【二分】Petrozavodsk Winter Training Camp 2017 Day 1: Jagiellonian U Contest, Monday, January 30, 2017 Problem A. The Catcher in the Rye
阿新 • • 發佈：2017-07-10
什麽 不同 stdin n) clas sqrt ios 這份 std
一個區域，垂直分成三塊，每塊有一個速度限制，問你從左下角跑到右上角的最短時間。
將區域看作三塊折射率不同的介質，可以證明，按照光路跑時間最短。
於是可以二分第一個入射角，此時可以推出射到最右側邊界上的位置，看什麽時候恰好射到右上角即可。
這份sb代碼貌似掛精度了。
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define EPS 0.0000000001
int T,h,a,b,c,va,vb,vc;
double t1,t2,t3;
bool check(double sina){
double sinb=sina*(double)vb/(double)va;
if(sinb-1.0>-EPS){
return 1;
}
double sinc=sinb*(double)vc/(double)vb;
if(sinc-1.0>-EPS){
return 1;
}
double y1=(double)a*sina/sqrt(1.0-sina*sina);
double y2=(double)b*sinb/sqrt(1.0-sinb*sinb);
double y3=(double)c*sinc/sqrt(1.0-sinc*sinc);
t1=y1/sina/(double)va;
t2=y2/sinb/(double)vb;
t3=y3/sinc/(double)vc;
return y1+y2+y3-(double)h>-EPS;
}
int main(){
// freopen("a.in","r",stdin);
scanf("%d",&T);
for(int zu=1;zu<=T;++zu){
scanf("%d%d%d%d%d%d%d",&h,&a,&b,&c,&va,&vb,&vc);
double l=EPS,r=1.0-EPS;
while(r-l>EPS){
double mid=(l+r)/2.0;
if(check(mid)){
r=mid;
}
else{
l=mid+EPS;
}
}
check(l);
printf("%.7f\n",t1+t2+t3);
}
return 0;
}【二分】Petrozavodsk Winter Training Camp 2017 Day 1: Jagiellonian U Contest, Monday, January 30, 2017 Problem A. The Catcher in the Rye