这题先推下公式，很容易发现圆锥的体积和半径满足一个二次函数，所以可以用三分
于是三分半径即可
代码：
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 10005;const double PI = acos(-1.0);struct Point {double x, y, z;void read() {scanf("%lf%lf%lf", &x, &y, &z);}} p[N];int n;double gao(double r) {double h = 0;for (int i = 0; i < n; i++) {double d = sqrt(p[i].x * p[i].x + p[i].y * p[i].y);h = max(h, p[i].z / (r – d) * r);}return h;}int main() {while (~scanf("%d", &n)) {double l = 0, r = 1e20;for (int i = 0; i < n; i++) {p[i].read();l = max(l, sqrt(p[i].x * p[i].x + p[i].y * p[i].y));}l += 1e-8;for (int i = 0; i < 200; i++) {double midl = (l * 2 + r) / 3;double midr = (l + r * 2) / 3;double h1 = gao(midl);double h2 = gao(midr);double s1 = PI * midl / 3 * midl * h1;double s2 = PI * midr / 3 * midr * h2;if (s1 > s2) l = midl;else r = midr;}printf("%.3f %.3f\n", gao(l), l);}return 0;}
，人创造奇迹常常是在瞬间，
