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>
#include <stdlib.h>
#include <math.h>
int main()
{
    int s,n,i,j,k,x,y;
    int a[2010][2010];
    int b[2010];

    scanf("%d",&n);
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
        {
            if (i==j) a[i][j]=0; else a[i][j]=abs(j-i);
        }
    scanf("%d%d",&x,&y);
    a[x][y]=1;
    for (i=1;i<=x;i++)
        for (j=x+1;j<=n-1;j++)
            for (k=j+1;k<=n;k++)
            {
                if (a[i][j]+a[j][k]<a[i][k]) a[i][k]=a[i][j]+a[j][k];
            }
    for (i=1;i<=n;i++)
        for (j=i+1;j<=n;j++)
        {
            b[a[i][j]]++;
        }
    s=n+1;
    while (b[s]==0)
        s--;
    for (i=1;i<=s;i++)
        printf("%d\n",b[i]);
    return 0;
}