ZOJ3550 Big Keng(三分)

题意：给定一个立体的图形，上面是圆柱，下面是圆台，圆柱的底面半径和圆台的上半径相等，然后体积的V时，问这个图形的表面积最小可以是多少。（不算上表面）。一开始拿到题以为可以YY出一个结果，就认为它是圆锥，赛后才知道原来要三分三分再三分。 就是对上下体积三分，对上半径和下半径三分。至于为什么是凸的貌似也不怎么好想，但是我后来确实发现单纯的圆锥肯定取不到最大值，这题就当作是学习三分的技巧啦－ －0

#pragma warning(disable:4996)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<cmath>
#define ll long long
#define eps 1e-5
using namespace std;

double r1, r2, V;
double v1, v2;
double pi = acos(-1.0);

int dcmp(double x)
{
	return (x > eps) - (x < -eps);
}

double cal(double x)
{
	r2 = x;
	double H = v1 / (pi*r1*r1);
	double h = 3 * v2 / (pi*r1*r1 + pi*r2*r2 + pi*r1*r2);
	double ans = 0;
	double mother = sqrt((r2 - r1)*(r2 - r1) + h*h);
	ans = pi*(r1 + r2)*mother + 2 * pi*r1*H + pi*r2*r2;
	return ans;
}

double lr(double x)
{
	v1 = x; v2 = V - x;
	double l = 0, r = r1;
	while (dcmp(r - l)>0)
	{
		double m1 = l + (r - l) / 3;
		double m2 = l + 2 * (r - l) / 3;
		double x1 = cal(m1);
		double x2 = cal(m2);
		if (x1 < x2) r = m2;
		else l = m1;
	}
	return cal(l);
}

double vol(double x)
{
	r1 = x;
	double l = 0, r = V;
	while (dcmp(r-l)>0)
	{
		double m1 = l + (r - l) / 3;
		double m2 = l + 2 * (r - l) / 3;
		double x1 = lr(m1),x2 = lr(m2);
		if (x1 < x2) r = m2;
		else l = m1;
	}
	return lr(l);
}

double solve()
{
	double l = 0, r = 10*V;
	while (dcmp(r - l)>0){
		double m1 = l + (r - l) / 3;
		double m2 = l + 2 * (r - l) / 3;
		double x1 = vol(m1), x2 = vol(m2);
		if (x1 < x2) r = m2;
		else l = m1;
	}
	return vol(l);
}

int main()
{
	while (cin >> V)
	{
		double ans = solve();
		printf("%.6lf\n", ans);
	}
	return 0;
}