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题目链接:https://www.nowcoder.com/acm/contest/141/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:

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#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;
}