如何使用给定的四个键打印A的最大数量

这是Google，Paytm和其他公司面试中提出的一个著名面试问题。
以下是问题陈述。

Imagine you have a special keyboard with the following keys: 
Key 1:  Prints 'A' on screen
Key 2: (Ctrl-A): Select 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. 

If you can only press the keyboard for N times (with the above four
keys), write a program to produce maximum numbers of A's. That is to
say, the input parameter is N (No. of keys that you can press), the 
output is M (No. of As that you can produce).

例子：

Input:  N = 3
Output: 3
We can at most get 3 A's on screen by pressing 
following key sequence.
A, A, A

Input:  N = 7
Output: 9
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

Input:  N = 11
Output: 27
We can at most get 27 A's on screen by pressing 
following key sequence.
A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V, Ctrl A, 
Ctrl C, Ctrl V, Ctrl V

以下是一些要注意的要点。
a)对于N <7，输出为N本身。
b)可以多次使用Ctrl V打印当前缓冲区(请参见上面的最后两个示例)。这个想法是通过使用简单的见解来计算N次击键的最佳字符串长度。产生最佳字符串长度的N次击键序列将以Ctrl-A，Ctrl-C后缀和仅Ctrl-V的后缀结尾。 (对于N> 6)
我们的任务是找出break = point，然后获得上述击键后缀。断点的定义是在此实例之后，我们只需要按一次Ctrl-A，Ctrl-C，然后再按一次Ctrl-V即可生成最佳长度。如果我们从N-3循环到1，并选择每个这些值作为断点，然后计算它们将产生的最佳字符串。一旦循环结束，我们将获得各种断点的最佳长度的最大值，从而为N次击键提供最佳长度。

下面是基于以上思想的实现。

C++14

/* A recursive C++ program to print maximum number of A's using
following four keys */
#include 
using namespace std;
 
// A recursive function that returns the optimal length string
// for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // Initialize result
    int max = 0;
 
    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    int b;
    for (b = N - 3; b >= 1; b--) {
 
        // If the breakpoint is s at b'th keystroke then
        // the optimal string would have length
        // (n-b-1)*screen[b-1];
        int curr = (N - b - 1) * findoptimal(b);
        if (curr > max)
            max = curr;
    }
    return max;
}
 
// Driver program
int main()
{
    int N;
 
    // for the rest of the array we will rely on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        cout << "Maximum Number of A's with " << N <<
        " keystrokes is " << findoptimal(N) << endl;
}
 
// This code is contributed by shubhamsingh10

C

/* A recursive C program to print maximum number of A's using
   following four keys */
#include 
 
// A recursive function that returns the optimal length string
// for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // Initialize result
    int max = 0;
 
    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    int b;
    for (b = N - 3; b >= 1; b--) {
        // If the breakpoint is s at b'th keystroke then
        // the optimal string would have length
        // (n-b-1)*screen[b-1];
        int curr = (N - b - 1) * findoptimal(b);
        if (curr > max)
            max = curr;
    }
    return max;
}
 
// Driver program
int main()
{
    int N;
 
    // for the rest of the array we will rely on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

/* A recursive Java program to print
  maximum number of A's using
  following four keys */
import java.io.*;
 
class GFG {
 
    // A recursive function that returns
    // the optimal length string  for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;
 
        // Initialize result
        int max = 0;
 
        // TRY ALL POSSIBLE BREAK-POINTS
        // For any keystroke N, we need to
        // loop from N-3 keystrokes back to
        // 1 keystroke to find a breakpoint
        // 'b' after which we will have Ctrl-A,
        // Ctrl-C and then only Ctrl-V all the way.
        int b;
        for (b = N - 3; b >= 1; b--) {
            // If the breakpoint is s at b'th
            // keystroke then the optimal string
            // would have length
            // (n-b-1)*screen[b-1];
            int curr = (N - b - 1) * findoptimal(b);
            if (curr > max)
                max = curr;
        }
        return max;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int N;
 
        // for the rest of the array we
        // will rely on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            System.out.println("Maximum Number of A's with keystrokes is " + N + findoptimal(N));
    }
}
 
// This code is contributed by vt_m.

Python3

# A recursive Python3 program to print maximum
# number of A's using following four keys
 
# A recursive function that returns
# the optimal length string for N keystrokes
 
def findoptimal(N):
     
    # The optimal string length is
    # N when N is smaller than
    if N<= 6:
        return N
 
    # Initialize result
    maxi = 0
 
    # TRY ALL POSSIBLE BREAK-POINTS
    # For any keystroke N, we need
    # to loop from N-3 keystrokes
    # back to 1 keystroke to find
    # a breakpoint 'b' after which we
    # will have Ctrl-A, Ctrl-C and then
    # only Ctrl-V all the way.
    for b in range(N-3, 0, -1):
        curr =(N-b-1)*findoptimal(b)
        if curr>maxi:
            maxi = curr
     
    return maxi
# Driver program
if __name__=='__main__':
     
 
# for the rest of the array we will
# rely on the previous
# entries to compute new ones
    for n in range(1, 21):
        print('Maximum Number of As with ', n, 'keystrokes is ', findoptimal(n))
        
# this code is contibuted by sahilshelangia

C#

/* A recursive C# program
to print maximum number
of A's using following
four keys */
using System;
 
class GFG {
    // A recursive function that
    // returns the optimal length
    // string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length
        // is N when N is smaller than 7
        if (N <= 6)
            return N;
 
        // Initialize result
        int max = 0;
 
        // TRY ALL POSSIBLE BREAK-POINTS
        // For any keystroke N, we need
        // to loop from N-3 keystrokes
        // back to 1 keystroke to find
        // a breakpoint 'b' after which
        // we will have Ctrl-A, Ctrl-C
        // and then only Ctrl-V all the way.
        int b;
        for (b = N - 3; b >= 1; b--) {
            // If the breakpoint is s
            // at b'th keystroke then
            // the optimal string would
            // have length (n-b-1)*screen[b-1];
            int curr = (N - b - 1) * findoptimal(b);
            if (curr > max)
                max = curr;
        }
        return max;
    }
 
    // Driver code
    static void Main()
    {
        int N;
 
        // for the rest of the array
        // we will rely on the
        // previous entries to compute
        // new ones
        for (N = 1; N <= 20; N++)
            Console.WriteLine("Maximum Number of A's with " + N + " keystrokes is " + findoptimal(N));
    }
}
 
// This code is contributed by Sam007

PHP

= 1; $b -= 1)
    {
            // If the breakpoint is s at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            $curr = ($N - $b - 1) * findoptimal($b);
            if ($curr > $max)
                $max = $curr;
    }
    return $max;
}
 
// Driver code
$N;
 
// for the rest of the array we will rely on the previous
// entries to compute new ones
for ($N = 1; $N <= 20; $N += 1)
    echo("Maximum Number of A's with"
        .$N."keystrokes is ".findoptimal($N)."\n");
 
// This code is contributed by aman neekhara
?>

Javascript

 
// A recursive Javascript program to print
// maximum number of A's using
// following four keys
     
// A recursive function that returns the
// optimal length string  for N keystrokes
function findoptimal(N)
{
     
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;
 
    // Initialize result
    let max = 0;
 
    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to
    // loop from N-3 keystrokes back to
    // 1 keystroke to find a breakpoint
    // 'b' after which we will have Ctrl-A,
    // Ctrl-C and then only Ctrl-V all the way.
    let b;
     
    for(b = N - 3; b >= 1; b--)
    {
         
        // If the breakpoint is s at b'th
        // keystroke then the optimal string
        // would have length
        // (n-b-1)*screen[b-1];
        let curr = (N - b - 1) * findoptimal(b);
         
        if (curr > max)
            max = curr;
    }
    return max;
}
 
// Driver code
let N;
 
// For the rest of the array we
// will rely on the previous
// entries to compute new ones
for(N = 1; N <= 20; N++)
    document.write("Maximum Number of A's with "+
                   N + " keystrokes is " +
                   findoptimal(N) + "<br>");
 
// This code is contributed by avanitrachhadiya2155
     

---

输出：

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324

上面的函数一次又一次地计算相同的子问题。通过存储子问题的解决方案并以自下而上的方式解决问题，可以避免对相同子问题进行重新计算。

以下是基于动态编程的C实现，其中使用辅助阵列屏幕[N]存储子问题的结果。

C++

/* A Dynamic Programming based C program to find maximum number of A's
that can be printed using four keys */
#include 
using namespace std;
 
 
// this function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result of subproblems
    int screen[N];
 
    int b; // To pick a breakpoint
 
    // Initializing the optimal lengths array for uptil 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom manner
    for (n = 7; n <= N; n++) {
        // Initialize length of optimal string for n keystrokes
        screen[n - 1] = 0;
 
        // For any keystroke n, we need to loop from n-3 keystrokes
        // back to 1 keystroke to find a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and then only ctrl-v all the way.
        for (b = n - 3; b >= 1; b--) {
            // if the breakpoint is at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            int curr = (n - b - 1) * screen[b - 1];
            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }
 
    return screen[N - 1];
}
 
// Driver program
int main()
{
    int N;
 
    // for the rest of the array we will rely on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        cout << "Maximum Number of A's with "<

C

/* A Dynamic Programming based C program to find maximum number of A's
   that can be printed using four keys */
#include 
 
// this function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result of subproblems
    int screen[N];
 
    int b; // To pick a breakpoint
 
    // Initializing the optimal lengths array for uptil 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom manner
    for (n = 7; n <= N; n++) {
        // Initialize length of optimal string for n keystrokes
        screen[n - 1] = 0;
 
        // For any keystroke n, we need to loop from n-3 keystrokes
        // back to 1 keystroke to find a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and then only ctrl-v all the way.
        for (b = n - 3; b >= 1; b--) {
            // if the breakpoint is at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            int curr = (n - b - 1) * screen[b - 1];
            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }
 
    return screen[N - 1];
}
 
// Driver program
int main()
{
    int N;
 
    // for the rest of the array we will rely on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

/* A Dynamic Programming based C
   program to find maximum number
   of A's that can be printed using
   four keys */
import java.io.*;
 
class GFG {
 
    // this function returns the optimal
    // length string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;
 
        // An array to store result
        // of subproblems
        int screen[] = new int[N];
 
        int b; // To pick a breakpoint
 
        // Initializing the optimal lengths
        // array for uptil 6 input strokes
        int n;
        for (n = 1; n <= 6; n++)
            screen[n - 1] = n;
 
        // Solve all subproblems in bottom manner
        for (n = 7; n <= N; n++) {
            // Initialize length of optimal
            // string for n keystrokes
            screen[n - 1] = 0;
 
            // For any keystroke n, we need
            // to loop from n-3 keystrokes
            // back to 1 keystroke to find
            // a breakpoint 'b' after which we
            // will have ctrl-a, ctrl-c and
            // then only ctrl-v all the way.
            for (b = n - 3; b >= 1; b--) {
                // if the breakpoint is
                // at b'th keystroke then
                // the optimal string would
                // have length
                // (n-b-1)*screen[b-1];
                int curr = (n - b - 1) * screen[b - 1];
                if (curr > screen[n - 1])
                    screen[n - 1] = curr;
            }
        }
 
        return screen[N - 1];
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int N;
 
        // for the rest of the array we will rely on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            System.out.println("Maximum Number of A's with keystrokes is " + N + findoptimal(N));
    }
}
 
// This article is contributed by vt_m.

Python3

# A Dynamic Programming based Python program
# to find maximum number of A's
# that can be printed using four keys
 
# this function returns the optimal
# length string for N keystrokes
def findoptimal(N):
 
    # The optimal string length is
    # N when N is smaller than 7
    if (N <= 6):
        return N
 
    # An array to store result of
    # subproblems
    screen = [0]*N
 
    # Initializing the optimal lengths
    # array for uptil 6 input
    # strokes.
     
    for n in range(1, 7):
        screen[n-1] = n
 
    # Solve all subproblems in bottom manner
    for n in range(7, N + 1):
     
        # Initialize length of optimal
        # string for n keystrokes
        screen[n-1] = 0
 
        # For any keystroke n, we need to
        # loop from n-3 keystrokes
        # back to 1 keystroke to find a breakpoint
        # 'b' after which we
        # will have ctrl-a, ctrl-c and then only
        # ctrl-v all the way.
        for b in range(n-3, 0, -1):
         
            # if the breakpoint is at b'th keystroke then
            # the optimal string would have length
            # (n-b-1)*screen[b-1];
            curr = (n-b-1)*screen[b-1]
            if (curr > screen[n-1]):
                screen[n-1] = curr
         
    return screen[N-1]
 
# Driver program
if __name__ == "__main__":
 
    # for the rest of the array we
    # will rely on the previous
    # entries to compute new ones
    for N in range(1, 21):
        print("Maximum Number of A's with ", N, " keystrokes is ",
                findoptimal(N))
 
# this code is contributed by
# ChitraNayal

C＃

/* A Dynamic Programming based C#
program to find maximum number
of A's that can be printed using
four keys */
using System;
public class GFG {
 
    // this function returns the optimal
    // length string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;
 
        // An array to store result
        // of subproblems
        int[] screen = new int[N];
 
        int b; // To pick a breakpoint
 
        // Initializing the optimal lengths
        // array for uptil 6 input strokes
        int n;
        for (n = 1; n <= 6; n++)
            screen[n - 1] = n;
 
        // Solve all subproblems in bottom manner
        for (n = 7; n <= N; n++) {
            // Initialize length of optimal
            // string for n keystrokes
            screen[n - 1] = 0;
 
            // For any keystroke n, we need
            // to loop from n-3 keystrokes
            // back to 1 keystroke to find
            // a breakpoint 'b' after which we
            // will have ctrl-a, ctrl-c and
            // then only ctrl-v all the way.
            for (b = n - 3; b >= 1; b--) {
                // if the breakpoint is
                // at b'th keystroke then
                // the optimal string would
                // have length
                // (n-b-1)*screen[b-1];
                int curr = (n - b - 1) * screen[b - 1];
                if (curr > screen[n - 1])
                    screen[n - 1] = curr;
            }
        }
 
        return screen[N - 1];
    }
 
    // Driver program
    public static void Main(String[] args)
    {
        int N;
 
        // for the rest of the array we will rely on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            Console.WriteLine("Maximum Number of A's with {0} keystrokes is {1}\n",
                              N, findoptimal(N));
    }
}
 
// This code is contributed by PrinciRaj1992

Java脚本

 
// A Dynamic Programming based javascript
// program to find maximum number
// of A's that can be printed using
// four keys
 
// This function returns the optimal
// length string for N keystrokes  
function findoptimal(N)
{
     
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result
    // of subproblems
    let screen = new Array(N);
    for(let i = 0; i < N; i++)
    {
        screen[i] = 0;
    }
     
    // To pick a breakpoint
    let b;
 
    // Initializing the optimal lengths
    // array for uptil 6 input strokes
    let n;
    for(n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom manner
    for(n = 7; n <= N; n++)
    {
         
        // Initialize length of optimal
        // string for n keystrokes
        screen[n - 1] = 0;
 
        // For any keystroke n, we need
        // to loop from n-3 keystrokes
        // back to 1 keystroke to find
        // a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and
        // then only ctrl-v all the way.
        for(b = n - 3; b >= 1; b--)
        {
             
            // If the breakpoint is
            // at b'th keystroke then
            // the optimal string would
            // have length
            // (n-b-1)*screen[b-1];
            let curr = (n - b - 1) * screen[b - 1];
             
            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }
    return screen[N - 1];
}
 
// Driver code
let N;
for(N = 1; N <= 20; N++)
    document.write("Maximum Number of A's with " +
                   N + " keystrokes is " +
                   findoptimal(N) + "<br>");
 
// This code is contributed by rag2127
 

输出：

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324

感谢Gaurav Saxena提供了上述方法来解决此问题。
随着A数量的增加，与再次按Ctrl-A，Ctrl-C和Ctrl-V相比，按Ctrl-V 3次以上的效果开始变得微不足道。因此，通过检查仅按Ctrl-V 1、2和3次的效果，可以使上面的代码更有效。

C++

// A Dynamic Programming based C++ program to find maximum
// number of A's that can be printed using four keys
 
#include 
using namespace std;
 
// This function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result of subproblems
    int screen[N];
 
    int b; // To pick a breakpoint
 
    // Initializing the optimal lengths array for uptil 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++) {
 
        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.
 
        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.
 
        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = max(2 * screen[n - 4],
                            max(3 * screen[n - 5],
4 * screen[n - 6]));
    }
 
    return screen[N - 1];
}
 
// Driver program
int main()
{
    int N;
 
    // for the rest of the array we will rely on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

// A Dynamic Programming based Java program
// to find maximum number of A's that
// can be printed using four keys
class GFG
{
 
// This function returns the optimal length
// string for N keystrokes
static int findoptimal(int N)
{
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result of subproblems
    int []screen = new int[N];
 
    int b; // To pick a breakpoint
 
    // Initializing the optimal lengths array
    // for uptil 6 input strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++)
    {
 
        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.
 
        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.
 
        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = Math.max(2 * screen[n - 4],
                        Math.max(3 * screen[n - 5],
                                 4 * screen[n - 6]));
    }
    return screen[N - 1];
}
 
// Driver Code
public static void main(String[] args)
{
    int N;
 
    // for the rest of the array we will rely
    // on the previous entries to compute new ones
    for (N = 1; N <= 20; N++)
        System.out.printf("Maximum Number of A's with" +
                               " %d keystrokes is %d\n",
                                     N, findoptimal(N));
    }
}
 
// This code is contributed by Princi Singh

Python3

# A Dynamic Programming based Python3 program
# to find maximum number of A's
# that can be printed using four keys
 
# this function returns the optimal
# length string for N keystrokes
def findoptimal(N):
 
    # The optimal string length is
    # N when N is smaller than 7
    if (N <= 6):
        return N
 
    # An array to store result of
    # subproblems
    screen = [0] * N
 
    # Initializing the optimal lengths
    # array for uptil 6 input
    # strokes.
     
    for n in range(1, 7):
        screen[n - 1] = n
 
    # Solve all subproblems in bottom manner
    for n in range(7, N + 1):
         
        # for any keystroke n, we will need to choose between:-
        # 1. pressing Ctrl-V once after copying the
        # A's obtained by n-3 keystrokes.
 
        # 2. pressing Ctrl-V twice after copying the A's
        # obtained by n-4 keystrokes.
 
        # 3. pressing Ctrl-V thrice after copying the A's
        # obtained by n-5 keystrokes.
        screen[n - 1] = max(2 * screen[n - 4],
                        max(3 * screen[n - 5],
                            4 * screen[n - 6]));
         
    return screen[N - 1]
 
# Driver Code
if __name__ == "__main__":
 
    # for the rest of the array we
    # will rely on the previous
    # entries to compute new ones
    for N in range(1, 21):
        print("Maximum Number of A's with ", N,
              " keystrokes is ", findoptimal(N))
 
# This code is contributed by ashutosh450

C＃

// A Dynamic Programming based C# program
// to find maximum number of A's that
// can be printed using four keys
using System;
 
class GFG
{
 
// This function returns the optimal length
// string for N keystrokes
static int findoptimal(int N)
{
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;
 
    // An array to store result of subproblems
    int []screen = new int[N];
 
    // Initializing the optimal lengths array
    // for uptil 6 input strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;
 
    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++)
    {
 
        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.
 
        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.
 
        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = Math.Max(2 * screen[n - 4],
                        Math.Max(3 * screen[n - 5],
                                4 * screen[n - 6]));
    }
    return screen[N - 1];
}
 
// Driver Code
public static void Main(String[] args)
{
    int N;
 
    // for the rest of the array we will rely
    // on the previous entries to compute new ones
    for (N = 1; N <= 20; N++)
        Console.Write("Maximum Number of A's with" +
                            " {0} keystrokes is {1}\n",
                                    N, findoptimal(N));
    }
}
 
// This code is contributed by PrinciRaj1992

输出：

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324