Area POJ

Area

 POJ - 1265

题意：起始点原点，给你线段横向占x点，纵向占y点，求该过程走完图形的面积，在线上格点上的点数和在图形内部的点数

思路：1.皮克定理，s=in+on/2+1，可以利用多边形面积公示先求出面积，

　　　2.线上的点数为gcd(x,y)

　　　3.再利用皮克定理求出内部的点数

// 
// Created by HJYL on 2020/2/4.
//
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps=1e-8;
const int maxn=1e6+5;
long long gcd(long long a,long long b) {
    while (b) {
        int r = b;
        b = a % b;
        a = r;
    }
    return a;
}
struct Point{
    int x,y;
    Point(int x=0,int y=0):x(x),y(y){}
    Point operator - (Point const &b)const
    {
        return Point(x-b.x ,y-b.y);
    }
    bool operator < (Point const &c)const{
        if(x==c.x)
            return y<c.y;
        return x<c.x;
    }
}p[maxn];
double Cross(Point A,Point B)
{
    return (A.x*B.y)-(A.y*B.x);
}
long long PolygonArea(Point* p, int n) {//p为端点集合，n为端点个数
    long long s = 0;
    for (int i = 1; i < n - 1; ++i)
        s += Cross(p[i] - p[0], p[i + 1] - p[0]);
    return s<0?-s:s;
}
int main()
{
    int T;
    scanf("%d",&T);
    int Case=1;
    while(T--)
    {
        int n;
        scanf("%d",&n);
        p[0].x=0,p[0].y=0;
        int x,y;
        long long on=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            p[i].x=p[i-1].x+x;
            p[i].y=p[i-1].y+y;
            on+=gcd(abs(x),abs(y));
        }
        long long s=PolygonArea(p,n+1);//s=on/2+in-1;
        long long in=(s+2-on)/2;
        printf("Scenario #%d:
",Case++);
        printf("%lld %lld %.1lf

",in,on,s/2.0);

    }
    return 0;
}