如何绑定网站域名,专业的镇江网站建设,福州网站建设专业公司,凡科快图软件下载键盘特殊Problem statement: 问题陈述#xff1a; Imagine you have a special keyboard with four types of keys: 想象一下#xff0c;您有一个特殊的键盘#xff0c;其中包含四种类型的键#xff1a; Key 1: Prints I on screen 按键1#xff1a;在屏幕上打印“ I”…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 Is 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 Is 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 Is 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 Is 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 Is
I I I // pressing key3 copies these three Is to buffer
I I I I I I // pressing key4 appends these three Is
I I I I I I I I I // pressing key4 again appends these three Is
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 n3 因此对于n 3 F(n) max(f(j)*(n-j-1)) for 1jn-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[n1] like following base value;
2) for i0 to n
dp[i]i;
3) Fill the DP table
for i4 to n
for ji-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 is got printed: specialKeyboard(n) endl;
return 0;
}
Output: 输出 Input n, number of times keys to be pressed: 7
max no of is got printed: 9
翻译自: https://www.includehelp.com/icp/special-keyboard.aspx键盘特殊
