原题链接在这里：

题目：

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

Example 2:

Input: N = 7Output: 9Explanation: 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 <= 50
2. Answers will be in the range of 32-bit signed integer.

题解：

DP问题.

初始化, dp[i] = i. i在[1,N]就是直接输入'A'.

状态转移, dp[i] = Math.max(dp[i], (i-j-1)*dp[j]), j在[1,i-3]区间内.  Ctrl A, Ctrl C, Ctrl V用了3个操作，所以j是到i-3. 本来已经有了d[j]个'A', 三个操作做完又加了dp[j]个'A', 还剩下i-3-j个操作，剩下的每一次都加dp[j]个'A'. 总共dp[j] + dp[j] + (i-3-j)*dp[j] = (i-j-1)*dp[j]. 

答案dp[N].

Time Complexity: O(N^2).

Space: O(N).

AC Java:

1 class Solution { 2     public int maxA(int N) { 3         int [] dp = new int[N+1]; 4         for(int i = 1; i<=N; i++){ 5             dp[i] = i; 6             for(int j = 1; j<=i-3; j++){ 7                 dp[i] = Math.max(dp[i], (i-j-1)*dp[j]); 8             } 9         }10         return dp[N];11     }12 }

类似.