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: Time Limit Exceeded)

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

int main(void){
	int n,x,y;
	int ans=0;
	int f;
	int min;
	scanf("%d %d %d",&n,&x,&y);
	for(int k=1; k<n; k++){
		ans=0;
		min=n;
			/*if(x>i){
				f=x-i;
			}
			else{
				f=i-x;
			}*/
		for(int i=1; i<n; i++){
			for(int j=i+1; j<=n; j++){
				min = fmin(abs(i-x)+abs(y-j)+1,j-i);
				if(k==min){
					ans++;
				}
				/*if(j>=y&&i<=x || (j-i)>(y-j+1+f)&&j<y&&j>x){
					if(j-i-(y-x-1)==k && j>=y && i<=x){
						ans++;
					//	printf("%d\t%d\n",i,j);
					}
					else if(y-j+1+f==k&&y>j&&y>x){
						ans++;
					//	printf("%d\t%d\n",i,j);
					}
				}
				else{
					if(j-i==k){
						ans++;
					//	printf("%d\t%d\n",i,j);
					}
				}*/
			}
		}
		printf("%d\n",ans);
	}

	return 0;	
}