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 <string.h>

	int dist[2020][2020];
	int k[2020];
int chmin(int a,int b){
	if(a>b)return b;
	else return a;
	
}
int main(){
	 int i,j,kk,N,X,Y,A[200200];
	 int max_len=-1,max_index,ans;

	scanf("%d%d%d",&N,&X,&Y);
	for(i=1;i<=N;i++)for(j=1;j<=N;j++)dist[i][j]=2020;
	
	//��
	for(i=1;i<=N-1;i++){
		dist[i][i+1]=1;
	}
	
	dist[X][Y]=1;
	
	for(i=2;i<=N;i++){
		for(j=1;j<i;j++){
			for(kk=1;kk<i;kk++){
				dist[j][i] = chmin(dist[j][kk]+dist[kk][i],dist[j][i]);
				dist[i][j] = dist[j][i];
			}
			//k[dist[j][i]]++;
			//printf("(%d,%d:%d)",j,i,dist[j][i]);
		}
		//printf("\n");
		
	}
		for(i=2;i<=N;i++){
		for(j=1;j<i;j++){
			for(kk=j+1;kk<=N;kk++){
				dist[j][i] = chmin(dist[j][kk]+dist[kk][i],dist[j][i]);
			}
			k[dist[j][i]]++;
			//printf("(%d,%d:%d)",j,i,dist[j][i]);
		}
		//printf("\n");
	}
	for(i=1;i<N;i++)
	printf("%d\n",k[i]);
 return 0;
 }