long long int i = 0; i < n; i++) {
    dis[i] = 0;
  }
  for (long long int i = 0; i < n - 1; i++) {
    long long int i_x = x - i;
    if (i_x < 0) i_x *= -1;
    long long int i_y = i_x + 1;
    for (long long int j = i + 1; j < n; j++) {
      if (i <= x) {
        if (j <= center_xy) {
          dis[j - i]++;
        } else if (j > center_xy && j < y) {
          dis[i_y + y - j]++;
        } else if (j >= y) {
          dis[i_y + j - y]++;
        }
      } else if (i > x && (center_xy + i_x) < y) {
        if (j <= (center_xy + i_x)) {
          dis[j - i]++;
        } else if (j > (center_xy + i_x) && j < y) {
          dis[i_y + y - j]++;
        } else if (j >= y) {
          dis[i_y + j - y]++;
        }
      } else {
        dis[j - i]++;
      }
    }

    /*
    for (long long int i = 1; i < n; i++) {
      printf("%lld ", dis[i]);
    }
    printf("\n");
    */
  }

  for (long long int i = 1; i < n; i++) {
    printf("%lld\n", dis[i]);
  }

  return 0;
}