SGU 204 Little Jumper（三分）

题目链接：http://acm.sgu.ru/problem.php?contest=0&problem=204

题意：给出下图中的参数。从A跳到B。在A起跳速度为v1，在中间起跳速度为v2。求min(max(v1,v2))。

思路：在两个木板之间的位置是单峰函数，三分。

const double EPS=1e-12;

double b1,t1,b2,t2,L,ds,df,g;





double cal(double x1,double y1,double x2)

{

    double a=y1/(x1*x1-x1*x2);

    return -g*a*x2*x2/2-g/(2*a);

}



double cal(double b1,double t1,double L1,double L2)

{

    double ans=min(cal(L1,b1,L1+L2),cal(L1,t1,L1+L2));

    double p=L1+L2,temp=g*p,y=L1*L2/p;

    if(y>=b1&&y<=t1) ans=temp;

    return ans;

}



double cal(double x)

{

    return max(cal(b1,t1,ds,x),cal(b2,t2,L-x,df));

}



int main()

{

    while(cin>>b1>>t1>>b2>>t2>>L>>ds>>df>>g)

    {

        double low=EPS,high=L,mid1,mid2,x,y;

        while(high-low>1e-12)

        {

            mid1=(low+high)*0.5;

            mid2=(mid1+high)*0.5;

            x=cal(mid1);

            y=cal(mid2);

            if(x<y) high=mid2;

            else low=mid1;

        }

        double ans=min(cal((low+high)*0.5),min(cal(low),cal(high)));

        printf("%.4lf\n",sqrt(ans));

    }

    return 0;

}