#include <iostream>#include<iomanip>#include <cstdio>#include <cmath>using namespacestd;#define eps 1e-8const 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;}

本題的解題方向在于得出圓錐表面積和體積的關(guān)系，得到R的上限，并借此通過(guò)二分法求得使圓錐體積最大的半徑，進(jìn)而得解，輸出時(shí)要注意格式。