这题只要根据题目，利用向量把点一个个求出来计算面积即可
不过据说有一种证明方法可以证明面积是1/7的原三角形
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int t;struct Point {double x, y;Point() {}Point(double x, double y) {this->x = x;this->y = y;}void read() {scanf("%lf%lf", &x, &y);}};typedef Point Vector;Vector operator + (Vector A, Vector B) {return Vector(A.x + B.x, A.y + B.y);}Vector operator – (Vector A, Vector 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);}double Cross(Vector A, Vector B) {return A.x * B.y – A.y * B.x;} //叉积double Area2(Point A, Point B, Point C) {return Cross(B – A, C – A);} //有向面积Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {Vector u = P – Q;double t = Cross(w, u) / Cross(v, w);return P + v * t;}int main() {scanf("%d", &t);while (t–) {Point A, B, C;A.read(); B.read(); C.read();Point D, E, F;D = (C – B) / 3 + B;E = (A – C) / 3 + C;F = (B – A) / 3 + A;Point R = GetLineIntersection(C, F – C, A, D – A);Point P = GetLineIntersection(B, E – B, A, D – A);Point Q = GetLineIntersection(C, F – C, B, E – B);printf("%d\n", (int)(fabs(Area2(P, Q, R)) / 2 + 0.5));}return 0;}
，问候不一定要慎重其事，但一定要真诚感人
