HDU 5979 Convex【计算几何】 (2016ACM/ICPC亚洲区大连站)

Convex

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

Total Submission(s): 294    Accepted Submission(s): 220

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）

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5979

题目大意：

　　N个点在一个半径为R的圆上，求这N个点的凸包面积。给出这N个点顺时针的夹角差值。

题目思路：

　　【计算几何】

　　水题。这题有点数学知识的应该都会。。求N个三角形的面积，S=0.5*a*b*sinc。

////by coolxxx/*#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>//#include<stdbool.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))*/#include<bits/stdc++.h>#pragma comment(linker,"/STACK:1024000000,1024000000")#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-8)#define J 10000#define mod 100000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#define N 40004using namespace std;typedef long long LL;double anss;LL aans;int cas,cass;int n,m,lll,ans;double s;int main(){#ifndef ONLINE_JUDGEfreopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;int x,y,z;//init();//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))while(~scanf("%d%d",&n,&m)){s=0;for(i=1;i<=n;i++){scanf("%d",&x);s+=0.5*m*m*sin(x/90.0*acos(0));}printf("%.3lf\n",s);}return 0;}/*////*/