题意:
一个水缸,人眼在其右斜上方,问水达到至少要多深才可以看见水底的某个点
分析:
赤果果的二分,二分水的深度,浅的能看见高的一定能看见嘛
然后就恰巧透过水缸的高度看到,算一下人眼能看到的位置,如果大于等于x,就是可以看到的嘛
代码:
//
// Created by TaoSama on 2015-12-09
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int w, h, x, xe, ye;
double miu;
double sq(double x) {return x * x;}
bool check(double d) {
double sinMPE = (xe - w) / (sqrt(sq(xe - w) + sq(ye - h)));
double sinCPN = sinMPE / miu;
double rig = (h - d) / (ye - h) * (xe - w);
double CN = d / sqrt(1 / sq(sinCPN) - 1);
double p = w - CN - rig;
return p >= x;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d%d%d%d%lf", &w, &h, &x, &xe, &ye, &miu);
double l = 0, r = h + 1;
for(int i = 1; i <= 100; ++i) {
double m = (l + r) / 2;
if(check(m)) r = m;
else l = m;
}
if(l <= h) printf("%.4f\n", l);
else puts("Impossible");
}
return 0;
}