【二分】Petrozavodsk Winter Training Camp 2017 Day 1: Jagiellonian U Contest, Monday, January 30, 2017 Problem A. The Catcher in the Rye
2017-07-09 21:52
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一个区域,垂直分成三块,每块有一个速度限制,问你从左下角跑到右上角的最短时间。
将区域看作三块折射率不同的介质,可以证明,按照光路跑时间最短。
于是可以二分第一个入射角,此时可以推出射到最右侧边界上的位置,看什么时候恰好射到右上角即可。
这份sb代码貌似挂精度了。
将区域看作三块折射率不同的介质,可以证明,按照光路跑时间最短。
于是可以二分第一个入射角,此时可以推出射到最右侧边界上的位置,看什么时候恰好射到右上角即可。
这份sb代码貌似挂精度了。
#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; }
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