Submission #1794463

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Source Code Expand

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <functional>
#include <numeric>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <fstream>
#include <bitset>
#include <time.h>
#include <tuple>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = 1e12;
const ll MOD = 1e9 + 7;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,N) for(int i=0;i<(N);i++)
#define ALL(s) (s).begin(),(s).end()
#define EQ(a,b) (abs((a)-(b))<EPS)
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
#define fi first
#define se second
#define N_SIZE (1LL << 20)
#define NIL -1
#define MAX_N 100100

ll sq(ll num) { return num*num; }
ll mod_pow(ll x, ll n) {
	if (n == 0)return 1;
	if (n == 1)return x%MOD;
	ll res = sq(mod_pow(x, n / 2));
	res %= MOD;
	if (n % 2 == 1) {
		res *= x;
		res %= MOD;
	}
	return res;
}
ll mod_add(ll a, ll b) { return (a + b) % MOD; }
ll mod_sub(ll a, ll b) { return (a - b + MOD) % MOD; }
ll mod_mul(ll a, ll b) { return a*b % MOD; }

int n;
Point p[2001];

int main() {
	cin >> n;
	rep(i, n) {
		double x, y;
		cin >> x >> y;
		p[i] = Point(x, y);
	}
	int r = 0, o = 0;
	rep(i, n) {
		//cout << "!" << i << endl;
		vector<double> vd;
		rep(j, n) {
			Point pp = p[j] - p[i];
			if (i != j) {
				vd.push_back(atan2(pp.imag(), pp.real()));
				vd.push_back(atan2(pp.imag(), pp.real()) + 2.0 * PI);
			}
		}
		sort(ALL(vd));
		//rep(j, vd.size())cout << vd[j] << endl;
		rep(j, vd.size() / 2) {
			auto it1 = upper_bound(ALL(vd), vd[j] + PI / 2.0 + EPS);
			auto it2 = lower_bound(ALL(vd), vd[j] + PI / 2.0 - EPS);
			auto ub = lower_bound(ALL(vd), vd[j] + PI - EPS);
			//cout << *it1 << " " << *it2 << " " << *ub << endl;
			r += (it1 - it2);
			o += (ub - it1);
		}
		//cout << "!!" << r << " " << o << endl;
	}
	cout << n*(n - 1)*(n - 2) / 6 - o - r << " " << r << " " << o << endl;
}

Submission Info

Judge Result