Codeforces Round #432 (Div. 2) C Five Dimensional Points

题目链接：点击打开链接

五维空间下的锐角判定（包括0）只需判断向量的数量积是否大于0即可

暴力枚举所有可能性

AC代码如下：

#include <iostream>#include <cmath>#include <cstring>using namespace std;struct point{    int x1, x2, x3, x4, x5;    point operator-(const point& o)    {        point t;        t.x1 = x1-o.x1;        t.x2 = x2-o.x2;        t.x3 = x3-o.x3;        t.x4 = x4-o.x4;        t.x5 = x5-o.x5;        return t;    }};point pPoint[1005];bool re[1005];bool judge(point a, point b){    double temp = a.x1*b.x1+a.x2*b.x2+a.x3*b.x3+a.x4*b.x4+a.x5*b.x5;    return  temp>0;}bool onLine(point a, point b){    if(a.x1*b.x2!=a.x2*b.x1)    return false;    if(a.x2*b.x3!=a.x3*b.x2)    return false;    if(a.x3*b.x4!=a.x3*b.x4)    return false;    if(a.x4*b.x5!=a.x5*b.x4)    return false;    return true;}int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int n;    while(cin>>n)    {        fill(re+1,re+1+n,1);        for(int i=1;i<=n;++i)        {            cin>>pPoint[i].x1>>pPoint[i].x2;            cin>>pPoint[i].x3>>pPoint[i].x4>>pPoint[i].x5;        }        int cnt=0;        for(int i=1;i<=n;++i)        {            int flag=0;            for(int j=1;j<=n;++j)            {                if(j==i)    continue;                for(int k=1;k<=n;++k)                {                    if(k==i||k==j)  continue;                    point a = pPoint[j]-pPoint[i];                    point b = pPoint[k]-pPoint[i];                    if(judge(a,b))    { flag=1; break; }                }                if(flag)     break;            }            if(flag)    {  cnt++; re[i]=0;}        }        cout<<n-cnt<<endl;        for(int i=1;i<=n;++i)            if(re[i])   cout<<i<<endl;    }    return 0;}