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: Runtime Error)

#include<stdio.h>

int main(void){
    int N;
    int X, Y;
    int i,j,k;
    scanf("%d", N, X, Y);

    int length[2000];
    for(i=0; i<2000; i++){
        length[i]=0;
    }

    for(i=0; i<N; i++){
        for(j=i+1; j<N; j++){
            if(i<=X && j <= X){
                length[j-i]++;
            }
            else if(i<=X && j<=(Y-X+1)/2){
                length[j-i]++;
            }
            else if(i <=X && j>(Y-X+1)/2){
                length[Y-j + 1 + X-i]++;
            }
            else if(i <= X && Y <= j){
                length[X-i+j-Y+1]++;
            }
            else if(X < i && i <= (Y-X+1)/2 &&j-i <= (Y-X+1)/2 && j <= Y){
                length[j-i]++;
            }
            else if(X < i && i <= (Y-X+1)/2 && Y-i <= (Y-X+1)/2 && Y < j){
                length[j-i]++;
            }
            else if(X < i && i <= (Y-X+1)/2 && j-i > (Y-X+1)/2 && j <= Y){
                length[Y-j+1+i-X]++;
            }
            else if(X < i && i <= (Y-X+1)/2 && Y-i > (Y-X+1)/2 && Y < j){
                length[j-Y+1+i-X]++;
            }
            else{
                length[j-i]++;
            }
        }
    }
    for(i=1;i<N-1;i++){
        printf("%d\n", length[i]);
    }
    printf("%d", length[N-1]);
    return 0;
}