C

题目链接：https://cn.vjudge.net/contest/281961#problem/C

题目大意：青蛙能从一个点跳到第三个点，如图，需要跳两次。问整个过程的最大起跳速度中的最小的。

具体思路：三分寻找最大值，具体证明：https://blog.csdn.net/thearcticocean/article/details/51533065

AC代码：

#include<iostream>
#include<stack>
#include<stdio.h>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
# define ll long long
const int maxn = 2e5+50000;
const int maxm = 1e6+100;
const double eps=1e-15;
double b1,t1,b2,t2,l,ds,df,g;
double cal(double L,double S,double T,double B)
{
    double h=S-S*S/L;
    if(h-T>eps)
    {
        double xx=g*S*(L-S)/2/T;
        double yy=xx/S/S*(T+0.5*g*S*S/xx)*(T+0.5*g*S*S/xx);
        return sqrt(xx+yy);
    }
    if(h-B<-eps)
    {
        double xx=g*S*(L-S)/2/B;
        double yy=xx/S/S*(B+0.5*g*S*S/xx)*(B+0.5*g*S*S/xx);
        return sqrt(xx+yy);
    }
    return sqrt(g*L);
}
double get(double x)
{
    double v1=cal(x+ds,ds,t1,b1),v2=cal(l-x+df,df,t2,b2);
    return v1-v2>eps?v1:v2;
}
int main()
{
    while(~scanf("%lf %lf",&b1,&t1))
    {
        scanf("%lf %lf %lf %lf %lf %lf",&b2,&t2,&l,&ds,&df,&g);
        double lt=0,rt=l,mid1,mid2;
        double y1,y2;
        int c=0;
        while(rt-lt>eps&&c++<100)
        {
            mid1=lt+(rt-lt)/3;
            mid2=rt-(rt-lt)/3;
            y1=get(mid1);
            y2=get(mid2);
            if(y1<y2)
                rt=mid2;
            else
                lt=mid1;
        }
        printf("%.6lf
",get(lt));
    }
    return 0;
}