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

    return ans;
}

int minSteps(int n) {
    if (n <= 1) return 0;
    if (n <= 5) return n;
    // n>1的操作，前两步骤都是copy & paste, 从'A'变成 'AA', 所以 cur_len == 2, res == 2, 上次copy的字符数量 == 1, 如果下一次要copy & paste, 要一次copy'AA' 2个字符  
    int cur_len = 2, res = 2, next_cp_num = 2, copy_num = 1;
    while (cur_len < n) {
        if ((n - cur_len) % next_cp_num == 0) {
           // copy + paste
           cur_len += next_cp_num;
           copy_num = next_cp_num;
           // copy + paste; 2 steps
           res += 2;
        } else {
            // paste the characters copied last time 
            cur_len += copy_num;
            // only paste, 1 step
            res += 1;
        }

        // 题目要求：Copy操作要Copy所有字符
        next_cp_num = cur_len;
    }

    return res;
}