description: Polycarp likes squares and cubes of positive integers. Here is the beginning of the sequence of numbers he likes: 1, 4, 8, 9, ....
  
  For a given number n, count the number of integers from 1 to n that Polycarp likes. In other words, find the number of such x that x is a square of a positive integer number or a cube of a positive integer number (or both a square and a cube simultaneously).
  
  Input
  
  The first line contains an integer t (1 ≤ t ≤ 20) — the number of test cases.
  
  Then t lines contain the test cases, one per line. Each of the lines contains one integer n (1 ≤ n ≤ 10^9).
  
  Output
  
  For each test case, print the answer you are looking for — the number of integers from 1 to n that Polycarp likes.
  
  Example
  
  Input
  
  
  6
  10
  1
  25
  1000000000
  999999999
  500000000
  
  
  Output
  
  
  4
  1
  6
  32591
  32590
  23125
solution: #include <bits/stdc++.h>
  using namespace std;
  inline long Length(long left, long right) { return right - left + 1; }
  template <class T, class V>
  bool Contains(const T& uoset, const V& value) {
    return uoset.find(value) != uoset.end();
  }
  using TPair = pair<long, long>;
  inline bool BySecond(const TPair& left, const TPair& right) {
    return left.second < right.second;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long T;
    cin >> T;
    if (T > 0)
      for (long i = 0; i <= T - 1; ++i) {
        long n;
        cin >> n;
        unordered_set<long> nums;
        long sq = 1;
        while (sq * sq <= n) {
          nums.insert(sq * sq);
          sq += 1;
        }
        long cu = 1;
        while (cu * cu * cu <= n) {
          nums.insert(cu * cu * cu);
          cu += 1;
        }
        cout << nums.size() << endl;
      }
    return 0;
  }
  