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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. #define PI acos(-1)
  5. int N;
  6.  
  7. double get_angle(int sz) {
  8.     int dot = (-sz * 2) + (sz - 1);
  9.     int len = (sz) + (sz) * (sz);
  10.     return acos(dot *1.0 / len*1.0);
  11. }
  12.  
  13. void solve(vector<vector<double> >& pts, int cur) {
  14.     if (cur == N+2) return;
  15.     double theta = get_angle(cur - 1) - PI / 2;
  16.     for (int i = 0; i < pts.size(); i++) {
  17.         for (int k = 0; k < pts[i].size(); k++) {
  18.             pts[i][k] *= cos(theta);
  19.         }
  20.         pts[i].push_back(-sin(theta));
  21.     }
  22.     vector<double> new_pt;
  23.     for (int i = 2; i < cur; i++) new_pt.push_back(0);
  24.     new_pt.push_back(1);
  25.     pts.push_back(new_pt);
  26.  
  27.     solve(pts, cur + 1);
  28. }
  29.  
  30. int main() {
  31.  
  32.     cout << fixed << setprecision(9);
  33.  
  34.     int T; cin >> T;
  35.     while (T) {
  36.         cin >> N;
  37.         if (N == 1) {
  38.             cout << 1 << "\n" << -1 << "\n";
  39.         } else {
  40.             vector<vector<double> > pts;
  41.             vector<double> p1, p2;
  42.             p1.push_back(1);
  43.             p2.push_back(-1);
  44.             pts.push_back(p1); pts.push_back(p2);
  45.             solve(pts, 3);
  46.  
  47.             for (int i = 0; i < pts.size(); i++) {
  48.                 for (int k = 0; k < pts[i].size(); k++) {
  49.                     cout << pts[i][k] << " ";
  50.                 } cout << "\n";
  51.             }
  52.         }
  53.         cout << "\n";
  54.         T--;
  55.     }
  56.  
  57.     return 0;
  58. }
Time limit exceeded #stdin #stdout 5s 65216KB
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https://ideone.com/8yR9v9
language:
C++14 (gcc 8.3)
created:
7 years ago
visibility:
secret

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