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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,x,y;
    cin>>n>>x>>y;
    for(int i=1;i<n;i++){
        if(i==1){
            cout<<n<<endl;
            continue;
        }
        int a=0;//経路数
        //近道を通らない
        for(int j=1;j<=n-i;j++){
            if(i<=abs(x-j)+abs(y-(j+i))+1)a++;
        }
        //近道を通り、x,yから合計i-1の長さの道を生やす
        for(int j=0;j<i;j++){//xからの距離
            int k=i-j-1;//yからの距離
            if(j && k){
                if(y+k-(x+j)>i && y+k<=n && x+j<=n)a++;
                if(y+k-(x-j)>i && y+k<=n && x-j>=1)a++;
                if(y-k-(x+j)>i && y-k>=1 && x+j<=n)a++;
                if(y-k-(x-j)>i && y-k>=1 && x-j>=1)a++;
            }
            if(j==0){
                if(y+k-x>i && y+k<=n)a++;
                if(y-k-x>i && y-k>=1)a++;
            }
            if(k==0){
                if(y-(x+j)>i && x+j<=n)a++;
                if(y-(x-j)>i && x-j>=1)a++;
            }
        }
        cout<<a<<endl;
    }
    return 0;
}