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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int main(){
    int n,key1,key2,i,j;scanf("%d%d%d",&n,&key1,&key2);
    int a[n+1][n+1]/*最短距離の行列*/,ans[n+1];

    for(i=1;i<=n-1;i++){
        ans[i]=0;
    }

    for(i=1;i<=n;i++){
        for(j=1;j<i;j++){
            if(i>=key2 && j<=key1){//いずれもkeyの外
                a[i][j]=i-j-(key2-key1-1);
            }else if(((key1<=i && i<=key2) && (j<=key1)) && (i-j>=key1+key2-i-j+1)){//大きい方がkeyの中
                a[i][j]=key1+key2-i-j+1;
            }else if(((key1<=j && j<=key2) && (i>=key2)) && i-j>=j-key1-key2+i+1){//小さい方がkeyの中
                a[i][j]=j-key1-key2+i+1;
            }else if(key1<=j && i<=key2 && i-j>=key2+j-i-key1+1){//どちらもkeyの中
                a[i][j]=key2+j-i-key1+1;
            }else{//それ以外
                a[i][j]=i-j;
            }
        }
    }
    for(i=1;i<=n;i++){
        for(j=1;j<i;j++){
            ans[a[i][j]]++;
        }
    }
    for(i=1;i<=n-1;i++){
        printf("%d\n",ans[i]);
    }
    return 0;
}