CFgym：Glyph Recognition（计算几何）

▶ Given some points in the plane, and 3 ≤ k ≤ 8, find two regular k-gons containing all the points between them.

▶  Each polygon is centered at the origin, i.e. all vertices have equal distance to (0, 0).

▶ Each polygon has a vertex on the positive x-axis.

▶ The inner polygon is the largest such polygon containing none of the sample points.

▶ The outer polygon is the smallest such polygon containing all of the sample points

▶ Over all values of k, maximize the ratio Sinner/Souter of polygon areas.

题意：给一堆点，要求用正3~8边形任意一种，去覆盖这些点，要满足上述条件，令内正多边形面积/外正多边形面积最大。

思路：

要求的是所有点的OC长度，先将点化成极坐标，根据对称，将所有点转移到第一份格子即[0,360/K]内。知道OA的长度和角度，从而根据三角函数轻易求出OB和OC的长度，取长度最小和最大的比一下即可。

# include <bits/stdc++.h># define pi acos(-1.0)using namespace std;struct node{    double dis, ang;}a[1003];double cal(int k, int n){    double f = 360.0/k*pi/180.0;    double imax = 0, imin = 1e18, tmp;    for(int i=0; i<n; ++i)    {        tmp = a[i].ang-(int)(a[i].ang/(360.0/k))*(360.0/k);        tmp = tmp*pi/180.0;        double len = cos(tmp-f/2)*(a[i].dis);        len = len/cos(f/2.0);        imax = max(imax, len);        imin = min(imin, len);    }    return imin*imin/(imax*imax);}int main(){    int n, id;    double x, y, ans = 0;    scanf("%d",&n);    for(int i=0; i<n; ++i)    {        scanf("%lf%lf",&x,&y);        a[i].dis = sqrt(x*x+y*y);        a[i].ang = atan(y/x)*180/pi;        if(x < 0) a[i].ang += 180;        if(a[i].ang < 0 && x >= 0) a[i].ang += 360;    }    for(int i=3; i<=8; ++i)    {        if(cal(i,n) > ans)        {            ans = cal(i, n);            id = i;        }    }    printf("%d %.10f\n",id,ans);    return 0;}