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: Wrong Answer)

#include <stdio.h>
struct number
{
    int i;
    int j;
    int distance;
};
int main()
{
    int n;
    scanf("%d", &n);
    int x, y;
    scanf("%d%d", &x, &y);
    int array[n];
    for(int i = 0;i<n;i++)
    {
        array[i] = i+1;
    }
    struct number differnces[n*(n-1)/2];
    int iter = 0;
    for(int i = 0;i<n-1;i++)
    {
        differnces[iter].i = i;

        for(int j = i+1;j<n;j++)
        {
            differnces[iter].j = j;
            if(array[i]==x && array[j]==y)
            {
                differnces[iter].distance = 1;
            }
            if(array[i]==x && array[j] > y)
            {
                differnces[iter].distance = 1+array[j]-y;
            }
            if(array[i]==x && array[j]<y)
            {
                differnces[iter].distance = array[j]-x;
            }
            if(array[i]<x && array[j] == y)
            {
                differnces[iter].distance = 1+x-array[i];
            }
            if(array[i]>x && array[j]==y)
            {
                differnces[iter].distance = y-array[i];
            }
            if(array[i]<x && array[j]>y)
            {
                differnces[iter].distance = 1+ x-array[i] + array[j]-y;
            } 
            if((array[i]<x && array[j]<y)||(array[i]<x && array[j]<x)||(array[i]>y && array[j]>y)||(array[i]>x && array[j]>y)||(array[i]>x && array[j]<y))
            {
                differnces[iter].distance = array[j]-array[i];
            }
            iter++;
        }
    }
    for(int k = 1;k<=n-1;k++)
    {
        int count  = 0;
        for(int i =0 ;i<(n*(n-1)/2);i++)
        {
            if(differnces[i].distance == k)
                count++;
        }
        printf("%d\n",count);
        
    }

}