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: Wrong Answer)

#include <inttypes.h>
#include <math.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    intmax_t N, X, Y;
    intmax_t tmp, d[(int)2e3] = {};

    scanf("%jd%jd%jd", &N, &X, &Y);

    for (size_t i = 1; i <= N; i++) {
        for (size_t j = i + 1; j <= N; j++) {
            tmp = j - i;
            if ((i > X) && (j < Y)) {
                if (tmp > ((Y - X) / 2))
                    tmp = Y - X;
            } else if ((i <= X) && (j >= Y)) {
                if (tmp > ((X - i) + (j - Y) + 1))
                    tmp = (X - i) + (j - Y) + 1;

            } else if ((i > X) && (j >= Y)) {
                if (tmp > (((Y - X) / 2) + (j - Y)))
                    tmp = ((Y - X) / 2) + (j - Y);
            } else {
                if (tmp > ((X - i) + ((Y - X) / 2)))
                    tmp = ((X - i) + (Y - X) / 2);
            }
            d[tmp]++;
        }
    }

    for (size_t i = 1; i < N; i++) {
        printf("%jd\n", d[i]);
    }

    return 0;
}