> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/l4dong-tai-gui-hua/4-keys-keyboard.md).

# 4 Keys Keyboard

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. 1 

   <

   \= N&#x20;

   <

   \= 50
2. Answers will be in the range of 32-bit signed integer.

分析

I可以直接print来，不copy, dp\[i] = i

j之前print, j之后都copy, j的 range = \[1,i-3] . dp\[i] = dp\[j]\*(i-j-1)

从j到i状态转换，去掉2次选择和复制操作，剩下的都是粘贴C，i的个数是j的C倍。

```
class Solution:
    def maxA(self, N: int) -> int:
        dp=[0]*(N+1)
        for i in range(1,N+1):                        
            dp[i] = i
            for j in range(1,i-3):
                dp[i] = max(dp[i], dp[j]*(i-j-1))

        return dp[N]
```