poj1954（pick公式：求多边形内点的个数）

#include<iostream>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
#include<math.h>

#define eps 1e-9
#define pi acos(-1.0)
#define P system("pause")
using namespace std;
struct point
{
     double x,y;       
};
double cross(point A,point B)
{
    return A.x*B.y-A.y*B.x;       
}

int gcd(int a,int b)
{
    if(b>a){int t;t=a;a=b;b=t;}
    if(b==0) return a;
    return  a%b==0?b:gcd(b,a%b);   
}
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
    point A,B,C;//面积 = 边界点数 ÷ 2 ＋ 内部点数 － 1
    while(scanf("%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y))
    {
        if(!A.x&&!B.x&&!C.x&&!A.y&&!B.y&&!C.y) break;
        int k=0;
       // k=gcd(fabs(A.x-B.x),fabs(A.y-B.y));
       k=  gcd(fabs(A.x-B.x),fabs(A.y-B.y))
         + gcd(fabs(A.x-C.x),fabs(A.y-C.y))+gcd(fabs(C.x-B.x),fabs(C.y-B.y));
        point v,w;
        w.x=A.x-B.x;w.y=A.y-B.y;
        v.x=A.x-C.x;v.y=A.y-C.y;
       // cout<<cross(w,v)/2<<" "<<k<<endl;;
        int area=fabs(cross(w,v))/2+1-k/2;
        printf("%d\n",area);
                                                                        
    }     

 //   P;                               
    return 0;    
}