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 N = 2e6 + 5;
  const long long inf = 0x3f3f3f3f;
  const long long mod = 1e9 + 7;
  long long n, m, k, a[N], b[N], c[N], y, ans, sum, l, r;
  string s;
  void init() {
    for (long long i = 1; i <= 100000; i++) {
      a[i] = i * i * i;
      b[i] = i * i;
    }
    for (long long i = 1; i <= 100000; i++) {
      if ((long long)sqrt(a[i]) * (long long)sqrt(a[i]) == a[i]) {
        c[y] = a[i];
        y++;
      }
    }
  }
  inline void run() {
    double n;
    cin >> n;
    long long sum = upper_bound(b + 1, b + 100000, n) - b;
    long long ans = upper_bound(a + 1, a + 100000, n) - a;
    long long p = upper_bound(c + 1, c + y, n) - c;
    p--;
    cout << sum + ans - p - 3 << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    long long T;
    init();
    for (cin >> T; T > 0; T--) run();
    return 0;
  }
  