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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:

int maxA(int N) {    if (N <= 6)return N;    vector
   
     dp(N + 1, 0);    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] = max(dp[i], dp[j] * (i - j - 1));        }    }    return dp[N];}
   

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