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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>

int myAbs(int a){
	if(a < 0) a = -a;
	return a;
}

int min(int a, int b){
	if(a < b){
		return a;
	}
	return b;
}

int minList(int a[ ], int size){
	int i = 0;
	int ret = a[i];
	for(i = 1; i < size; i++){
		ret = min(ret, a[i]);
	}
	return ret;
}

int main(){
	int N, X, Y;
	scanf("%d %d %d", &N, &X, &Y);
	int i, j;
	int a[3];
	int b[N];
	for(i = 0; i < N; i++){
		b[i] = 0;
	}
	for(i = 1; i < N; i++){
		for(j = i+1; j <= N; j++){
			a[0] = myAbs(j - i);
			a[1] = myAbs(X - i) + 1 + myAbs(j - Y);
			a[2] = myAbs(Y - i) + 1 + myAbs(j - X);
			b[minList(a, 3)]++; 
		}
	}
	for(i = 1; i < N; i++){
		printf("%d\n", b[i]);
	}
	return 0;
}