POJ 1265 Area （有向面積， pick 定理）

<span style="font-family: Arial, Helvetica, sans-serif;"></span><pre name="code" class="cpp"><span style="font-family: Arial, Helvetica, sans-serif;">pick 定理： S = a + b /2 - 1</span>

其中，S 爲格點多邊形面積，a 爲多邊形內部整點數，b 爲多邊形邊上整點數。

某條邊（A,B）上的整點數 = gcd (|xA - xB|, |yA - yB|) (不包括 A 點)

<span style="font-family: Arial, Helvetica, sans-serif;">#include <cstdio></span>

#include <cstring>
#include <cmath>
using namespace std;

struct Point
{
    int x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    //向量
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    //叉積
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    //點積
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}

int main()
{
    int t;

    scanf("%d", &t);
    for (int ca = 1; ca <= t; ca++) {
        Point cur(0,0), next;
        int n;
        int area = 0, onedge = 0, inside = 0;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            int dx, dy;
            scanf("%d%d", &dx, &dy);
            next.x = cur.x + dx;
            next.y = cur.y + dy;
            area += cur ^ next;
            onedge += gcd(abs(cur.x-next.x), abs(cur.y-next.y));
            cur = next;
        }
        area += cur ^ Point(0,0);
        onedge += gcd(abs(cur.x), abs(cur.y));
        inside = (area - onedge + 2) / 2;
        printf("Scenario #%d:\n%d %d %.1lf\n\n", ca, inside, onedge, area / 2.0);
    }

    return 0;
}