Subtask 1

Since ~1 \leq N \leq 10^6~, we can just simulate the cookies.

Time Complexity: ~O(N)~

Subtask 2

Simulating for ~1 \leq N \leq 10^{18}~ panels will TLE. Instead, we can work backwards and use a recursive approach:

Let ~c(x)~ be the number of cookies on panel ~x~. Notice that the only previous panels that contribute to the total sum are ~c(\frac x 2)~, ~c(\frac {x-1} 2)~, and ~c(\frac x 3)~. Therefore, ~c(x) = c(\frac x 2) + c(\frac {x-1} 2) + c(\frac x 3)~, with ~c(x) = 0~ for noninteger ~x~. It turns out that ~c(x)~ is fast enough to AC!

#include <stdio.h>

int f(long x) {
  if (x < 2) {
    return x;
  } else {
    return f(x / 2) + (x % 3 == 0 ? f(x / 3) : 0);
  }
}

int main() {
  long n;
  scanf("%ld", &n);
  printf("%d\n", f(n));
}

(note how ~c(\frac x 2) + c(\frac {x-1} 2)~ is collapsible into ~c(\lfloor \frac x 2 \rfloor)~.)

Time Complexity: ~O(log N)~, probably