题意：给你三角形的三条边，求图中DEF的面积和阴影部分的面积。
题解：一些模板，三角形的旁切圆半径：
、
与
三旁心为
#include<set> #include<map> #include<queue> #include<stack> #include<cmath> #include<vector> #include<string> #include<cstdio> #include<cstring> #include<iomanip> #include<stdlib.h> #include<iostream> #include<algorithm> using namespace std; #define eps 1E-8 /*注意可能会有输出-0.000*/ #define Sgn(x) (x<-eps? -1 :x<eps? 0:1)//x为两个浮点数差的比较,注意返回整型 #define Cvs(x) (x > 0.0 ? x+eps : x-eps)//浮点数转化 #define zero(x) (((x)>0?(x):-(x))<eps)//判断是否等于0 #define mul(a,b) (a<<b) #define dir(a,b) (a>>b) typedef long long ll; typedef unsigned long long ull; const int Inf=1<<28; const ll INF=1ll<<60; const double Pi=acos(-1.0); const int Mod=1e9+7; const int Max=10010; struct point { double x,y; }poi[Max]; double ans1,ans2; double Jud(double a,double b,double c) { //cout << a*a+b*b-c*c << " " << 2*a*b << endl; return acos((a*a+b*b-c*c)/(2*a*b)); } double Dis(point p1,point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } void Solve(double a,double b,double c) { double x1,y1,x2,y2,x3,y3,href; href=(a+b+c)/2; double S=sqrt(href*(href-a)*(href-b)*(href-c)); x3=y3=x2=0; y2=a; y1=(a*a+b*b-c*c)/(2*a); x1=sqrt(b*b-y1*y1); poi[0].x=(-a*x1+b*x2+c*x3)/(-a+b+c); poi[0].y=(-a*y1+b*y2+c*y3)/(-a+b+c); poi[1].x=(a*x1-b*x2+c*x3)/(a-b+c); poi[1].y=(a*y1-b*y2+c*y3)/(a-b+c); poi[2].x=(a*x1+b*x2-c*x3)/(a+b-c); poi[2].y=(a*y1+b*y2-c*y3)/(a+b-c); double aa=Dis(poi[1],poi[2]); double bb=Dis(poi[0],poi[1]); double cc=Dis(poi[0],poi[2]); href=(aa+bb+cc)/2; ans1=sqrt(href*(href-aa)*(href-bb)*(href-cc)); double r1=(2*S)/(-a+b+c); double r2=(2*S)/(a-b+c); double r3=(2*S)/(a+b-c); //cout << aa << " " << bb << " " << cc <<endl; ans2=(Jud(bb,cc,aa)/2)*r1*r1; ans2+=(Jud(aa,bb,cc)/2)*r2*r2; ans2+=(Jud(cc,aa,bb)/2)*r3*r3; return; } int main() { int a,b,c; int coun=0; while(~scanf("%d %d %d",&a,&b,&c)) { if(!a&&!b&&!c) break; Solve((double)a,(double)b,(double)c); printf("Case %d: %.2f %.2f ",++coun,ans1,ans2); } return 0; }