题目大意:某个逗逼用网球拍打苍蝇,苍蝇是个标准的圆,网球拍是这样(中心对称):
输入如图的数据,计算打中苍蝇的概率。
题解:
将网球拍的线和边框宽度全部扩大f,计算此时的空档面积与总面积的比(相当于苍蝇的圆心只能处于空档中),即为概率。因为球拍中心对称,所以只用计算四分之一就行了。
如何计算空档面积:
- 正常:
正常计算:g*g - 少两个点
分成弓形和梯形 - 少一个点
先算大正方形,减去小三角形,加上弓形。 - 少三个点
弓形加三角形
以球拍圆心建坐标系,枚举空档坐标,判断它的点是否在圆内(点到圆心的距离≤半径)来判断它的形状。
弓形面积计算:(θ是弧度)
角度θ计算:
枚举空档时就枚举上面的x,y。
代码:
#include<cstdio>
#include<cmath>
using namespace std;
#define EPS 1e-100
inline int dcmp(double a,double b)
{
if(a-b<-EPS)
return -1;
else if(a-b<=EPS)
return 0;
return 1;
}
inline int inround(double a,double b,double R)
{
return dcmp(a*a+b*b,R*R)!=1;
}
inline double gg(double a,double b)
{
return sqrt(a*a+b*b);
}
inline double ngg(double a,double b)
{
return sqrt(a*a-b*b);
}
inline double Trapezoid(double a,double b,double h)
{
return h*(a+b)/2;
}
inline double CircleSegment(double r,double th)
{
return r*r*(th-sin(th))/2;
}
inline double Triangle(double a,double h)
{
return a*h/2;
}
int main()
{
int N;
scanf("%d",&N);
for(int Case=1;Case<=N;Case++)
{
double area1=0.0,area2;
double f,R,t,r,g;
scanf("%lf%lf%lf%lf%lf",&f,&R,&t,&r,&g);
area2=R*R*M_PI/4;
R-=t+f;
r+=f;
g-=2*f;
if(dcmp(R,0.0)!=1||dcmp(g,0.0)!=1)
{printf("Case #%d: %.6lf\n",Case,1.0);continue;}
for(double x1=r;dcmp(x1,R)==-1;x1+=g+2*r)
for(double y1=r;dcmp(y1,R)==-1;y1+=g+2*r)
{
if(!inround(x1,y1,R))continue;
double x2=x1+g,y2=y1+g;
if(inround(x2,y1,R)&&inround(x1,y2,R)&&inround(x2,y2,R))
{
area1+=g*g;
continue;
}
if(!inround(x2,y1,R)&&inround(x1,y2,R)&&!inround(x2,y2,R))
{
area1+=Trapezoid(ngg(R,y1)-x1,ngg(R,y2)-x1,g);
area1+=CircleSegment(R,asin(y2/R)-asin(y1/R));
continue;
}
if(!inround(x2,y1,R)&&!inround(x1,y2,R)&&!inround(x2,y2,R))
{
area1+=Triangle(ngg(R,x1)-y1,ngg(R,y1)-x1);
area1+=CircleSegment(R,acos(x1/R)-asin(y1/R));
continue;
}
if(inround(x2,y1,R)&&inround(x1,y2,R)&&!inround(x2,y2,R))
{
area1+=g*g-Triangle(y2-(ngg(R,x2)),x2-(ngg(R,y2)));
area1+=CircleSegment(R,asin(y2/R)-acos(x2/R));
continue;
}
if(inround(x2,y1,R)&&!inround(x1,y2,R)&&!inround(x2,y2,R))
{
area1+=Trapezoid(ngg(R,x1)-y1,ngg(R,x2)-y1,g);
area1+=CircleSegment(R,acos(x1/R)-acos(x2/R));
continue;
}
}
printf("Case #%d: %.6lf\n",Case,(area2-area1)/area2);
}
return 0;
}