#include using namespace std; struct Point { int x, y; }; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); int N; cin >> N; int total = 2 * N; vector pts(total); for (int i = 0; i < total; i++){ cin >> pts[i].x >> pts[i].y; } // 尝试竖直直线:按 x 轴排序(当 x 相同时按 y 排序) { vector temp = pts; sort(temp.begin(), temp.end(), [](const Point &a, const Point &b){ return a.x == b.x ? a.y < b.y : a.x < b.x; }); int x_low = temp[N-1].x, x_high = temp[N].x; if(x_low < x_high){ // 若能选整数 t 满足 x_low < t < x_high if(x_low + 1 < x_high){ int t = x_low + 1; cout << 1 << " " << 0 << " " << -t << "\n"; return 0; } else { // x_low 与 x_high 连续,取 t = (2*x_low + 1)/2, 即输出 2x - (2*x_low+1)=0 int t = 2 * x_low + 1; cout << 2 << " " << 0 << " " << -t << "\n"; return 0; } } } // 竖直直线失败则尝试水平直线:按 y 轴排序(当 y 相同时按 x 排序) { vector temp = pts; sort(temp.begin(), temp.end(), [](const Point &a, const Point &b){ return a.y == b.y ? a.x < b.x : a.y < b.y; }); int y_low = temp[N-1].y, y_high = temp[N].y; if(y_low < y_high){ if(y_low + 1 < y_high){ int t = y_low + 1; cout << 0 << " " << 1 << " " << -t << "\n"; return 0; } else { int t = 2 * y_low + 1; cout << 0 << " " << 2 << " " << -t << "\n"; return 0; } } } // 若以上两种均不行,则尝试以斜率1构造直线:令 s = x+y { vector s(total); for (int i = 0; i < total; i++){ s[i] = (long long)pts[i].x + pts[i].y; } sort(s.begin(), s.end()); long long s_low = s[N-1], s_high = s[N]; if(s_low < s_high){ if(s_low + 1 < s_high){ long long t = s_low + 1; cout << 1 << " " << 1 << " " << -t << "\n"; return 0; } else { long long t = 2 * s_low + 1; cout << 2 << " " << 2 << " " << -t << "\n"; return 0; } } } // 最后尝试以斜率 -1 构造直线:令 d = x-y { vector d(total); for (int i = 0; i < total; i++){ d[i] = (long long)pts[i].x - pts[i].y; } sort(d.begin(), d.end()); long long d_low = d[N-1], d_high = d[N]; if(d_low < d_high){ if(d_low + 1 < d_high){ long long t = d_low + 1; cout << 1 << " " << -1 << " " << -t << "\n"; return 0; } else { long long t = 2 * d_low + 1; cout << 2 << " " << -2 << " " << -t << "\n"; return 0; } } } // 若万一都不行,输出一个默认直线(其实理论上不会到这一步) cout << "1 0 0\n"; return 0; }