ZOJ 3203 Light Bulb (三分) |
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ZOJ 3203 Light Bulb (三分)
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Light Bulb Time Limit: 1 Second Memory Limit: 32768 KB
Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is wandering in his incommodious house, thinking of how to earn more money. One day, he found that the length of his shadow was changing from time to time while walking between the light bulb and the wall of his house. A sudden thought ran through his mind and he wanted to know the maximum length of his shadow. Input The first line of the input contains an integer T (T = 10-2. Output For each test case, output the maximum length of mildleopard's shadow in one line, accurate up to three decimal places.. Sample Input 3 2 1 0.5 2 0.5 3 4 3 4
Sample Output 1.000 0.750 4.000
Author: GUAN, Yao Source: The 6th Zhejiang Provincial Collegiate Programming Contest
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3203
题意：求人的影子在灯光下投影到地面和墙上的最大值。 思路：设x表示人和灯之间的距离，用k表示人投影到墙上的长度，通过三角形相似可得公式：（H-h）/（H-k）= X / D => k = H - (H-h)*D/x；则L = k + D - x； 上界r = D， 下界l = D*(H-h)/H #include #include #include #include using namespace std; const double eps = 1e-6; int t; double H,h,D; double val(double x){ return H - (H-h)*D/x + D-x; } int main(){ scanf("%d",&t); while(t --){ scanf("%lf%lf%lf",&H,&h,&D); double l = D - h*D/H, r = D; while(l + eps < r){ double lm = l + (r - l) / 3; double rm = r - (r - l) / 3; if(val(lm) < val(rm)) l = lm; else r = rm; } printf("%.3lf\n",val(r)); } return 0; }
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