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;
#define FOR(i,n) for(int i=0;i<(n);i++)
#define MOD 1000000007
#define all(n) (n).begin(),(n).end()
typedef long long ll;
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}else{return 0;}}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}else{return 0;}}
template <class T> inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template <class T> inline T LCM(T a,T b){return a*b/GCD(a,b);}

int main(){
	int n;
	cin >> n;
	int x,y;
	cin >> x >> y;
	vector<int> g[2010];
	FOR(i,n-1){
		g[i+2].push_back(i+1);
		g[i+1].push_back(i+2);
	}
	g[x].push_back(y);
	g[y].push_back(x);

	int dis[2010][2010];
	FOR(i,2010)FOR(j,2010)dis[i][j]=0;
	bool reached[2010];

	for(int i=1;i<=n;i++){
		
		FOR(j,2010)reached[j]=0;
		queue<int> q;
		q.push(i);
		reached[i]=1;

		while(!q.empty()){
			int x=q.front();
			q.pop();
			FOR(j,g[x].size()){
				if(!reached[g[x][j]]){
					dis[i][g[x][j]]=dis[i][x]+1;
					reached[g[x][j]]=1;
					q.push(g[x][j]);
				}
			}

		}
	}

	int ans[2010];
	FOR(i,2010)ans[i]=0;

	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			ans[dis[i][j]]++;
		}
	}

	for(int i=1;i<=n-1;i++){
		cout << ans[i]/2 << endl;
	}
	
}