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>
int main(void)
{
    int N,X,Y;
    scanf("%d %d %d",&N, &X, &Y);
    X -= 1;
    Y -= 1;
    int gyoretu[N][N];
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if(i==j){
                gyoretu[i][j]=0;
            }
            else if(i-j==1 || j-i==1)
            {
                gyoretu[i][j]=1;
            }
            else
            {
                gyoretu[i][j]=2000;
            }
        }
    }
    gyoretu[X][Y]=1;
    gyoretu[Y][X]=1;

    for(int k=0;k<N;k++){
        for(int i=0;i<N;i++){
            for(int j=0;j<N;j++){
                if (gyoretu[i][j]> gyoretu[i][k] + gyoretu[k][j]){
                    gyoretu[i][j] =gyoretu[i][k] + gyoretu[k][j];
                }
            }
        }
    }
    int count[N-1];
    for (int i=0;i<N;i++)
    {
        count[i]=0;
    }
    for(int i=0;i<N;i++){
        for(int j=i;j<N;j++){
            //printf("%d ",gyoretu[i][j]);
            count[gyoretu[i][j]]+=1;
        }
        //printf("\n");
    }
    for (int i=1;i<N;i++)
    {
        printf("%d\n",count[i]);
    }
    
}