[LeetCode] 4 Keys Keyboard 四键的键盘

Imagine you have a special keyboard with the following keys:

Key 1: (A): Print one 'A' on screen.

Key 2: (Ctrl-A): Select the whole screen.

Key 3: (Ctrl-C): Copy selection to buffer.

Key 4: (Ctrl-V): Print buffer on screen appending it after what has already been printed.

Now, you can only press the keyboard for N times (with the above four keys), find out the maximum numbers of 'A' you can print on screen.

Example 1:

Input: N = 3
Output: 3
Explanation: 
We can at most get 3 A's on screen by pressing following key sequence:
A, A, A

Example 2:

Input: N = 7
Output: 9
Explanation: 
We can at most get 9 A's on screen by pressing following key sequence:
A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V

Note:

1 <= N <= 50
Answers will be in the range of 32-bit signed integer.

这道题给了四个操作，分别是打印A，全选，复制，粘贴。每个操作都算一个步骤，给了我们一个数字N，问N个操作最多能输出多个A。分析题目中的例子可以发现，N步最少都能打印N个A出来，因为可以每步都是打印A。那么能超过N的情况肯定就是使用了复制粘贴，这里由于全选和复制要占用两步，所以能增加A的个数的操作其实只有N-2步，那么如何确定打印几个A，剩下都是粘贴呢，其实是个 trade off，A打印的太多或太少，都不会得到最大结果，所以打印A和粘贴的次数要接近，最简单的方法就是遍历所有的情况然后取最大值，打印A的次数在 [1, N-3] 之间，粘贴的次数为 N-2-i，加上打印出的部分，就是 N-1-i了，参见代码如下：

解法一：

class Solution {
public:
    int maxA(int N) {
        int res = N;
        for (int i = 1; i < N - 2; ++i) {
            res = max(res, maxA(i) * (N - 1 - i));
        }
        return res;
    }
};

这道题也可以用 DP 来做，用一个一维数组 dp，其中 dp[i] 表示步骤总数为i时，能打印出的最多A的个数，初始化为 N+1 个，然后来想状态转移方程怎么求。对于 dp[i] 来说，求法其实跟上面的方法一样，还是要遍历所有打印A的个数，然后乘以粘贴的次数加1，用来更新 dp[i]，参见代码如下：

解法二：

class Solution {
public:
    int maxA(int N) {
        vector<int> dp(N + 1, 0);
        for (int i = 0; i <= N; ++i) {
            dp[i] = i;
            for (int j = 1; j < i - 2; ++j) {
                dp[i] = max(dp[i], dp[j] * (i - j - 1));
            }
        }
        return dp[N];
    }
};