牛客网暑期ACM多校训练营（第三场）I

http://tokitsukaze.live/2018/07/27/2018niuke3.I/

题意：
给一个三角形，让你在三角形内随机选n个点，问这n个点在凸包上的期望是多少。

题解：
显然三角形长什么样，对答案没有影响。
注意到n<=10，所以做法就是：在三角形内随机n个点，求凸包，然后求多少个点在凸包上。随机次数越多，答案是准确率越高。最后打个表交上去就行了。

代码：

#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define mem(a,b) memset((a),(b),sizeof(a))
#define MP make_pair
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
using namespace __gnu_cxx;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef vector<int> VI;
typedef vector<ll> VL;
const int INF=0x3f3f3f3f;
const ll LLINF=0x3f3f3f3f3f3f3f3f;
const double PI=acos(-1.0);
const double eps=1e-6;
const int MAX=1e5+10;
const ll mod=1e9+7;
/*********************************  head  *********************************/
int sgn(double x)
{  
    if(fabs(x)<eps) return 0;
    else return x>0?1:-1;  
}
struct Point
{  
    double x,y;
    Point(){}
    Point(double a,double b)
    {
        x=a;
        y=b;
    }
    void input()
    {
        scanf("%lf%lf",&x,&y);
    }
};
typedef Point Vector;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}  
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}  
Vector operator *(Vector a,double p){return Vector(a.x*p,a.y*p);}  
Vector operator /(Vector a,double p){return Vector(a.x/p,a.y/p);}
bool operator <(Point a,Point b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator ==(Point a,Point b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}
double cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
vector<Point> graham(vector<Point> p)
{  
    int n,m,k,i;
    sort(p.begin(),p.end());
    p.erase(unique(p.begin(),p.end()),p.end());
    n=p.size();
    m=0;
    vector<Point> res(n+1);  
    for(i=0;i<n;i++)
    {  
        while(m>1&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<=0) m--;  
        res[m++]=p[i];  
    }  
    k=m;  
    for(i=n-2;i>=0;i--)
    {  
        while(m>k&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<=0) m--;  
        res[m++]=p[i];  
    }  
    if(n>1) m--;  
    res.resize(m);
    return res;
}
Point randp()
{
    while(1)
    {
        Point p(rand(),rand());
        if(p.x+p.y<=32766) return p;
    }
    return Point(0,0);
}
void gao()
{
    srand(time(0));
    for(int i=4;i<=10;i++)
    {
        int t=20000000;
        double ans=0;
        while(t--)
        {
            vector<Point> p;
            for(int j=0;j<i;j++) p.pb(randp());
            p=graham(p);
            ans+=sz(p);
        }
        ans/=20000000;
        printf("%.6f,",ans);
    }
    puts("");
}
int main()
{
//    gao();
    double ans[]={0,0,0,3,3.666724,4.166653,4.566743,4.900092,5.185803,5.435999,5.658361};
    int n,i;
    Point p;
    for(i=0;i<3;i++) p.input();
    scanf("%d",&n);
    printf("%.4f\n",ans[n]);
    return 0;
}