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: Accepted)

#include <stdio.h>

int max(int p0,int p1){
  return((p0>p1)?p0:p1);
}

int main(){
  int n,x,y,k,m,mid,l,p;
  scanf("%d",&n);
  scanf("%d",&x);
  scanf("%d",&y);
  for(k=1;k<n;k++){
    mid=(x+y)/2;
    l=y-x+1;
    m=max(mid-k,0)-max(mid-x-k,0); //AA,AB
    m+=max(n-mid-k,0)-max(y-mid-1-k,0); //CD,DD
    m+=max(x+y-mid-k,0)-max(x+1-k,0)-max(y-mid-k,0); //AC
    m+=max(n+x-y+1-k,0)-max(x-k,0)-max(n-y+1-k,0); //AD
    m+=max(mid-x+n-y+2-k,0)-max(n-y+2-k,0)-max(mid-x+1-k,0)-(((x+y)%2==0)&(mid-x+1<=k)&(mid-x+1+n-y>=k)); //BD
    p=1+(l>k*2)-(l<k*2);
    m+=p*l/2-p*2+(k==1); //BB,BC,CC
    m+=((x+y)%2==0)&(y-mid<=k)&(mid+k<=n); //MD
    printf("%d\n",m);
  }
  return(0);
}