POJ-3737 UmBasketella题解及体会

  #include
<iostream>
#include<iomanip>
#include
<cstdio>
#include
<cmath>
using namespace
std;
#define eps 1e-8
const double pi=acos(-1.0);

double S;
double
cal(double x){

    return(1.0/3.0)*x*sqrt(S*(S-(2*pi*x*x)));

}
int
main(){

    while(cin>>S)

    {

        double left=0.1,right=sqrt(S/2.0/pi);

        while(right-left>eps)

        {

            double middle=(left+right)/2;

            double middleright=(middle+right)/2;

            if(cal(middle)>cal(middleright))

                right=middleright;

            else

                left=middle;

        }

        cout<<setprecision(2)<<fixed<<cal(left)<<endl;

        cout<<setprecision(2)<<fixed<<sqrt(S*(S-2*pi*left*left))/pi/left<<endl;

        cout<<setprecision(2)<<fixed<<left<<endl;

    }

    return 0;

}

本题的解题方向在于得出圆锥表面积和体积的关系，得到R的上限，并借此通过二分法求得使圆锥体积最大的半径，进而得解，输出时要注意格式。