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 = 1e5 + 10;
  long long a[N], b[1050], c[N];
  int cnt = 0;
  void init() {
    cnt = 0;
    for (int i = 1; i < N - 50; i++) a[cnt++] = i * i;
    cnt = 0;
    for (int i = 1; i < 1010; i++) b[cnt++] = i * i * i;
    int u = 0, t = 0;
    cnt = 0;
    while (a[u] && b[t]) {
      if (a[u] < b[t]) {
        c[cnt++] = a[u++];
      } else if (a[u] == b[t]) {
        c[cnt++] = a[u++];
        t++;
      } else
        c[cnt++] = b[t++];
    }
  }
  int main() {
    init();
    int T;
    cin >> T;
    while (T--) {
      int n;
      scanf("%d", &n);
      int i = 0;
      while (c[i] <= n) {
        i++;
      }
      printf("%d\n", i);
    }
    return 0;
  }
  