标签:style blog http io color os ar sp div
题意:
给出三角形ABC顶点的坐标,DEF分别是三边的三等分点。求交点所形成的三角形PQR的面积。
分析:
根据梅涅劳斯定理,我们得到:
所以AR : RP : PD = 3 : 3 : 1
同理,BE和CF也被分成这样比例的三段。
△ADC = (2/3)△ABC
△CDR = (4/7)△ADC
△CPR = (3/4)△CDR
△PQR = (1/2)△CPR
所以:△PQR = (1/7)△ABC
1 #include <cstdio> 2 #include <cmath> 3 struct Point 4 { 5 double x, y; 6 Point(double x=0, double y=0):x(x), y(y) {} 7 }; 8 typedef Point Vector; 9 Point operator - (const Point& a, const Point& b) 10 { 11 return Point(a.x-b.x, a.y-b.y); 12 } 13 double Cross(Vector a, Vector b) 14 { 15 return a.x*b.y - a.y*b.x; 16 } 17 double area(const Point& a, const Point& b, const Point& c) 18 { 19 return fabs(Cross(b-a, c-a)/2); 20 } 21 22 int main() 23 { 24 //freopen("11437in.txt", "r", stdin); 25 int T; 26 scanf("%d", &T); 27 while(T--) 28 { 29 Point a, b, c; 30 scanf("%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y); 31 double ans = area(a, b, c) / 7; 32 printf("%d\n", (int)floor(ans+0.5000)); 33 } 34 return 0; 35 }
UVa 11437 (梅涅劳斯定理) Triangle Fun
标签:style blog http io color os ar sp div
原文地址:http://www.cnblogs.com/AOQNRMGYXLMV/p/4055962.html