題意：給你初始一個A，再給你一串A，每次可以選擇兩種操作：復制當前所有字符，粘貼之前復制的字符。問你從一個給到指定個數(shù)的A最短需要多少步驟。
解題思路：本質就是求一個數(shù)的因子和。
假設最終串S有n個A，C代表拷貝，P代表粘貼，且最終串 = CPPCPCPPP，那么拆分一下操作步驟其實就是（CPP）（CP）（CPPP），在執(zhí)行最后一個括號串操作之前，串的情況是S1 = （CPP）（CP），就是說S是S1的4倍，那么從S1到S需要操作4次，且4是n的一個因子。再往前S2 = CPP，就是說S1是S2的2倍，從S2到S1 需要操作2次，2同樣是n的一個因子，依次類推，最少的操作步驟，其實就是n的所有因子和。
官方解釋:
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.

class Solution {
public:
    int minSteps(int n) {
        int ans = 0, d = 2;
        while(n > 1){
            while(n % d == 0){
                ans += d;
                n /= d;
            }
            d++;
        }
        return ans;
    }
};

時間復雜度：O（n）.