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;
  const int dx[9] = {-1, 0, 1, 0, -1, -1, 1, 1, 0};
  const int dy[9] = {0, 1, 0, -1, -1, 1, -1, 1, 0};
  long long n;
  void solve() {
    long long a, b, c;
    {
      long long left = 0, right = 1e5;
      while (left < right) {
        long long mid = left + right + 1 >> 1;
        long long y = mid * mid;
        if (y <= n)
          left = mid;
        else
          right = mid - 1;
      }
      a = left;
    }
    {
      long long left = 0, right = 1e5;
      while (left < right) {
        long long mid = left + right + 1 >> 1;
        long long y = mid * mid * mid;
        if (y <= n)
          left = mid;
        else
          right = mid - 1;
      }
      b = left;
    }
    {
      long long left = 0, right = 1e3;
      while (left < right) {
        long long mid = left + right + 1 >> 1;
        long long y = mid * mid * mid * mid * mid * mid;
        if (y <= n)
          left = mid;
        else
          right = mid - 1;
      }
      c = left;
    }
    cout << a + b - c << "\n";
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
      cin >> n;
      solve();
    }
    return 0;
  }
  