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>
#include<stdlib.h>
int main(){
  int N,X,Y,i,j,k,min;
  scanf("%d%d%d",&N,&X,&Y);
  int a[N+2][N+2];
  for(i=1;i<N;i++)
    for(j=i+1;j<=N;j++)
      a[i][j]=j-i;
  if(X==Y){
    printf("%d\n",N-1);
    for(k=2;k<N;k++)
      printf("0\n");
  }
  else if(X>Y){
    i=X;
    X=Y;
    Y=i;
  }
  for(i=1;i<N;i++){
    for(j=i+1;j<=N;j++){
      min=abs(X-i)+1+abs(Y-j);
      if(min<a[i][j])
        a[i][j]=min;
    }
  }  
//出力
  int count;
  for(k=1;k<N;k++){
    count=0;
    for(i=1;i<N;i++){
      for(j=i+1;j<=N;j++){
        if(a[i][j]==k)
          count++;
      }
    }
    printf("%d\n",count);
  }
  return 0;
}