This is a greedy problem. First thing to notice is that exam which is occurring earlier compared to others should be prepared first, i.e. exam which is having less d_i should be prepared earlier(if value of s allows to do so).

Sort the exams in the increasing order of d_i. If we go through solution we can see that it gives optimum solution.

Let us take an exam from the sorted order = {s_i , d_i, c_i}. Since we have sorted exams , 1 ≤ j ≤ n AND j ≠ i; either d_i ≤ d_j or j^th exam is completed(We are processing in increasing order of d_i). Now we have to prepare c_i days for exam i in the interval s_i and d_i(and obviously we are forced to do so, so that's an optimal step). Now since all other exams(for which preparation is not done yet) are after i^th exam, what is the best way to choose c_i days in the interval [s_i, d_i] such that we leave better opportunities for the preparation of other exams?If we iterate from s_i to d_i and choose first c_i days which are free for preparation. Hence our solution is optimal.

Worst case time complexity is : O(m*n).

We can see that all numbers in given sequence are distinct. since all numbers in sequence are of the form . Hence if a_i = a_j then = = 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 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 and 2 * a_i exists in sequence. by little bit of tricks this = , 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 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 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:   ≠  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 and 2 * a_i (remember only one of them is present so once you find it you print it).