90分，求调

#include <iostream>
#include <cmath>
using namespace std;

long long n;
bool check(double a)
{
    return floor(a * a * a) <= n;
}

int main()
{
    cin >> n;
    long long l = 0, r = n + 1, ans = 0;
    while(l <= r)
    {
        int m = (l + r)/2;
        if(check(m))
        {
            l = m + 1;
            ans = m;
        }
        else r = m - 1;
    }
    cout << ans;
    return 0;
}