#include <bits/stdc++.h>

using namespace std;


int f(int N) {
    int answer = 0;
    while (N > 1) {
        N = sqrt(N);
        answer++;
    }

    return answer;
}

signed main() {
    cout << 0 << " " << f(0) << "\n";

    for (int i = 1; ; i++) {
        if (f(i) != f(i - 1)) {
            cout << i << " " << f(i) << "\n";
        }
    }

    return 0;
}

0 0
2 1
4 2
16 3
256 4
65536 5

We can see that result for $$${2 ^ {2 ^ n}}$$$ is $$$n + 1$$$. Result for $$${2 ^ {2 ^ n}} \leqslant k \leqslant {2 ^ {2 ^ n + 1}}$$$ is $$$n + 1$$$. So the formula for $$$n$$$ is $$$\log(\log(n)) + 1$$$. In C++ such function as __builtin_clz has O(1) complexity.

Code (code works properly for int type, for 64-bit integers use 63 - __builtin_clzll(x)):

int N;
cin >> N;

int answer = 31 - __builtin_clz(31 - __builtin_clz(i)) + 1;
cout << answer;