学步园

此题就是个坑题， 卡精度，eps  1e-12  才可过  并且最终答案 f(left) 不可行。。  只能是 最终left right mid 中的小的值。。。

话不多说，代码

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <cstring>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <assert.h>
#include <queue>
#define REP(i,n) for(int i=0;i<n;i++)
#define TR(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)
#define ALLL(x) x.begin(),x.end()
#define SORT(x) sort(ALLL(x))
#define CLEAR(x) memset(x,0,sizeof(x))
#define FILLL(x,c) memset(x,c,sizeof(x))
using namespace std;
const double eps = 1e-12;
double b1, t1, b2, t2, l, ds, df,g;
double f(double t,double b , double l1 ,double l2){
     double vv2 = g*(l1+l2);
     double h = l1*l2/(l1+l2);
     if(h<=t && h >=b){
     	   return sqrt(vv2);
     }
     double a = g*(l1+l2)*l1/2 - l1*l1*g/2;
     double vx = sqrt(a/t);
     double vy = g*(l1+l2)/(2*vx);
     double v1 = sqrt(vx*vx+vy*vy);
     
     double vx2 = sqrt(a/b);
     double vy2 = g*(l1+l2)/(2*vx2);
     double v2 = sqrt(vx2*vx2+vy2*vy2);
     return min(v1,v2);
}
double f2(double a){
	   return max(f(t1,b1,ds,a),f(t2,b2,l-a,df));
}
void solve(){
	 double left = eps;
	 double right = l;
	 while(right - left > eps){
	 	   double lmid = (left +right)/2;
	 	   double rmid = (lmid +right)/2;
	 	   if(f2(lmid)>f2(rmid)){
	 	   	    left =lmid;
	 	   }else{
	 	   	    right =rmid;
	 	   }
	 }
	 double ans =  min(f2(left), min(f2(right),f2((left+right)/2)));
	 printf("%.4lf\n",ans);
}
int main(){
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&b1, &t1, &b2, &t2, &l, &ds, &df,&g)){
    	    solve();
    }
    return 0;
}