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>
int main(void){
	int n, x, y;
	int r, c;
	int *results;
	int distance;
	scanf("%d", &n);
	scanf("%d", &x);
	scanf("%d", &y);
	results = (int *)malloc(sizeof(int) * n);
	for(int i=0; i<n; i++){
		results[i] = 0;
	}
	for(int j=0; j<n; j++){
		for(int k=j; k<n; k++){
			distance = abs(j-k);
			if(distance > abs(j-x+1)+abs(k-y+1)+1){
				distance = abs(j-x+1)+abs(k-y+1)+1;
			}
			if(distance > abs(j-y+1)+abs(k-x+1)+1){
				distance = abs(j-y+1)+abs(k-x+1)+1;
			}
			results[distance]++;
		}
	}
	for(int l=1; l<n; l++){
		printf("%d\n", results[l]);
	}
	return 0;
}