UVa11437

题目链接

分析：
基本操作的训练

//这里写代码片#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#define LD doubleusing namespace std;const LD eps=1e-8;struct node{    LD x,y;    node (LD xx=0,LD yy=0)    {        x=xx;y=yy;    }};node operator + (const node &a,const node &b){return node(a.x+b.x,a.y+b.y);}node operator - (const node &a,const node &b){return node(a.x-b.x,a.y-b.y);}node operator * (const node &a,const double &b){return node(a.x*b,a.y*b);}node operator / (const node &a,const double &b){return node(a.x/b,a.y/b);}node operator < (const node &a,const node &b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}node operator == (const node &a,const node &b){return a.x==b.x&&a.y==b.y;}int dcmp(LD x){    if (fabs(x)<eps) return 0;    else if (x>0) return 1;    else return -1;}LD Cross(const node &x,const node &y){return x.x*y.y-x.y*y.x;}LD Dot(const node &x,const node &y){return x.x*y.x+x.y*y.y;}node XiangJiao(node P,node v,node Q,node w){    node u=P-Q;    LD t=Cross(w,u)/Cross(v,w);    return P+v*t;}LD SS(node A,node B,node C){    LD ans=Cross(B-A,C-A);    return fabs(ans)/2;}int main(){    node A,B,C,D,E,F,P,R,Q;    int T;    scanf("%d",&T);    while (T--)    {        scanf("%lf%lf",&A.x,&A.y);        scanf("%lf%lf",&B.x,&B.y);        scanf("%lf%lf",&C.x,&C.y);        D=B+(C-B)/3;        E=C+(A-C)/3;        F=A+(B-A)/3;        P=XiangJiao(B,E-B,A,D-A);        Q=XiangJiao(C,F-C,E,E-B);        R=XiangJiao(A,D-A,C,F-C);        LD ans=SS(P,Q,R);        printf("%d\n",(int)(ans+0.5));    }    return 0;}