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C. View Angle
time limit per test
memory limit per test
input
output
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
n (1 ≤ n ≤ 105)
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
10 - 6.
Examples
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
Note
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:
大意：已知各点坐标，求能够把所有点覆盖，且以坐标原点为顶点的最小角度是多少
思路：要求黄色区域的最小角度，就是求其余白色部分的最大角度，然后用 360减它即可。而白色部分的最大角的两条边定是分别以两个相邻的点组成的。所以将所有由相邻边的构成的角排序，求出最大的角度即为白色部分的最大角度。atan2(x，y) 函数也可以求 y 为 0 时的角度，而 atan(x) 就不能了
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define PI acos(-1.0)
using namespace std;
int n;
struct node
{
double x,y,angle;
}point[100010];
bool cmp(node a,node b)
{
return a.angle<b.angle;
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&point[i].x,&point[i].y);
point[i].angle=atan2(point[i].y,point[i].x);
if(point[i].angle<0)
point[i].angle+=2*PI;
}
if(n==1) // 特殊情况
{
puts("0.00000000");
continue;
}
sort(point,point+n,cmp);
double ans=point[0].angle-point[n-1].angle+2*PI; // 这个也是一个夹角
for(int i=1;i<n;i++)
{
ans=max(ans,point[i].angle-point[i-1].angle);
}
printf("%.8lf\n",(2*PI-ans)*180.0/PI);
}
return 0;
}
