algo-aha

要输入n个字符，求最少击键次数

• https://leetcode.com/problems/2-keys-keyboard/

int minSteps(int n) {
    // 一组操作是C后带若干个P，CP、CPP、...，最终n=len(CP)*len(CPP)*...
    // 这题就是把n因子分解，求所有>=2因子的和
    int ans = 0, factor = 2;
    while (n > 1) {
        while (n % factor == 0) {
            ans += factor;
            n /= factor;
        }
        factor++;
    }
    return ans;
}

击键n次，最多能输入多少字符

• https://leetcode.com/problems/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 <= 50
2. Answers will be in the range of 32-bit signed integer.

int maxA(int N) {
    // dp[i]表示按i次键盘时A的最大个数
    // 一种是按1次A，dp[i]=dp[i-1]+1
    // 一种是成组按3<=k<=i次，ctrlA+ctrlC+(k-2)ctrlV，字符数 *= (k-1)
    //  dp[i]=max{ dp[i-k] * (k-1) }
    vector<int> dp(N + 1, 0);
    for (int i = 1; i <= N; i++) {
        dp[i] = dp[i-1] + 1;
        for (int k = 3; k <= i; k++) {
            dp[i] = max(dp[i], dp[i-k] * (k-1));
        }
    }
    return dp[N];
}