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：
1 #include<iostream> 2 #include<stack> 3 #include<stdio.h> 4 #include<cstring> 5 #include<string> 6 #include<cmath> 7 #include<queue> 8 #include<algorithm> 9 using namespace std; 10 # define ll long long 11 const int maxn = 2e5+50000; 12 const int maxm = 1e6+100; 13 const double eps=1e-15; 14 double b1,t1,b2,t2,l,ds,df,g; 15 double cal(double L,double S,double T,double B) 16 { 17 double h=S-S*S/L; 18 if(h-T>eps) 19 { 20 double xx=g*S*(L-S)/2/T; 21 double yy=xx/S/S*(T+0.5*g*S*S/xx)*(T+0.5*g*S*S/xx); 22 return sqrt(xx+yy); 23 } 24 if(h-B<-eps) 25 { 26 double xx=g*S*(L-S)/2/B; 27 double yy=xx/S/S*(B+0.5*g*S*S/xx)*(B+0.5*g*S*S/xx); 28 return sqrt(xx+yy); 29 } 30 return sqrt(g*L); 31 } 32 double get(double x) 33 { 34 double v1=cal(x+ds,ds,t1,b1),v2=cal(l-x+df,df,t2,b2); 35 return v1-v2>eps?v1:v2; 36 } 37 int main() 38 { 39 while(~scanf("%lf %lf",&b1,&t1)) 40 { 41 scanf("%lf %lf %lf %lf %lf %lf",&b2,&t2,&l,&ds,&df,&g); 42 double lt=0,rt=l,mid1,mid2; 43 double y1,y2; 44 int c=0; 45 while(rt-lt>eps&&c++<100) 46 { 47 mid1=lt+(rt-lt)/3; 48 mid2=rt-(rt-lt)/3; 49 y1=get(mid1); 50 y2=get(mid2); 51 if(y1<y2) 52 rt=mid2; 53 else 54 lt=mid1; 55 } 56 printf("%.6lf\n",get(lt)); 57 } 58 return 0; 59 }
