uva 11178

题意：根据A,B,C三点的位置确定D,E,F三个点的位置。

贴模板

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
using namespace std;
const int N=100010;
const int inf=0x3f3f3f3f;
const double eps = 1e-10;

struct Point
{
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { }
};

typedef Point Vector;

Vector operator + (Vector A, Vector B)
{
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Point A, Point B)
{
    return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p)
{
    return Vector(A.x * p, A.y * p);
}

Vector operator / (Vector A, double p)
{
    return Vector(A.x / p, A.y / p);
}

bool operator < (const Point& a, const Point& b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b)
{
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B)//点乘
{
    return A.x * B.x + A.y * B.y;    
}

double Length(Vector A) //向量的模
{
    return sqrt(Dot(A, A));   
}

double Angle(Vector A, Vector B)//两个向量的夹角
{
    return acos(Dot(A, B) / Length(A) / Length(B));    
}

double Cross(Vector A, Vector B)//叉乘
{
    return A.x * B.y - A.y * B.x;    
}

double Area(Point A, Point B, Point C)//三个点组成的三角形的面积
{
    return Cross(B - A, C - A);    
}

Vector Rotate(Vector A, double rad)   //向量A逆时针旋转rad弧度后的坐标
{
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}


Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)//求直线交点
{
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

Point getD(Point A, Point B, Point C)
{
    Vector v1 = C - B;
    double a1 = Angle(A-B, v1);
    v1 = Rotate(v1, a1/3);

    Vector v2 = B - C;
    double a2 = Angle(A-C, v2);
    v2 = Rotate(v2, -a2/3);

    return GetLineIntersection(B, v1, C, v2);
}
int main()
{
    int T;
    Point A, B, C, D, E, F;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
        D = getD(A, B, C);
        E = getD(B, C, A);
        F = getD(C, A, B);
        printf("%lf %lf %lf %lf %lf %lf
", D.x, D.y, E.x, E.y, F.x, F.y);
    }
    return 0;
}