题意：

  　　给你一个多边形，然后给你一个在多边形里面的点，让你求出以这个点为圆心半径r,使得该多边形与该圆的相交的面积占多边形的（1-p/q）

思路：

用求多边形与圆相交面积的几何模板，二分半径求得答案

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>

using namespace std;

const double eps = 1e-9;
const double PI = acos(-1.0);


int dcmp(double x)
{
    if( x > eps ) return 1;
    return x < -eps ? -1 : 0;
}

struct Point
{
    double x,y;
    Point()
    {
        x = y = 0;
    }
    Point(double a,double b)
    {
        x = a;
        y = b;
    }
    inline void input()
    {
        scanf("%lf%lf",&x,&y);
    }
    inline Point operator-(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    inline Point operator+(const Point &b)const
    {
        return Point(x + b.x,y + b.y);
    }
    inline Point operator*(const double &b)const
    {
        return Point(x * b,y * b);
    }
    inline double dot(const Point &b)const
    {
        return x * b.x + y * b.y;
    }
    inline double cross(const Point &b,const Point &c)const
    {
        return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);
    }
    inline double Dis(const Point &b)const
    {
        return sqrt((*this-b).dot(*this-b));
    }
    inline bool InLine(const Point &b,const Point &c)const  //三点共线
    {
        return !dcmp(cross(b,c));
    }
    inline bool OnSeg(const Point &b,const Point &c)const  //点在线段上，包括端点
    {
        return InLine(b,c) && (*this - c).dot(*this - b) < eps;
    }
    int operator^(const Point &b) const
    {
        return y*b.x-x*b.y;
    }
};

inline double min(double a,double b)
{
    return a < b ? a : b;
}
inline double max(double a,double b)
{
    return a > b ? a : b;
}
inline double Sqr(double x)
{
    return x * x;
}
inline double Sqr(const Point &p)
{
    return p.dot(p);
}

Point LineCross(const Point &a,const Point &b,const Point &c,const Point &d)
{
    double u = a.cross(b,c), v = b.cross(a,d);
    return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v));
}

double LineCrossCircle(const Point &a,const Point &b,const Point &r,
                       double R,Point &p1,Point & p2)
{
    Point fp = LineCross(r, Point(r.x+a.y-b.y, r.y+b.x-a.x), a, b);
    double rtol = r.Dis(fp);
    double rtos = fp.OnSeg(a, b) ? rtol : min(r.Dis(a), r.Dis(b));
    double atob = a.Dis(b);
    double fptoe = sqrt(R * R - rtol * rtol) / atob;
    if( rtos > R - eps ) return rtos;
    p1 = fp + (a - b) * fptoe;
    p2 = fp + (b - a) * fptoe;
    return rtos;
}

double SectorArea(const Point &r,const Point &a,const Point &b,double R)  //不大于180度扇形面积，r->a->b逆时针
{
    double A2 = Sqr(r - a), B2 = Sqr(r - b), C2 = Sqr(a - b);
    return R * R * acos( (A2 + B2 - C2) * 0.5 / sqrt(A2) / sqrt(B2)) * 0.5;
}

double TACIA(const Point &r,const Point &a,const Point &b,double R)
{
    double adis = r.Dis(a), bdis = r.Dis(b);
    if( adis < R + eps && bdis < R + eps )
        return r.cross(a, b) * 0.5;
    Point ta, tb;
    if( r.InLine(a,b) ) return 0.0;
    double rtos = LineCrossCircle(a, b, r, R, ta, tb);
    if( rtos > R - eps )
        return SectorArea(r, a, b, R);
    if( adis < R + eps )
        return r.cross(a, tb) * 0.5 + SectorArea(r, tb, b, R);
    if( bdis < R + eps )
        return r.cross(ta, b) * 0.5 + SectorArea(r, a, ta, R);
    return r.cross(ta, tb) * 0.5 + SectorArea(r, tb, b, R) + SectorArea(r, a, ta, R);
}

const int MAXN  = 505;
Point p[MAXN];

double SPICA(int n,Point r,double R)
{
    int i;
    double ret = 0, if_clock_t;
    for( i = 0 ; i < n ; ++i )
    {
        if_clock_t = dcmp(r.cross(p[i], p[(i + 1) % n]));
        if( if_clock_t < 0 )
            ret -= TACIA(r, p[(i + 1) % n], p[i], R);
        else ret += TACIA(r, p[i], p[(i + 1) % n], R);
    }
    return fabs(ret);
}


double ComputePolygonArea(int n)
{
    double sum=0;
    for(int i=1;i<=n-1;i++)
        sum+=(p[i]^p[i-1]);
    sum+=(p[0]^p[n-1]);
    return fabs(sum/2);
}

int main()
{
    int n,m;
    scanf("%d",&n);///多边形n个顶点
    for(int i = 0 ; i < n ; ++i )///顶点坐标
        p[i].input();
    double polyArea = ComputePolygonArea(n);///计算多边形面积
    scanf("%d",&m);
    while(m--)
    {

        Point circle;
        circle.input(); ///圆心坐标
        int pp,qq;
        scanf("%d%d",&pp,&qq);
        double area = (1.0-(double)pp/qq)*polyArea;

        ///二分圆的半径
      //  printf("%f\n",area);
        double l =0, r=1e18;
        ///固定二分次数
        for(int i=1;i<300;i++){
            double mid = (l+r)/2.0;
            double insection = SPICA(n,circle,mid); ///圆与多边形交的面积
            if(insection>area){
                r = mid-eps;
            }else{
                l = mid;
            }
        }
        printf("%.10lf\n",r);
    }
    return 0;
}