UVA 12901 Refraction 几何/大雾题

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);

    }

}