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 <math.h>
#define N 2000

int main(void){
	
	int n,x,y;
	scanf("%d %d %d",&n,&x,&y);
	int g[N];
	for(int i = 1; i < n; i++){
		g[i] = n-i;
	}
	int limit = (y-x)/2;
	
	//printf("limit = %d\n", limit);
	
	for(int i = 1; i <= x; i++){
		for(int j = 1; j < limit; j++){
			g[x-i+1+j]++;
			g[y-j-i]--;
			//printf("distace between %d to %d --> from %d to %d\n",i,y-j,y-j-i,x-i+1+j);
		}
		for(int j = 0; y+j <= n; j++){
			g[x-i+1+j]++;
			g[y+j-i]--;
			//printf("distace between %d to %d --> from %d to %d\n",i,y+j,y+j-i,x-i+1+j);
		}
	}
	for(int i = x+1; i < x+limit; i++){
		for(int j = 1; j+i-x < limit; j++){
			g[i-x+1+j]++;
			g[y-j-i]--;
			//printf("distace between %d to %d --> from %d to %d\n",i,y-j,y-j-i,i-x+1+j);
		}
		for(int j = 0; y+j <= n; j++){
			g[i-x+1+j]++;
			g[y+j-i]--;
			//printf("distace between %d to %d --> from %d to %d\n",i,y+j,y+j-i,i-x+1+j);
		}
	}
	for(int i = 1; i < n; i++){
		printf("%d",g[i]);
		if( i < n ) printf("\n");
	}
	
}