CF851 C 暴力

给出n个5维下的点，求点a不与其它任意的b,c重合，向量ab,ac的夹角都为钝角，这样的点个数，并打印它们。

转换二维下的求角度的函数为五维的，而且由于要求角度大于90度，在二维情况下最多有4个点，也就是象限的数量，那么推导到5维就有$2^5$个象限，所以实际上只需要判断这么多个点就能退出了，并不会TLE

/** @Date    : 2017-09-04 23:12:31
  * @FileName: C.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
const double Pi = acos(-1.0);
//减少精度问题
int sign(double x)
{
	if(fabs(x) < eps)
		return 0;
	if(x < 0)
		return x;
	else return x;
}

struct yuu
{
	double a, b, c, d, e;
	yuu(){}
	yuu(double aa, double bb, double cc, double dd, double ee):a(aa),b(bb),c(cc),d(dd),e(ee){}
	yuu operator -(const yuu &y) const
	{
		//cout << y.a << y.b << y.c << y.d << y.e<<endl;
		return yuu(sign(y.a-a) , sign(y.b-b) , sign(y.c-c) , sign(y.d-d) , sign(y.e-e));
	}
	double operator *(const yuu &y)
	{
		double ans = (a*y.a) + (b*y.b) + (c*y.c)+ (d*y.d) + (e*y.e);
		//cout << ans << endl;
		return sign(ans);
	}
};

double leng(yuu x, yuu y)
{
	if(sign(x*y) == 0)
		return -1;
	double ans = sqrt(sign(x*y));
	return ans;
}

yuu p[1010];
int main()
{
	int n;
	while(cin >> n)
	{
		for(int i = 0; i < n; i++)
		{
			double  a, b, c, d, e;//double 写错int
			scanf("%lf%lf%lf%lf%lf",&a, &b, &c, &d, &e);
			p[i] = yuu(a, b, c, d, e);
			//cout << p[i].a << endl;
		}
		vector<int>q;
		for(int i = 0; i < n; i++)
		{
			int flag = 0;
			for(int j = 0; j < n && !flag; j++)
			{
				for(int k = j + 1; k < n; k++)
				{
					if(i == j || j == k || i == k)
						continue;
					yuu ij = p[i] - p[j];
					yuu ik = p[i] - p[k];
					/*if(leng(ij,ij) == -1 || leng(ik,ik) == -1)
						continue;*/
					double agl = acos(ij * ik / (leng(ij, ij) * leng(ik, ik)));
					/*if(i == 1 && j == 2 && k == 3)
						printf("%.lf %.lf\n", leng(ij, ij) , leng(ik, ik));*/
					//cout << i << j<< k<< "~"<<agl * 180.00 / Pi<< endl;
					if(agl * 2.00000 + eps < Pi)
					{
						flag = 1;
						break;
					}
				}
			}
			if(!flag)
				q.PB(i);
		}
		sort(q.begin(), q.end());
		printf("%d\n", q.size());
		for(int i = 0; i < q.size(); i++)
			printf("%d%s", q[i] + 1, i==q.size() - 1?"\n":" ");
	}
    return 0;
}