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>

#define rep(i,l,n) for(int (i)=(l);(i)<(n);(i)++)
#define min(a,b) (a)<(b)?(a):(b)
#define max(a,b) (a)>(b)?(a):(b)
#define abs(a) (a)>0?(a):-(a)

int main(){
	int N,X,Y; scanf("%d %d %d",&N,&X,&Y);
	X--; Y--;
	int ans[N];
	rep(i,0,N) ans[i]=0;
	rep(i,0,N){	// i:距離
		int dis[N];
		rep(j,0,N) dis[j]=abs(i-j);
		if( i <= Y ){
			if( dis[X] < dis[Y] ){
				dis[Y] = dis[X]+1;
				rep(k,1,Y-X){
					if( dis[Y-k] > dis[Y]+k ){
						dis[Y-k] = dis[Y]+k;
					}else{
						break;
					}
				}
				rep(k,1,N-Y){
					if( dis[Y+k] > dis[Y]+k ){
						dis[Y+k] = dis[Y]+k;
					}else{
						break;
					}
				}
			}
		}
		rep(j,i,N) ans[dis[j]]++;
	}
	rep(i,1,N) printf("%d\n",ans[i]);
	return 0;
}