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;
  using LL = long long;
  constexpr int N = 1e5;
  constexpr int INF = 0x3f3f3f3f;
  vector<LL> seq;
  unordered_set<LL> st;
  inline void init() {
    for (LL i = 1; i <= N; ++i) {
      if (!st.count(i * i)) {
        st.insert(i * i);
        seq.push_back(i * i);
      }
      if (!st.count(i * i * i)) {
        st.insert(i * i * i);
        seq.push_back(i * i * i);
      }
    }
    sort(seq.begin(), seq.end());
  }
  void solve() {
    int n;
    cin >> n;
    int l = 0, r = seq.size() - 1;
    while (l < r) {
      int mid = l + r + 1 >> 1;
      if (seq[mid] <= n)
        l = mid;
      else
        r = mid - 1;
    }
    cout << l + 1 << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  