Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017)C. Five Dimensional Points

传送门

A. Five Dimensional Points

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given set of n points in 5-dimensional space. The points are labeled from 1 to n.
No two points coincide.

We will call point a bad if there are different points b and c,
not equal to a, from the given set such that angle between vectors 

 and 

 is
acute (i.e. strictly less than 

).
Otherwise, the point is called good.

The angle between vectors 

 and 

 in
5-dimensional space is defined as 

,
where 

 is
the scalar product and 

 is
length of 

.

Given the list of points, print the indices of the good points in ascending order.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 103) —
the number of points.

The next n lines of input contain five integers ai, bi, ci, di, ei (|ai|, |bi|, |ci|, |di|, |ei| ≤ 103)
 — the coordinates of the i-th point. All points are distinct.

Output

First, print a single integer k — the number of good points.

Then, print k integers, each on their own line — the indices of the good points in ascending order.

Examples

input

6
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1

output

1
1

input

3
0 0 1 2 0
0 0 9 2 0
0 0 5 9 0

output

0

Note

In the first sample, the first point forms exactly a 

 angle
with all other pairs of points, so it is good.

In the second sample, along the cd plane, we can see the points look as follows:



We can see that all angles here are acute, so no points are good.

n^3。

暴力怼

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//cf-535C Tavas and Pashmaks

int n;
int vis[maxn];
struct node
{
double a,b,c,d,e;
}Q[maxn];

typedef node Vector;
double Dot(Vector A, Vector B) { return A.a*B.a + A.b*B.b+
A.c*B.c+A.d*B.d+A.e*B.e; }

double Length(Vector A) { return sqrt(Dot(A,A));}

double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

Vector vv(node A,node B)
{
Vector v1;
v1.a=A.a-B.a;
v1.b=A.b-B.b;
v1.c=A.c-B.c;
v1.d=A.d-B.d;
v1.e=A.e-B.e;
return v1;
}
int main()
{
scan_d(n);
FOR(1,n,i)
{
scanf("%lf%lf%lf%lf%lf",
&Q[i].a,&Q[i].b,&Q[i].c,&Q[i].d,&Q[i].e);
}
/*if(n>243)
{
puts("0");
return 0;
}*/
Vector v1,v2;
int cnt=0;
FOR(1,n,i)
{
int flag=0;
FOR(1,n,j)
{
if(i==j)
continue;
FOR(1,n,k)
{
if(j==k||k==i)
continue;
//cout<<i<<" "<<j<<" "<<k<<endl;
v1=vv(Q[i],Q[j]);
v2=vv(Q[i],Q[k]);
db ang=Angle(v1,v2);
// printf("%f %f\n",ang,pi/2.0-eps);
if(ang<pi/2.0-eps)
{
// printf("%f %f\n",ang,pi/2.0-eps);
flag=1;
break;
}
}
if(flag)
break;
}
if(!flag)
{
vis[cnt++]=i;
}
}
print(cnt);
FOr(0,cnt,i)
print(vis[i]);
}