键盘特殊

Problem statement:

问题陈述：

Imagine you have a special keyboard with four types of keys:

想象一下，您有一个特殊的键盘，其中包含四种类型的键：

Key 1: Prints 'I' on screen

按键1：在屏幕上打印“ I”

Key 2: Select whatever printed in screen

键2：选择屏幕上打印的任何内容

Key 3: Copy selection to buffer

关键3：将选择复制到缓冲区

Key 4: Append buffer on-screen after what has already been printed. If you can only press the keyboard for N times (with the above four keys), write a program to produce maximum numbers of I's possible to be printed on the screen.

关键4：在已打印的内容之后在屏幕上追加缓冲区。 如果您只能按键盘N次(使用上述四个键)，请编写一个程序以产生最大数量的I(可以在屏幕上打印)。

Input:

输入：

Input is N, number of times keys can be pressed in total.

输入为N ，总共可以按键的次数。

Output:

输出：

Print maximum number of I's possible to print

打印我可以打印的最大数量

Constraints:

限制条件：

1 ≤ N ≤ 75

Example:

例：

Input:
2
Output:
2
Explanation:
We can at most get 2 I's on screen by pressing 
following key sequence Key1, key1.
Pressing other keys have no effect. 
Like key 1, key2 will produce only one I on screen. 
Input:
7
Output:
9
Explanation:
We can at most get 9 I's on screen by pressing 
following key sequence.
Key1, Key1, Key1, Key2, Key3, key4, Key4
I //after pressing key1
I I  //again pressing key 1
I I I //again pressing key1
I I I //pressing key2 selects three I's
I I I // pressing key3 copies these three I's to buffer
I I I I I I // pressing key4 appends these three I's 
I I I I I I I I I // pressing key4 again appends these three I's 

Solution Approach:

解决方法：

Basically,

基本上，

Two things need to be understood to solve this problem

解决此问题需要了解两点

Key4 appends whatever is printed already on screen before 3 key pressing

按下3键之前， Key4会附加屏幕上已经打印的内容

That means at moment 4,

这意味着在第四时刻

You can append whatever was printed while moment 1 as to print in moment 4, you need to press key2 at moment 2 and key3 at moment 3.

您可以在第1时刻添加要在第4时刻打印的内容，然后在第2时刻按key2 ，在第3时刻按key3 。

So, the recursive function can be written as

因此，递归函数可以写成

Let,

让，

F(n) = max number of I’s printed on screen

F(n) =我在屏幕上打印的最大数量

So, for n>3

因此，对于n> 3

F(n) = max(f(j)*(n-j-1)) for 1<=j<=n-3
Where, 
F(j) = already printed characters up to moment j 
and (n-j-1) is number of appending possible   

So, now we need to convert the recursion into DP.

因此，现在我们需要将递归转换为DP。

1)  Initialize dp[n+1] like following base value;
2)  for i=0 to n
dp[i]=i;
3)  Fill the DP table
for i=4 to n
for j=i-3 to 1,j--
dp[i]=max(dp[i],dp[j]*(i-j-1));
end for
End for
4)  Return dp[n]

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int specialKeyboard(int n)
{
int dp[n + 1];
for (int i = 0; i <= n; i++)
dp[i] = i;
for (int i = 6; i <= n; i++) {
for (int j = i - 3; j >= 1; j--) {
dp[i] = std::max(dp[i], dp[j] * (i - j - 1));
}
}
return dp[n];
}
int main()
{
int t, n, item;
cout << "Input n, number of times keys to be pressed: ";
scanf("%d", &n);
cout << "max no of i's got printed: " << specialKeyboard(n) << endl;
return 0;
}

Output:

输出：

Input n, number of times keys to be pressed: 7
max no of i's got printed: 9

翻译自: https://www.includehelp.com/icp/special-keyboard.aspx

键盘特殊