Refraction

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83008#problem/E

Description

HINT

题意

上面有一个眼镜，要看水池里的东西，考虑折射，问你水面最少得多高？

题解：

解方程，普通的大雾题，解解方程就好了

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************double w,h,x,xe,ye,p;int main()
{
    int t=read();
    while(t--)
    {
        cin>>w>>h>>x>>xe>>ye>>p;
        double tan1=(xe-w)/(ye-h);
        double tan2=tan(asin(sin(atan(tan1))/p));        double x1=w-tan1*h;
        double ans=(x-x1)/(tan1-tan2);
        if(ans<=)
            cout<<"0.0000"<<endl;
        else if(ans>h)
            cout<<"Impossible"<<endl;
        else
            printf("%.4f\n",ans);
    }
}