C - Little Jumper （三分）

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+;
 const int maxm = 1e6+;
 const double eps=1e-;
 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)//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)//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=,rt=l,mid1,mid2;
         double y1,y2;
         int c=;
         while(rt-lt>eps&&c++<)
         {
             mid1=lt+(rt-lt)/;
             mid2=rt-(rt-lt)/;
             y1=get(mid1);
             y2=get(mid2);
             if(y1<y2)
                 rt=mid2;
             else
                 lt=mid1;
         }
         printf("%.6lf\n",get(lt));
     }
     return ;
 }