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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

#define pb push_back
#define sort(s) sort(s.begin(),s.end())
#define reverse(s) reverse(s.begin(),s.end())
#define rep(i,n) for(ll (i)=0;(i)<(n);(i)++)

const ll mod = 1e9+7;

//最大公約数
ll gcd(ll a, ll b) {
    return b ? gcd(b, a%b) : a;
}

//最小公倍数
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

//素数判定
bool isPrime(ll x){
    ll i;
    if(x < 2)return 0;
    else if(x == 2) return 1;
    if(x%2 == 0) return 0;
    for(i = 3; i*i <= x; i += 2) if(x%i == 0) return 0;
    return 1;
}

//桁和
int digsum(ll n) {
    int res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

//桁数
int dignum(ll n) {
    int res = 0;
    while(n > 0) {
        res++;
        n /= 10;
    }
    return res;
}

//文字列中の特定の文字カウント
ll stringcount(string s,char c){
    return count(s.cbegin(),s.cend(),c);
}

//階乗
ll ka(ll i){
    ll res=1;
    while(i>0){
        res=res*i;
        i--;
    }
    return res;
}
 
//ncr
ll ncr(ll x,ll y){
    ll a = x;
    ll b = 1;
    if(y>(x/2)){
        y=x-y;
    }
    for(ll i=1;i<y;i++){
        a*=x-i;
        b*=i+1;
    }
    return a/b;
}

//フィボナッチ数列 -92
ll f(ll x){
    ll dp[x+1];
    dp[1]=1;
    dp[2]=1;
    for(ll i=3;i<=x;i++){
        dp[i]=dp[i-1]+dp[i-2];
    }
    return dp[x];
}



int main(){
    int n,x,y;
    cin >> n >> x >> y;
    int max = n-(y-x);
    map<int,int> ans;
    rep(i,n-1) ans[i]++;

    for(int i=1;i<n;i++){
        for(int k=i+1;k<=n;k++){
            int s = k-i;
            if(i<=x&&k<=y){
                s = min(s,x-i+1+y-k);
            }else if(i>=x&&k>=y){
                s = min(s,i-x+1+k-y);
            }else if(i<x&&k>y){
                s = s - (y-x-1) ;
            }
            ans[s-1]++;
        }
    }
    
    for(int i=0;i<n-1;i++){
        ans[i]--;
        cout << ans[i] << endl;
    }
}