Refract Facts (二分) (UVALive 7292) Regionals 2015--North America
Refract Facts
A submarine is using a communications laser to send a message to a jet cruising overhead. The sea surface is flat. The submarine is cruising at a depth d below the surface. The jet is at height h above the sea surface, and a horizontal distance x from the sub. The submarine turns toward the jet before starting communications, but needs to know the angle of elevation, φ, at which to aim the laser. When the laser passes from the sea into the air, it is refracted (its path is bent). The refraction is described by Snell’s law, which says that light approaching the horizontal surface at an angle θ 1 , measured from the vertical, will leave at an angle θ 2 , given by the formula
sin θ 1
n 1
=
sin θ 2
n 2
where n 1 and n 2 are the respective refraction indices of the water and air.
(The refraction index of a material is inversely proportional to how fast light
can travel through that material.)
Input
Each test case consists of a single line of input containing 5 space-separated floating point numbers:
• d, the depth of the submarine (specifically, of the laser emitter) in feet, 1 ≤ d ≤ 800
• h, the height of the plane in feet, 100 ≤ h ≤ 10, 000
• x, the horizontal distance from the sub to the plane in feet, 0 ≤ x ≤ 10, 000
• n 1 , the refractive index of water, 1.0 < n 1 ≤ 2.5
• n 2 , the refractive index of air, 1.0 ≤ n 2 < n 1
Input ends with a line containing 5 zeroes (0 0 0 0 0).
Output
For each test case, print a single line containing the angle of elevation φ at which the submarine should
aim its laser to illuminate the jet.
The angle should be displayed in degrees and rounded to the closest 1/100 of a degree. Exactly two
digits after the decimal point should be displayed.90
1200 4000 1.5 1.01
100 10000 2.5 1.01
0 0 0
Sample Output
44.37
11.51
2.30
思路： 這個題比賽的時候我們推公式沒有推出來，是一個一元四次的方程，然後我們當時就放棄了，後來才發現這個題原來是個二分，要求的角度是在0到90度之間的，然後對0到90不斷二分，我是根據mid求出來的H，然後和給定的h比較，如果H大於h，就說明mid大了，反之亦然。
#include<iostream> #include<algorithm> #include<cstdlib> #include<cstdio> #include<string> #include<cstring> #include<set> #include<cmath> using namespace std; #define PI acos(-1) typedef long long LL; double d,h,x,n1,n2; double cc(double l,double r) { while(r-l>=0.001) { double mid=(l+r)/2.0; double m=d/(tan(mid*PI/180.0)); double w=x-m; double ct1=90.0-mid; double sinct2=sin(ct1*PI/180.0)*n2/n1; double hct2=asin(sinct2); double ct2=hct2*180.0/PI; double hh=w/tan(ct2*PI/180.0); if(hh>h) r=mid; else l=mid; } return l; } int main() { while(cin>>d>>h>>x>>n1>>n2) { if(d==0&&h==0&&x==0&&n1==0&&n2==0) break; double l=0.0,r=90.0; double ans=cc(l,r); printf("%.2lf\n",ans); } return 0; }