Big Keng

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题目描述：

一个有着共同底的圆柱套一个圆台.具体的图在http://acm.bnu.edu.cn/v3/contest_show.php?cid=6413#problem/B
它的r,R,h1,h2都不知道.现在给出体积.求最小的表面积.注意是一个桶,有很多面没有.

题解：

大大大暴力的三分.三分3次. 关键是注意精度的把握.

重点：

(1)几何可以用三分做+long double做
(2)三分的边界开始的设定可能和输入有关.

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(int i = a;i < b;i++)#define REP_D(i, a, b) for(int i = a;i <= b;i++)typedef long long ll;using namespace std;const double Pi = acos(-1.0);double v;double h ;double calcu(double R, double r, double H){    double tmpv = v - Pi*R*R*H;    h = 3.0*tmpv/(Pi*(R*R+R*r+r*r));    return Pi*r*r + Pi*(R+r)*sqrt((R-r)*(R-r)+h*h) + 2.0*Pi*R*H;}double calu_r(double R, double H){    double rl = 1e-2*v, rr = R;    double mid, midmid, ansmid, ansmidmid;    while(fabs(rl-rr) > 1e-7)    {        mid = (rl + rr)/2;        midmid = (rl + mid)/2;        ansmid = calcu(R, mid, H);        ansmidmid = calcu(R, midmid,H);        if(ansmid < ansmidmid)        {            rl = midmid;        }        else        {            rr = mid;        }    }    return ansmid;}double solve(double H){    double Rl = 1e-2*v,Rr = 1e2*v;    double mid, midmid, ansmid, ansmidmid;    while(fabs(Rl-Rr) > 1e-7)    {        mid = (Rl+Rr)/2;        midmid = (mid+Rl)/2;        ansmid = calu_r(mid, H);        ansmidmid = calu_r(midmid, H);        if(ansmid < ansmidmid)        {            Rl = midmid;        }        else        {            Rr = mid;        }    }    return ansmid;}double calcu_H(){    double Hl = 1e-2*v,Hr = 1e2*v;//边界和v有关.    double mid, midmid, ansmid, ansmidmid;    while(fabs(Hl-Hr) > 1e-7)    {        mid = (Hl+Hr)/2;        midmid = (mid+Hl)/2;        ansmid = solve(mid);        ansmidmid = solve(midmid);        if(ansmid < ansmidmid)        {            Hl = midmid;        }        else        {            Hr = mid;        }    }    return ansmid;}int main(){    //freopen("2Bin.txt", "r", stdin);    //freopen("3Bout.txt", "w", stdout);    //printf("dsf");    while(scanf("%lf", &v) != EOF)    {        double key = calcu_H();        printf("%f\n", key);    }    return 0;}