Answer: 40257

				k = (y2-y1)/(x2-x1);
           		b = (y1*x2-y2*x1)/(x2-x1);

import java.util.Set;
import java.util.TreeSet;

public class Main{
    public static void main(String[] args){
        int res = 20+21; // The slope does not exist the situation of the X-axis in the X-axis
        Set<Node> set = new TreeSet<>();
        Node[] nodes = new Node[20*25];
        int cnt = 0;
        for(int i=0;i<=19;i++){
            for(int j=0;j<=20;j++){
                nodes[cnt++] = new Node(1.0*i,1.0*j);
            }
        }
        for(int i=0;i<cnt;i++){
            for(int j=i+1;j<cnt;j++){
                double x1 = nodes[i].x;
                double y1 = nodes[i].y;
                double x2 = nodes[j].x;
                double y2 = nodes[j].y;
                if(x1==x2||y1==y2) continue;
                double k = (y2-y1)/(x2-x1);
                double b = (y1*x2-y2*x1)/(x2-x1);
                set.add(new Node(k,b));
            }
        }
        System.out.println(res+set.size());
    }
}
class Node implements Comparable<Node>{
    double x,y;

    public Node(double x, double y) {
        this.x = x;
        this.y = y;
    }



    @Override
    public int compareTo(Node o) {
        if(Double.compare(this.x,o.x)!=0) return Double.compare(this.x,o.x);
        return Double.compare(this.y,o.y);
    }
}

Answer: 2430
Idea: Test Division acquires all the amounts of approximately 2021041820210418 and then decomposes the decomposition of the decomposition into two products, obtains three counting products, and sequencing three numbers, if there is no such alumnus, then Cumulatory answers based on the three numbers

import java.util.Arrays;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        Set<Long> set = new TreeSet<>();// Store all the amounts
        for(long i=1;i<=n/i;i++){
            if(n%i==0) {
                set.add(i);
                set.add(n/i);
            }
        }
        Set<Node> res = new TreeSet<>();// Record the answer
        Object[] objects = set.toArray();
        int ans = 0;
        for(int i=0;i<objects.length;i++){
            Long a = (Long)objects[i];
            for(long j=1;j<=a/j;j++){// Decompose a product of two to two about
                if(a%j==0){
                    long[] arr = new long[3];
                    arr[0] = j;
                    arr[1] = a/j;
                    arr[2] = n/a;
                    Arrays.sort(arr,0,3);
                    if(res.contains(new Node(arr[0],arr[1],arr[2]))) continue;
                    if(arr[0]==arr[1]&&arr[1]==arr[2]){
                        ans += 1;
                    }else if(arr[0]==arr[1]||arr[1]==arr[2]){
                        ans += 3;
                    }else{
                        ans += 6;
                    }
                    res.add(new Node(arr[0],arr[1],arr[2]));
                }
            }
        }
        System.out.println(ans);
    }
}
class Node implements Comparable<Node>{
    long a,b,c;

    public Node(long a, long b, long c) {
        this.a = a;
        this.b = b;
        this.c = c;
    }

    @Override
    public int compareTo(Node o) {
        if(Long.compare(this.a,o.a)!=0) return Long.compare(this.a,o.a);
        if(Long.compare(this.b,o.b)!=0) return Long.compare(this.b,o.b);
        return Long.compare(this.c,o.c);
    }
}

Idea