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;
  template <typename T>
  void inline read(T &x) {
    long long f = 1;
    x = 0;
    char s = getchar();
    while (s<'0' | s> '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0')
      x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
    x *= f;
  }
  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() {
    long long n;
    read(n);
    long long a = sqrt(n), b = pow(n, 1 / 3.0);
    long long c = pow(n, 1 / 6.0);
    while ((b + 1) * (b + 1) * (b + 1) <= n) b++;
    while ((c + 1) * (c + 1) * (c + 1) * (c + 1) * (c + 1) * (c + 1) <= n) c++;
    printf("%d\n", a + b - c);
  }
  signed main() {
    long long T;
    init();
    for (cin >> T; T > 0; T--) run();
    return 0;
  }
  