Convex
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 294 Accepted Submission(s): 220Problem DescriptionWe 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
InputThere 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.
OutputFor 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 Input4 1 90 90 90 90 6 1 60 60 60 60 60 60
Sample Output2.000 2.598
Source
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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。
1 // 2 //by coolxxx 3 /* 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 #include<math.h> 19 //#include<stdbool.h> 20 #define min(a,b) ((a)<(b)?(a):(b)) 21 #define max(a,b) ((a)>(b)?(a):(b)) 22 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 23 */ 24 #include<bits/stdc++.h> 25 #pragma comment(linker,"/STACK:1024000000,1024000000") 26 #define abs(a) ((a)>0?(a):(-(a))) 27 #define lowbit(a) (a&(-a)) 28 #define sqr(a) ((a)*(a)) 29 #define mem(a,b) memset(a,b,sizeof(a)) 30 #define eps (1e-8) 31 #define J 10000 32 #define mod 100000007 33 #define MAX 0x7f7f7f7f 34 #define PI 3.14159265358979323 35 #define N 40004 36 using namespace std; 37 typedef long long LL; 38 double anss; 39 LL aans; 40 int cas,cass; 41 int n,m,lll,ans; 42 double s; 43 int main() 44 { 45 #ifndef ONLINE_JUDGE 46 freopen("1.txt","r",stdin); 47 // freopen("2.txt","w",stdout); 48 #endif 49 int i,j,k; 50 int x,y,z; 51 // init(); 52 // for(scanf("%d",&cass);cass;cass--) 53 // for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 54 // while(~scanf("%s",s)) 55 while(~scanf("%d%d",&n,&m)) 56 { 57 s=0; 58 for(i=1;i<=n;i++) 59 { 60 scanf("%d",&x); 61 s+=0.5*m*m*sin(x/90.0*acos(0)); 62 } 63 printf("%.3lf\n",s); 64 } 65 return 0; 66 } 67 /* 68 // 69 70 // 71 */
HDU 5979 Convex【计算几何】 (2016ACM/ICPC亚洲区大连站)
原文：http://www.cnblogs.com/Coolxxx/p/6272328.html
