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 <stdlib.h>
#include <strings.h>

#ifdef TEST
struct test {
    int ans;
} td[] = {
    {},
};
#endif

int m[2000][2000];
void
solver(int n, int x, int y)
{
    int i, j, k;
    int a;

    bzero(m, sizeof(m));
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            if(abs(j - i) == 1) {
                m[i][j] = 1;
                m[j][i] = 1;
            }
        }
        m[x - 1][y - 1] = 1;
        m[y - 1][x - 1] = 1;
    }
    for(k = 0; k < n; k++) {
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                if(m[i][k] == 0 || m[k][j] == 0)
                    continue;
                if(m[i][j] == 0 || m[i][j] > m[i][k] + m[k][j]) {
                    m[i][j] = m[i][k] + m[k][j];
                }
            }
        }
    }

#if 0
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            printf("%d ", m[i][j]);
        }
        printf("\n");
    }
#endif

    printf("%d\n", n);
    for(k = 2; k < n; k++) {
        a = 0;
        for(i = 0; i < n; i++) {
            for(j = i + 1; j < n; j++) {
                if(m[i][j] == k)
                    a++;
            }
        }
        printf("%d\n", a);
    }
}

int
main(int argc, char *argv[])
{
#ifdef TEST
    int tt;
    for(tt = 0; tt < sizeof(td) / sizeof(struct test); tt++) {
        solver();
        printf("%d\n", td[tt].ans);
    }
#endif
    int N, X, Y;
    scanf("%d%d%d", &N, &X, &Y);
    solver(N, X, Y);
    return 0;
}