poj 3737 UmBasketella(三分+求导)

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【题目大意】：给出一个圆锥体的表面积，求最大的体积，并输出其半径和高。

【解题思路】：下午在比赛的时候是直接求导推的公式做的。

晚上回来想想其实三分极值可以做，但是一直wa，不知道为什么。

我原来三分是这么写的 mid1=(low+high)/2.0; mid2=(mid+high)/2.0; 但是一直过不了，后来改成mid1=low+(high-low)/3.0; mid2=high-(high-low)/3.0；就过了~~不知道什么原因。留着思考....

【代码】：

//三分

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <string>#include <cctype>#include <map>#include <iomanip>                   using namespace std;                   #define eps 1e-10#define pi acos(-1.0)#define inf 1<<30#define linf 1LL<<60#define pb push_back#define lc(x) (x << 1)#define rc(x) (x << 1 | 1)#define lowbit(x) (x & (-x))#define ll long longdouble s;double solve_v(double r){    return  pi*r*r*sqrt(pow((s-pi*r*r)/(pi*r),2)-r*r)/3;}int main(){while (~scanf("%lf",&s)){double v=0.0;double low=0,high=sqrt(s/pi);while (low+eps<high){double mid1=low+(high-low)/3.0;double mid2=high-(high-low)/3.0;double v1,v2;            v1=solve_v(mid1);            v2=solve_v(mid2);            if (v1>v2) v=v1,high=mid2;else v=v2,low=mid1;}    double h=sqrt(pow((s-pi*high*high)/(pi*high),2)-high*high);    printf("%.2f\n%.2f\n%.2f\n", v,h,high);        }return 0;}

//求导

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <string>#include <cctype>#include <map>#include <iomanip>                   using namespace std;                   #define eps 1e-8#define pi acos(-1.0)#define inf 1<<30#define linf 1LL<<60#define pb push_back#define lc(x) (x << 1)#define rc(x) (x << 1 | 1)#define lowbit(x) (x & (-x))#define ll long longdouble s,r,h,v,l;void solve(){    r=sqrt(s/(4*pi));    l=s/(pi*r)-r;    h=sqrt(l*l-r*r);    v=1.0/3*pi*r*r*h;}int main() {    while (~scanf("%lf",&s)){        solve();         printf ("%.2f\n",v);        printf ("%.2f\n",h);        printf ("%.2f\n",r);    }     return 0;}