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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include "time.h"
#include "unistd.h"
#include "string.h"
#include <math.h>
int main(int argc, char* argv[]){
  long int a=0,b=0,c=0,ans=0;
  char s[7];

  scanf("%ld %ld",&a,&b);
  long int d[b],e[b];
  for (int i = 0; i < b; i++) {
    scanf("%ld",&d[i]);
    e[i]=d[i];
  }
  for (int i = 1; i < b; i++) {
    e[i]=d[i]-d[i-1];
  }
  e[0]=a-d[b-1]+d[0];
  long int max=0,l=0;
  for (int i = 0; i < b; i++) {
    if (e[i]>max) {
      max=e[i];
      l=i;
    }
  }
  if (l!=0) {
    printf("%d",a-d[l]+d[l-1]);
  }else{
    printf("%d",d[b-1]-d[0]);
  }
  return 0;
}