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>

int main(){
	int n, x, y;
	int ans[2005] = {0};
	int i, j;

	scanf("%d %d %d", &n, &x, &y);

	for(i=1; i<n; i++){
		for(j=i+1; j<=n; j++){
			if(i <= x){
				if(j <= x){
					ans[j-i]++;
				}else if(j >= x && j < y){
					if(j-x < y-j+1){
						ans[j-i]++;
					}else{
						ans[x-i + y-j+1]++;
					}
				}else{
					ans[j-i-(y-x-1)]++;
				}
			}else if(i > x && i < y){
				if(j >= x && j < y){
					if(j-i < i-x + y-j + 1){
						ans[j-i]++;
					}else{
						ans[i-x + y-j + 1]++;
					}
				}else{
					if(y-i < i-x+1){
						ans[j-i]++;
					}else{
						ans[j-y + i-x+1]++;
					}
				}
			}else{
				ans[j-i]++;
			}

//			printf("%d %d:", i, j);
//			int k;
//			for(k=1; k<n; k++){
//				printf("%d ", ans[k]);
//			}
//			printf("\n");
		}
	}

	for(i=1; i<n; i++){
		printf("%d\n", ans[i]);
	}

	return 0;
}