[POJ3737]UmBasketella（三分法）

题目描述

传送门

题解

三分法
设圆锥底面半径为r，高为h
$V=\sqrt {(s^2-2s\pi r^2)r^2}$，类似于一个关于r的二次函数单峰
三分r
计算$h=\sqrt {{s^2-2s\pi r}\over \pi^2r^2}$

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const double pi=acos(-1.0);
const double eps=1e-6;
int dcmp(double x)
{
    if (x<=eps&&x>=-eps) return 0;
    return (x>0)?1:-1;
}
double s,r,h,v;

double calc(double r)
{
    double h=sqrt((s*s-2*s*pi*r*r)/(pi*pi*r*r));
    return (1.0/3.0)*pi*r*r*h;
}
double find()
{
    double l=0,r=sqrt(s/pi),mid1,mid2,ans;
    while (dcmp(r-l)>0)
    {
        mid1=l+(r-l)/3.0;
        mid2=r-(r-l)/3.0;
        if (calc(mid1)<=calc(mid2)) ans=mid2,l=mid1;
        else r=mid2;
    }
    return ans;
}
int main()
{
    while (~scanf("%lf",&s))
    {
        r=find();
        h=sqrt((s*s-2*s*pi*r*r)/(pi*pi*r*r));
        v=calc(r);
        printf("%.2lf\n%.2lf\n%.2lf\n",v,h,r);
    }
}