The straight line of the daily real questions of the Blue Bridge Cup

topic source

2021 Blue Bridge Cup Provincial

Topic link: http://acm.mangata.ltd/p/P1485

test center

math, violence

Video explanation

https://www.bilibili.com/video/BV1BY411J795

ideas

Because the selected points are all integer coordinates, so we select at most $21\times 20=420points$ , then we can select each two points as a straight line, but in this case, there may be repeated selection of a line, so we need to de-duplicate, so how to de-duplicate? We know that the standard equation for a line is:$Ax+By+C=0$ Then that means we only need to determine$A,B,C$ is fine, but here you need to pay attention to$A,B,C$ need to be simplified, otherwise we will not be able to achieve the effect of de-duplication. For the de-duplication of the structure of three or more elements, we can choose to write a structure by ourselves, but we can also use it directlyvector. Then throw eachvectorinto setit$What\; about\; A,B,C$ ? , we all know the equation of the line determined by two points:$\frac{(and-{and}_1)}{Y_1-{and}_2}=\frac{x-x_1}{x_1-x_2}$Then we can get: $({and}_2-Y_1)x+(x_1-x_2)and+(x_2{Y}_1-x_1{Y}_2)=0$

We can get: $A=Y_2-Y_1、B=x_1-x_2、C=x_2{Y}_1-x_1{Y}_2$, then we remove the common factors of the three and put them vectorin , and then put them in setto complete the calculation of a line. Finally, we only need to output setthe size of the container.

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define endl "\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

const double eps = 0.0000001;

int main()
{
    
    
	vector<PII> Point;
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n = 20,m = 21;
//	cin>>n>>m;

	for(int x = 0;x < n; ++x)
		for(int y = 0;y < m; ++y)
			Point.push_back({
    
    x,y});

	int l = Point.size();
	set<vector<int>> ans;
	for(int i = 0;i < l; ++i) {
    
    
		for(int j = i + 1;j < l; ++j) {
    
    
			vector<int> t;
			int x1 = Point[i].first,y1 = Point[i].second;
			int x2 = Point[j].first,y2 = Point[j].second;
			int a,b,c;
			a = y2 - y1;
			b = x1 - x2;
			c = x2 * y1 - x1 * y2;
			int d = __gcd(__gcd(a,b),c);
			t.push_back(a/d);
			t.push_back(b/d);
			t.push_back(c/d);
			ans.insert(t);
		}
	}
	cout<<ans.size()<<endl;
	return 0;
}
/*
ans = 40257
*/