Codeforces 977D Editorial

977D — Divide by three, multiply by two

We can see that all numbers in given sequence are distinct. since all numbers in sequence are of the form $\text{[math]}$ . Hence if a_i = a_j then $\text{[math]}$ = $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ = $\text{[math]}$ $\text{[math]}$ 2^x - m = 3^y - n which is not possible because any power of 2 will be an even number and any power of 3 will be an odd number.
if there exists $\text{[math]}$ in sequence then 2 * a_i can not be in sequence and vice versa. We can prove it using contradiction. Suppose there is a number a_i such that both $\text{[math]}$ and 2 * a_i exists in sequence. by little bit of tricks this $\text{[math]}$ $\text{[math]}$ = $\text{[math]}$ , this again is not possible by same argument as above, we just have to change the order of exponents.
Hence for each a_i in sequence we see if $\text{[math]}$ or 2 * a_i is present(remember that only one of them can be present). Now if there is any a_i such that both $\text{[math]}$ and 2 * a_i is not in sequence then this should be a_n. if there is any such a_i that for all 0 ≤ j ≤ n: $\text{[math]}$ ≠ a_i AND 2 * a_i ≠ a_i, then this is a₁.
Once you have got a₁, you keep on producing sequence just by doing binary search for $\text{[math]}$ and 2 * a_i (remember only one of them is present so once you find it you print it).