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 double pi = acos(-1.0);
  int mi2(int x1) {
    long long sum = 0;
    int cnt = 1;
    while (sum <= x1) {
      cnt++;
      sum = cnt * cnt;
    }
    return cnt - 1;
  }
  int mi3(int x1) {
    long long sum = 0;
    int cnt = 1;
    while (sum <= x1) {
      cnt++;
      sum = cnt * cnt * cnt;
    }
    return cnt - 1;
  }
  int mi6(int x1) {
    long long sum = 0;
    int cnt = 1;
    while (sum <= x1) {
      cnt++;
      sum = cnt * cnt * cnt * cnt * cnt * cnt;
    }
    return cnt - 1;
  }
  int main(void) {
    int t;
    int n;
    int put;
    cin >> t;
    for (int i = 1; i <= (t); i++) {
      cin >> n;
      int x1, x2, x3;
      x1 = int(pow(n + 0.5, 1.0 / 2));
      x2 = int(pow(n + 0.5, 1.0 / 3));
      x3 = int(pow(n + 0.5, 1.0 / 6));
      put = mi2(n) + mi3(n) - mi6(n);
      cout << x1 + x2 - x3 << "\n";
    }
  }
  