2016 ICPC大连赛区 hdu5974 Convex

Convex

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

Total Submission(s): 170    Accepted Submission(s): 145


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 1
90 90 90 90
6 1
60 60 60 60 60 60

 

Sample Output

2.000
2.598

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

 

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讲道理，大连的题目质量怎么这么低，水题也太多了吧。

题意：已知一个凸包，每个点距离原点的距离为定值d，然后给你两两边的夹角，求凸包面积。

思路：将凸包拆成n个三角形，然后套公式就得了。下面给代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
typedef long long LL;
using namespace std;
#define maxn 1005
#define ll l,mid,now<<1
#define rr mid+1,r,now<<1|1
#define lson l1,mid,l2,r2,now<<1
#define rson mid+1,r1,l2,r2,now<<1|1
#define pi 3.14159
int main(){
int n,d;
while(~scanf("%d%d",&n,&d)){
double sum=0;
while(n--){
int num;
scanf("%d",&num);
sum+=sin(num*pi/180);
}
printf("%.3lf\n",d*d*sum/2);
}
}