651. 4 Keys Keyboard

Imagine you have a special keyboard with the following keys:

Key 1: (A): Prints 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.

思路：刚开始BFS，TLE，是用DP无疑，但是最后还是要看答案啊。。。。

感觉老是差一截。。。。

多写写前几个DP数组的值，多思考思考dp[n]与之前的dp[0..n-1]之间的关系

ref：http://www.geeksforgeeks.org/how-to-print-maximum-number-of-a-using-given-four-keys/

/* * DP * 最重要是多写前几个DP的值发现规律 */public class Solution {    public int maxA(int n) {        if(n <= 6)return n; // 多写几个发现下规律嘛                // dp[i]表示i次可以得到的最大的A，所以第i次肯定不是复制，因为那还不如按个A        int[] dp = new int[1+n];        for(int i=1; i<=6; i++)dp[i]=i;                // dp[n]怎么由之前的求出来呢？        // 可以是dp[0..n-1]中的一直按A或者Ctrl_V而来的        for(int i=7; i<=n; i++) {        dp[i] = dp[i-1] + 1;        for(int j=i-3; j>0; j--) {        dp[i] = Math.max(dp[i], dp[j]*(i-j-1));        }        }                return dp[n];    }}