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>
#define REP(i,n) for(int i = 0; i < n; i++)
#define REPR(i,n) for(int i = n; i >= 0; i--)
#define FOR(i,m, n) for(int i = m-1; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define chmax(x,y) (x = max(x,y))
#define chmin(x,y) (x = min(x,y))
#define FILL2D(A,c) fill(A[0], A[0] +sizeof(A)/sizeof(**A),c)
#define FILL3D(A,c) fill(A[0][0], A[0][0] +sizeof(A)/sizeof(***A),c)
//vector<vector<要素の型>> 変数名(要素数1, vector<要素の型>(要素数2, 初期値))
//vector<intなど> 変数名(要素数,初期値)
using namespace std;
typedef long long ll;
const int MAX = 510000;
const int INF =2000000000;
const int MOD = 1000000007;
//gcd,lcmセット
ll gcd(ll a, ll b){
   if (a%b == 0) return b;
   else return gcd(b, a%b);
}
ll lcm(ll a, ll b){	
   return a*b/gcd(a, b);
}
int pow_kai(int a, int n){//aのn乗を計算します。
  int x = 1;
  while(n > 0){//全てのbitが捨てられるまで。
    if(n&1){//1番右のbitが1のとき。
      x = x*a;
    }
    a = a*a;
    n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
  }
  return x;
}
// a^n mod を計算する
long long modPow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
// a^{-1} mod を計算する
ll modInv(ll a, ll mod) {
    return modPow(a, mod - 2, mod);
}

long long fac[MAX], finv[MAX], inv[MAX];

// combinationのテーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
 
bool isPrime(ll x){
    int i;
    if(x < 2)return false;
    else if(x == 2) return true;
    if(x%2 == 0) return false;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return false;
    return true;
}
//dfsではvector<vector<int>> g(n);で◎
int main() {
  	int a[2020][2020];
	REP(i,2020) REP(j,2020) a[i][j]=INF;
    //COMinit();
    int n,x,y;
    cin>>n>>x>>y;
    for(int i=1;i<=n;i++) {
      for(int j=i+1;j<=n;j++){
        if(j==i+1) a[i][j]=1;
        else if(j>=y&&i<=x) a[i][j]=j-y+x-i+1;
        else if(j<y&&i>x) a[i][j]=min(j-i,y-x-(j-i)+1);
        else a[i][j]=min(abs(j-y)+1+abs(i-x),j-i);
    	//cout<<"i:"<<i<<"j:"<<j<<"|"<<a[i][j]<<endl;
    }}
    int k[n];
    REP(i,n) k[i]=0;
    for(int i=1;i<=n;i++)for(int j=i+1;j<=n;j++) k[a[i][j]]++;
    for(int i=1;i<n;i++) cout<<k[i]<<endl;
}