原题链接：http://codeforces.com/gym/100523/attachments/download/2798/20142015-ct-s02e07-codeforces-trainings-season-2-episode-7-en.pdf

题意

给你一堆三维点，问你他们是否共面

题解

模板题，套版就好，需要注意的是共线

代码

#include<iostream>
#include<cmath>
#include<cstdio>
#define MAX_N 100005
using namespace std;

const double eps=1e-10;

inline double Sqrt(double a) {
    return a <= 0 ? 0 : sqrt(a);
}

inline double Sqr(double a) {
    return a * a;
}

class Point_3 {
public:
    double x, y, z;

    Point_3() { }

    Point_3(double xx, double yy, double zz) : x(xx), y(yy), z(zz) { }

    Point_3 operator-(Point_3 a) {
        return Point_3(x - a.x, y - a.y, z - a.z);
    }

    double Length() const {
        return Sqrt(Sqr(x) + Sqr(y) + Sqr(z));
    }
};

Point_3 Det(const Point_3 &a,const Point_3 &b) {
    return Point_3(a.y * b.z - a.z * b.y, a.z * b.x - a.x * b.z, a.x * b.y - a.y * b.x);
}

double Dot(const Point_3 &a,const Point_3 &b) {
    return a.x * b.x + a.y * b.y + a.z * b.z;
}

double vlen(Point_3 P){return P.Length();}

int dot_inline(Point_3 p1,Point_3 p2,Point_3 p3) {
    return vlen(Det(p1 - p2, p2 - p3)) < eps;
}

bool zero(double x) {
    return fabs(x) < eps;
}

Point_3 pvec(Point_3 s1,Point_3 s2,Point_3 s3) {
    return Det((s1 - s2), (s2 - s3));
}

int dots_onplane(Point_3 a,Point_3 b,Point_3 c,Point_3 d) {
    return zero(Dot(pvec(a, b, c), d - a));
}

int n;
Point_3 point3[MAX_N];

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        point3[i] = Point_3(a, b, c);
    }
    if (n <= 3) {
        cout << "TAK" << endl;
        return 0;
    }
    int p = -1;
    for (int i = 2; i < n; i++) {
        if (dot_inline(point3[0], point3[1], point3[i]) == 0) {
            p = i;
            break;
        }
    }
    if (p == -1) {
        cout << "TAK" << endl;
        return 0;
    }
    for (int i = 2; i < n; i++) {
        if (i == p)continue;
        if (dots_onplane(point3[0], point3[1], point3[p], point3[i]) == 0) {
            cout << "NIE" << endl;
            return 0;
        }
    }
    cout << "TAK" << endl;
    return 0;
}