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Nevtr4l

Divisores del futuro

Mar 11th, 2025 (edited)
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C++ 0.59 KB | Source Code | 0 0
  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. void solve() {
  6.   long long N; cin >> N;
  7.  
  8.   vector<long long> divisores;
  9.   for (long long div = 1; div * div <= N; div++) {
  10.     if (N % div == 0) {
  11.       divisores.push_back(div);
  12.       if (N / div != div) {
  13.         divisores.push_back(N / div);
  14.       }
  15.     }
  16.   }
  17.  
  18.   sort(divisores.begin(), divisores.end());
  19.  
  20.   cout << divisores.size() << endl;
  21.   for (int i = 0; i < divisores.size(); i++) {
  22.     cout << divisores[i] << " ";
  23.   }
  24.   cout << endl;
  25. }
  26.  
  27. int main() {
  28.   int t; cin >> t;
  29.   while (t--) {
  30.     solve();
  31.   }
  32.  
  33.   return 0;
  34. }
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