HDU 4033【数学几何余弦定理+二分求解方程】

题目：Regular Polygon

题意：

给一个正N边形内一点到各顶点的距离，求正N边形的边长。

解题思路:

听了题意后，马上想到的是二分答案，然后想到是根据面积相等构造等式，于是杯具了，直接导成都无望入线T_T，原因是根据面积相等构造的函数单调性难以判断，很麻烦。赛后一天cxyue跟我说根据内角度之和为360度就很容易二分了，原因是当边长与内角大小成正比，无疑是单调关系。但是提交总是WA，错误出在精度上，XX**，处理这些精度问题最讨厌了。精度调到1e-9TLE，调到1e-7WA，只有1e-8AC,妹！

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8 #include <cmath>
 9 
10 using namespace std;
11 
12 const int MAX = 100 + 10;
13 const double INF = 10000000.0f;
14 const double PI = acos(-1.0);
15 int N;
16 double Edge[MAX];
17 void setRange(double& min, double& max)
18 {
19     Edge[N + 1] = Edge[1];
20     min = -1.0, max = INF;
21     for(int i = 1; i <= N; ++i)
22     {
23         if(fabs(Edge[i] - Edge[i + 1]) > min)
24             min = fabs(Edge[i] - Edge[i + 1]);
25         if(Edge[i] + Edge[i + 1] < max)
26             max = Edge[i] + Edge[i + 1];
27     }
28 }
29 
30 double getArc(double a, double b, double c)
31 {
32     return acos((b * b + c * c - a * a) / (2.0 * b * c));
33 }
34 
35 double func(double edge)
36 {
37     double angle = 0.0;
38     for(int i = 1; i <= N; ++i)
39         angle += getArc(edge, Edge[i], Edge[i + 1]);
40     return angle - PI * 2;
41 }
42 double solve(double left, double right)
43 {
44     double mid, funcM, funcL, funcR;
45     while(left < right)
46     {
47         mid = (right + left) / 2.0;
48         funcM = func(mid);
49         funcL = func(left);
50         funcR = func(right);
51         if(fabs(funcM) < 1e-8)
52             return mid;
53         if(funcM * funcL < 0)
54             right = mid;
55         else if(funcM * funcR < 0)
56             left = mid;
57         else
58             return -1.0;
59     }
60 }
61 int main()
62 {
63     freopen("in.txt","r",stdin);
64     int T;
65     scanf("%d", &T);
66     for(int t = 1; t <= T; ++t)
67     {
68         scanf("%d", &N);
69         for(int i = 1; i <= N; ++i)
70             scanf("%lf", &Edge[i]);
71         double min = -1.0, max = INF;
72         setRange(min, max);
73         double ans = solve(min + 1e-8, max - 1e-8);
74         printf("Case %d: ", t);
75         if(min < max && ans != -1.0)
76             printf("%.3f\n", ans);
77         else
78             printf("impossible\n");
79     }
80     return 0;
81 }