HDU 5979 Convex（数学）

Convex

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 50    Accepted Submission(s): 42

Problem Description

We have a special convex that all points have the same distance to origin point.
As you know we can get N segments after linking the origin point and the points on the convex. We can also get N angles between each pair of the neighbor segments.
Now give you the data about the angle, please calculate the area of the convex

 

Input

There are multiple test cases.
The first line contains two integer N and D indicating the number of the points and their distance to origin. (3 <= N <= 10, 1 <= D <= 10)
The next lines contain N integers indicating the angles. The sum of the N numbers is always 360.

 

Output

For each test case output one float numbers indicating the area of the convex. The printed values should have 3 digits after the decimal point.

 

Sample Input

4 190 90 90 906 160 60 60 60 60 60

 

Sample Output

2.0002.598

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

题意：一个凸多边形每个顶点到其原点的距离相等，且能把该多边形分成几个部分，给定每个部分的角度，求多边形面积。

思路：一开始没读懂题，以为每个角度都相同WA了一次，但是考虑到了三角形要额外处理，因为三角形有可能会有一个角度超过180。还有就是要掌握角度和弧度的转化。

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#include <math.h>using namespace std;#define pi 3.1415926int ang[4];int main(){int n,d;while(~scanf("%d %d",&n,&d)){double area=0;for(int i=0;i<n;i++){int angle;scanf("%d",&angle);if(angle<=180)area+=0.5*d*d*sin(angle*1.0*pi/180.0);else{angle=360-angle;area-=0.5*d*d*sin(angle*1.0*pi/180.0);}} printf("%.3lf\n",area);}return 0;}