题解 | #画三角#

求一下面积即可，如果面积小于0，则是顺时针，否则为逆时针。

/*
 * Created by 孙书仪.
 * 诸神天佛，佑我上分！
*/
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const int N = 2e5 + 10;
const int M = 1e6 + 12;
const int inf = 0x7fffffff;
const ll INF = 1e18;
const ll mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1);
int sgn(double x) {
	if(fabs(x) < eps)return 0;
	return x < 0 ? -1 : 1;
}

struct pot {
	double x,y;
	pot(){}
	pot(double x,double y):x(x),y(y){}
	pot operator-(const pot& a) const {return pot(x - a.x,y - a.y);}

	bool operator==(const pot& a) const {return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;}
	bool operator<(const pot& a)const {return sgn(x - a.x) == 0 ? sgn(y - a.y) < 0 : sgn(x - a.x) < 0;}

};

struct line {
	pot p1,p2;
	line(){}
	line(pot p1,pot p2):p1(p1),p2(p2){}
};

double det(pot a,pot b){return a.x * b.y - a.y * b.x;}
double dot(pot a,pot b){return a.x * b.x + a.y * b.y;}

pot p[N];
void solve() {
	int n;cin >> n;
	for(int i = 0;i < n;i++) {
		cin >> p[i].x >> p[i].y;
	}

	double ans = 0;
	for(int i = 0;i < n;i++) {
		ans += det(p[i],p[(i + 1) % n]);
	}

	if(ans < 0) {
		cout << "clockwise" << '\n';
	}
	else cout << "counterclockwise" << '\n';

}

int main() {
	ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
	//cout << fixed << setprecision(3);
	int t = 1;
	//cin >> t;
	while(t--) {
		solve();
	}
	return 0;
}