POJ 3737 UmBasketella

//POJ 3737 UmBasketella//AC by warteac//2013-7-23/*    对锥形地面半径三分找到体积最大    求高度判断高度是否可能存在；精度1e-3 WA,1e-4AC*/#include<iostream>#include<cstdio>#include<cmath>#include<iomanip>using namespace std;const double pi = acos(-1.0);const double eps = 1e-4;double s;//计算高度double getH(double r){    double l = s/r/pi - r;    double h = l*l - r*r;    if(h < 0) return 0;    else return sqrt(h);}//计算体积double getV(double r){    double h = getH(r);    if(h)    return pi*r*r*h/3;    else return 0;}//获得1/3处的点double getMid1(double l, double r){    return (2*l + r)/3;}//获得2/3处的点double getMid2(double l, double r){    return (l + 2*r)/3;}//三分double computing(){    double l = 0, r = sqrt(s/pi);    double mid1, mid2,v1,v2;    while(fabs(l - r) > eps){        mid1 = getMid1(l,r);        mid2 = getMid2(l,r);        v1 = getV(mid1);        v2 = getV(mid2);        if(v1 < v2) l = mid1;        else r = mid2;    }    return getMid1(l,r);}int main(){    while(cin >> s){        double r = computing();        cout <<setiosflags(ios::fixed);          cout << setprecision(2);                cout << getV(r) << endl;        cout << getH(r) << endl;                cout << r << endl;    }    return 0;}