【HDOJ5979】Convex（三角函数）

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

n<=10，R<=10

思路：S=1/2*sinθ*a*b

角度转弧度再用sin

C++有点小毛病，叫队友改了下

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define fi first
#define se second
#define MP make_pair
#define N   1100000
#define MOD 1000000007
#define eps 1e-8
#define pi  acos(-1)


int n;
double L;


int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

int main()
{
    while(scanf("%d%lf",&n,&L)!=EOF)
    {
         double ans=0;
         for(int i = 1; i <= n; ++i)
         {
             double x;
             scanf("%lf",&x);
             x/=180*1.0;
             x*=pi*1.0;
             ans+=1.0/2*L*L*sin(x);
         }
         printf("%.3lf
",ans);
    }
    return 0;
}