CodeForces 850A(105/600)

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.

Example
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

思维题
要搞清楚形成钝角的条件
想清楚一个空间分割成几块就没哟纯钝角了

#include
#include
#include
using namespace std;
int tu[1001][6];
int cao[1001];
int main()
{
    int n;
    cin >> n;
    for (int a = 1; a <= n; a++)
    {
        for (int b = 1; b <= 5; b++)cin >> tu[a][b];
    }
    if (n > 11)
    {
        cout << 0 << endl;
        return 0;
    }
    int dan = 0;
    for(int a = 1; a <= n; a++)
    {
        int jc = 0;
        for (int b = 1; b <= n; b++)
        {
            if (a == b)continue;
            for (int c = 1; c <= n; c++)
            {
                if (a == c || b == c)continue;
                int zs = 0;
                for (int d = 1; d <= 5; d++)zs += (tu[b][d] - tu[a][d])*(tu[c][d] - tu[a][d]);
                if (zs > 0)jc = 1;
            }
        }
        if (jc)continue;
        cao[++dan] = a;
    }
    cout << dan << endl;;
    for (int a = 1; a <= dan; a++)cout << cao[a] << " ";
}