UVa 11731 Ex-circles 与旁切圆有关的计算

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题目地址:pdf版本

额，题目还是纯高中数学知识，多一点轮换性计算。利用余弦定理，面积法求旁切圆半径。

代码：

#include<iostream>#include<cstdio>#include<cmath>using namespace std;const double PI=acos(-1);double a,b,c;double A,B,C;double getAngle(double a,double b,double c){    double C=acos((a*a+b*b-c*c)/(2*a*b));        return C;}double Area(double a,double b,double c){    double p=(a+b+c)/2;    return sqrt(p*(p-a)*(p-b)*(p-c));}int main(){    int cas=0;    double s1,s2;    while(cin>>a>>b>>c)    {        if(a==0) break;        C=getAngle(a,b,c);        A=getAngle(b,c,a);        B=getAngle(c,a,b);                double ra=2*Area(a,b,c)/(b+c-a);        double rb=2*Area(a,b,c)/(c+a-b);        double rc=2*Area(a,b,c)/(a+b-c);                        double sa=ra*a/2;        double sb=rb*b/2;        double sc=rc*c/2;                s1=sa+sb+sc+Area(a,b,c);                double AA=PI/2-A/2;        double BB=PI/2-B/2;        double CC=PI/2-C/2;                s2=AA*ra*ra+BB*rb*rb+CC*rc*rc;                s2/=2;                        printf("Case %d: %.2lf %.2lf\n",++cas,s1,s2);    }}

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