CodeForces 257C View Angle(求角度）

View Angle
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces
257C

Appoint description:
System Crawler (Oct 17, 2016 1:45:35 PM)

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: xi, yi (|xi|, |yi| ≤ 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

3
2 0
0 2
-2 2

Output

135.0000000000

Input

4
2 0
0 2
-2 0
0 -2

Output

270.0000000000

Input

2
2 1
1 2

Output

36.8698976458

Hint

Solution for the first sample test is shown below:



Solution for the second sample test is shown below:



Solution for the third sample test is shown below:



Solution for the fourth sample test is shown below:



给出n点坐标，求将这n个坐标全部覆盖的最小角度。

在C语言的math.h或C++中的cmath中有两个求反正切的函数atan(double x)与atan2(double y,double x) 他们返回的值是弧度 要转化为角度再自己处理下。

前者接受的是一个正切值（直线的斜率）得到夹角，但是由于正切的规律性本可以有两个角度的但它却只返回一个，因为atan的值域是从-90~90 也就是它只处理一四象限，所以一般不用它。

第二个atan2(double y,double x) 其中y代表已知点的Y坐标 同理x ,返回值是此点与远点连线与x轴正方向的夹角，这样它就可以处理四个象限的任意情况了，它的值域相应的也就是-180~180了

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
double a[100010];
int n;
int main()
{
while(~scanf("%d",&n))
{
double x,y;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&x,&y);
a[i]=atan2(y,x);
}
sort(a,a+n);
a
=a[0]+2*acos(-1.0);
double ans=2*acos(-1.0);
for(int i=0;i<n;i++)
ans=min(ans,2*acos(-1.0)-fabs(a[i+1]-a[i]));
printf("%lf\n",ans*180/acos(-1.0));

}

}