Approach #1: Prime Factorization [Accepted]

Intuition

We can break our moves into groups of (copy, paste, ..., paste). Let C denote copying and P denote pasting. Then for example, in the sequence of moves CPPCPPPPCP, the groups would be [CPP][CPPPP][CP].

Say these groups have lengths g_1, g_2, .... After parsing the first group, there are g_1 'A's. After parsing the second group, there are g_1 * g_2 'A's, and so on. At the end, there are g_1 * g_2 * ... * g_n 'A's.

We want exactly N = g_1 * g_2 * ... * g_n. If any of the g_i are composite, say g_i = p * q, then we can split this group into two groups (the first of which has one copy followed by p-1 pastes, while the second group having one copy and q-1 pastes).

Such a split never uses more moves: we use p+q moves when splitting, and pq moves previously. As p+q <= pq is equivalent to 1 <= (p-1)(q-1), which is true as long as p >= 2 and q >= 2.

Algorithm By the above argument, we can suppose g_1, g_2, ... is the prime factorization of N, and the answer is therefore the sum of these prime factors.

Complexity Analysis

  • Time Complexity: . When N is the square of a prime, our loop does steps.

  • Space Complexity: , the space used by ans and d.


Analysis written by: @awice.