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 <stdbool.h>
//#include <limits.h>
//#include <math.h>

#define Min(a, b) ((a < b) ? a : b)

int main() {
  int n, x, y;
  scanf("%d%d%d", &n, &x, &y);
  x--;
  y--;
  int ans[n-1];
  for (int i=0; i<n; i++)
    ans[i] = 0;
  for (int i=0; i<n-1; i++) {
    for (int j=i+1; j<n; j++) {
      if (i<x) {
        if (j<x)
          ans[j-i]++;
        else if (j>=y)
          ans[x-i+j-y+1]++;
        else
          ans[x-i+Min(j-x, y-j+1)]++;
      } else if (i>=y) {
        ans[j-i]++;
      } else {
        if (j>=y) {
          ans[j-y+Min(i-x+1, y-i)]++;
        } else {
          ans[Min(j-i, y-j+i-x+1)]++;
        }
      }
    }
  }
  for (int i=1; i<n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}