原题链接在这里：https://leetcode.com/problems/4-keys-keyboard/description/
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 = 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.
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:
class Solution {
public int maxA(int N) {
int [] dp = new int[N+1];
for(int i = 1; i<=N; i++){
dp[i] = i;
for(int j = 1; j<=i-3; j++){
dp[i] = Math.max(dp[i], (i-j-1)*dp[j]);
}
}
return dp[N];
}
}