에 대하여
#include <iostream>
#include <utility>
#include <cmath>
#define P pair<double, double>
using namespace std;
int N;
double answer = 0;
P dot[10000];
void input()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> N;
for (int i = 0; i < N; ++i)
cin >> dot[i].first >> dot[i].second;
}
double ccw(P a, P b, P c)
{
double outer = a.first * b.second + b.first * c.second + c.first * a.second;
outer -= (b.first * a.second + c.first * b.second + a.first * c.second);
return outer / 2;
}
void solve()
{
for (int i = 1; i < N; ++i)
answer += ccw(dot[0], dot[i - 1], dot[i]);
cout << fixed;
cout.precision(1);
cout << abs(answer);
}
int main()
{
input();
solve();
return 0;
}
#include <iostream>
using namespace std;
int outer;
int p1[2], p2[2], p3[2];
void input()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> p1[0] >> p1[1];
cin >> p2[0] >> p2[1];
cin >> p3[0] >> p3[1];
}
void ccw()
{
outer = p1[0] * p2[1] + p2[0] * p3[1] + p3[0] * p1[1];
outer -= p1[1] * p2[0] + p2[1] * p3[0] + p3[1] * p1[0];
}
void solve()
{
ccw();
if (outer > 0)
cout << 1;
else if (outer < 0)
cout << -1;
else
cout << 0;
}
int main()
{
input();
solve();
return 0;
}