**View Angle
2016-10-23 20:37
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View Angle
Crawling in process...
Crawling failed
Time Limit:2000MS
Memory Limit:262144KB
64bit IO Format:%I64d & %I64u
Submit
Status
Practice
CodeForces 257C
uDebug
Description
Input
Output
Sample Input
Sample Output
Hint
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
Output
Input
Output
Input
Output
Input
Output
Sample Output
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:
找相邻点间最大的角
Crawling in process...
Crawling failed
Time Limit:2000MS
Memory Limit:262144KB
64bit IO Format:%I64d & %I64u
Submit
Status
Practice
CodeForces 257C
uDebug
Description
Input
Output
Sample Input
Sample Output
Hint
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
Sample Output
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:
找相邻点间最大的角
#include<stdio.h> #include<math.h> #include<algorithm> #define M acos(-1.0) using namespace std; struct dian { double x; double y; double angle; }p[100001]; bool cmp(dian a,dian b) { return a.angle<b.angle; } int main() { int n; while(~scanf("%d",&n)) { for(int i=0;i<n; i++) { scanf("%lf%lf",&p[i].x,&p[i].y); p[i].angle=atan2(p[i].y,p[i].x); } sort(p,p+n,cmp); p .angle=p[0].angle+2*M; double ans=2*M; for(int i=0;i<n;i++) { ans=min(2*M-fabs(p[i+1].angle-p[i].angle),ans); } printf("%lf\n",ans*180.0/M); } return 0; }
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