UVALive 6092 Catching Shade in Flatland

​​https://icpcarchive.ecs.baylor.edu/external/60/6092.pdf​​

坐标上有很多相离圆，问一条可以绕原点旋转的线段与这些圆相交的最长弦长之和。

直接模拟即可。注意是线段。

#include<iostream>  
#include<algorithm>
#include<math.h>
#include<cstdio>
using namespace std;

int main()
{
  int n, a[500], b[500], r[500];
  while (cin >> n, n)
  {
    for (int i = 0; i < n; i++) cin >> a[i] >> b[i] >> r[i];
    double maxn = 0;
    for (int i = 0; i < 24 * 60; i++)
    {
      double k = asin(1.0)*(i / 360.0), sum = 0;
      for (int j = 0; j < n; j++)
      if ((i / 360 == 0 || i / 360 == 3) && a[j] + b[j] * tan(k)>0 || (i / 360 == 1 || i / 360 == 2) && a[j] + b[j] * tan(k) < 0)
      {
        double dis = -2 * tan(k) *a[j] * b[j] + b[j] * b[j] - a[j] * a[j];
        dis = a[j] * a[j] + dis / (tan(k)*tan(k) + 1);
        if (dis < r[j] * r[j]) sum += 2 * sqrt(r[j] * r[j] - dis);
      }
      maxn = max(maxn, sum);
    }
    printf("%.3lf\n", maxn);
  }
  return 0;
}