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>

int min(int x,int y){
    if(x>y){
        return y;
    }else{
        return x;
    }
}

int main(){
    int n,x,y;
    scanf("%d %d %d",&n,&x,&y);
    int ans[n];
    for(int i=0;i<n+1;i++){
        ans[i]=0;
    }
    for(int i=1;i<n+1;i++){
        for(int j=i;j<n+1;j++){
            if(i<=x && x<=j && i<=y && y<=j){
                ans[x-i+1+j-y]++;
            }else if(i<=x && x<=j){
                ans[min(j-i,x-i+1+y-j)]++;
            }else if(i<=y && y<=j){
                ans[min(j-i,i-x+1+j-y)]++;
            }else{
                ans[j-i]++;
            }
        }
    }
    for(int i=1;i<n;i++){
        printf("%d\n",ans[i]);
    }
    return 0;
}