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)

//#define GBN_USE_C
#ifdef GBN_USE_C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#else
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
#endif

#define rep(i,n) for(int i = 0; i < (n); ++i)
 
typedef long long ll;
typedef char str255[256];

#define BIG_PRIME 1'000'000'007
#define BP_remainder(x) ((x) % BIG_PRIME)

const int INF = 1 << 30;
const int INF_NEG = INF * -1;
const ll INF_LONG = 1LL << 62;
const ll INF_LONG_NEG = INF_LONG * -1;

#define MAX_N 2020

int main()
{
    int N, X, Y; cin >> N >> X >> Y;

    int **matrix = new int*[MAX_N];
    rep(i, MAX_N) matrix[i] = new int[MAX_N];
    
    for (int i = 1; i <= N; ++i) matrix[i][i] = 0;

    for (int row = 1; row <= X; ++row){ // スタートが枝分かれ前
        for (int cmn = row + 1; cmn <= N; ++cmn){
            if (cmn <= X) matrix[row][cmn] = matrix[cmn][row] = cmn - row;
            else if (cmn >= Y) matrix[row][cmn] = matrix[cmn][row] = cmn - row - (Y - X - 1);
            else { // ゴールがが枝の中
                matrix[row][cmn] = matrix[cmn][row] = 
                    min(cmn - row, (X + 1 - row) + (Y - cmn)); // X + 1 - row: Y までの最短距離
            }
        }
    }
    for (int row = X + 1; row <= Y - 1; ++row){ // スタートが枝の中
        for (int cmn = row + 1; cmn <= N; ++cmn){
            if (cmn < Y){ // ゴールが枝の中
                matrix[row][cmn] = matrix[cmn][row] = 
                        min(cmn - row, Y - cmn + 1 + row - X);
            } else {
                matrix[row][cmn] = matrix[cmn][row] = 
                        min(cmn - row, cmn - Y + 1 + row - X);
            }
        }
    }
    for (int row = Y; row <= N; ++row){ // 
        for (int cmn = row + 1; cmn <= N; ++cmn){
            matrix[row][cmn] = matrix[cmn][row] = cmn - row;
        }
    }
//#define GBN_DEBUG
#ifdef GBN_DEBUG
    for (int row = 1; row <= N; ++row){
        for (int cmn = 1; cmn <= N; ++cmn){
            cout << matrix[row][cmn];
        }
        puts("");
    }
#endif

    int ans[N]; rep(i, N + 1) ans[i] = 0;
            
    for (int i = 1; i <= N; ++i){
        for (int j = 1; j <= N; ++j){
            ++ans[matrix[i][j]];
        }
    }
    for (int i = 1; i < N; ++i)
        cout << ans[i] / 2 << endl;

    rep(i, MAX_N) delete[] matrix[i];
    delete[] matrix;
    return 0;
}