cf View Angle

最新推荐文章于 2025-04-29 14:14:11 发布

转载最新推荐文章于 2025-04-29 14:14:11 发布 · 171 阅读

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原文链接：http://www.cnblogs.com/123tang/p/5980563.html

本文介绍了一种新的驾驶员视力检查算法，该算法通过计算平面上多个假人模型所构成的最小角度来评估驾驶员的视觉范围。利用atan2函数确保数值稳定性，并通过排序和遍历找到最小覆盖角。

View Angle

Description

Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.

As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of mannequins.

Next n lines contain two space-separated integers each: x_i, y_i (|x_i|, |y_i| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.

Output

Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10^- 6.

Sample Input

Input

2
2 0
0 2

Output

90.0000000000

Input

Output

135.0000000000

Input

Output

270.0000000000

Input

2
2 1
1 2

Output

36.8698976458

Hint

【题解】

atan2函数与 atan 的不同

atan2 比 atan 稳定。

如：atan(y/x)，当 y 远远大于 x 时，计算结果是不稳定的。

atan2(y,x)的做法：当 x 的绝对值比 y 的绝对值大时使用 atan(y/x)；反之使用 atan(x/y)。这样就保证了数值稳定性。

本题必须用atan2，否则不能得出正确答案。

AC代码：

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 #define pai acos(-1.0)
 7 #define maxn 100005
 8 
 9 int main()  
10 {  
11     int n;
12     double a[maxn], b;  
13     scanf("%d", &n);  
14     for(int i=0; i<n; i++)  
15     {  
16         scanf("%lf %lf", &a[i], &b);
17         a[i] = atan2(b, a[i]);    
18     }  
19     sort(a, a+n);  
20     a[n] = a[0] + 2 * pai;  
21     double ans = 2*pai;  
22     for(int i=0; i<n; i++)  
23     {  
24         ans = min(ans, 2 * pai - fabs(a[i+1] - a[i]));  
25     }  
26     printf("%lf\n", ans*180.0/pai);  
27     return 0;  
28 }