我们需要在屏幕上写入N个相同的字符，并且每次我们可以插入一个字符，删除最后一个字符并复制并粘贴所有已写入的字符，即在复制操作后，总写入字符将变为两倍。现在我们有时间进行插入，删除和复制。我们需要输出最少的时间来使用这些操作在屏幕上写入N个字符。
例子：
Input : N = 9
insert time = 1
removal time = 2
copy time = 1
Output : 5
N character can be written on screen in 5 time units as shown below,
insert a character
characters = 1 total time = 1
again insert character
characters = 2 total time = 2
copy characters
characters = 4 total time = 3
copy characters
characters = 8 total time = 4
insert character
characters = 9 total time = 5我们可以使用动态编程解决此问题。在手工解决了一些示例之后，我们可以观察到一种模式，在编写每个字符，我们有两种选择：通过插入获取或通过复制获取，以较少的时间为准。现在相应地写关系
假设dp [i]是在屏幕上写入i个字符的最佳时间，
If i is even then,
dp[i] = min((dp[i-1] + insert_time),
(dp[i/2] + copy_time))
Else (If i is odd)
dp[i] = min(dp[i-1] + insert_time),
(dp[(i+1)/2] + copy_time + removal_time)在奇数的情况下，由于要复制(i + 1)/ 2个字符会增加删除时间，因为屏幕上将需要删除一个额外的字符。
C++
// C++ program to write characters in
// minimum time by inserting, removing
// and copying operation
#include
using namespace std;
// method returns minimum time to write
// 'N' characters
int minTimeForWritingChars(int N, int insert,
int remove, int copy)
{
if (N == 0)
return 0;
if (N == 1)
return insert;
// declare dp array and initialize with zero
int dp[N + 1];
memset(dp, 0, sizeof(dp));
// first char will always take insertion time
dp[1] = insert;
// loop for 'N' number of times
for (int i = 2; i <= N; i++)
{
/* if current char count is even then
choose minimum from result for (i-1)
chars and time for insertion and
result for half of chars and time
for copy */
if (i % 2 == 0)
dp[i] = min(dp[i-1] + insert,
dp[i/2] + copy);
/* if current char count is odd then
choose minimum from
result for (i-1) chars and time for
insertion and
result for half of chars and time for
copy and one extra character deletion*/
else
dp[i] = min(dp[i-1] + insert,
dp[(i+1)/2] + copy + remove);
}
return dp[N];
}
// Driver code
int main()
{
int N = 9;
int insert = 1, remove = 2, copy = 1;
cout << minTimeForWritingChars(N, insert,
remove, copy);
return 0;
} Java
// Java program to write characters in
// minimum time by inserting, removing
// and copying operation
public class GFG{
// method returns minimum time to write
// 'N' characters
static int minTimeForWritingChars(int N, int insert,
int remove, int copy)
{
if (N == 0)
return 0;
if (N == 1)
return insert;
// declare dp array and initialize with zero
int dp[] = new int [N + 1];
// first char will always take insertion time
dp[1] = insert;
// loop for 'N' number of times
for (int i = 2; i <= N; i++)
{
/* if current char count is even then
choose minimum from result for (i-1)
chars and time for insertion and
result for half of chars and time
for copy */
if (i % 2 == 0)
dp[i] = Math.min(dp[i-1] + insert, dp[i/2] + copy);
/* if current char count is odd then
choose minimum from
result for (i-1) chars and time for
insertion and
result for half of chars and time for
copy and one extra character deletion*/
else
dp[i] = Math.min(dp[i-1] + insert,
dp[(i+1)/2] + copy + remove);
}
return dp[N];
}
// Driver code to test above methods
public static void main(String []args)
{
int N = 9;
int insert = 1, remove = 2, copy = 1;
System.out.println(minTimeForWritingChars(N, insert,remove, copy));
}
// This code is contributed by Ryuga
}Python3
# Python3 program to write characters in
# minimum time by inserting, removing
# and copying operation
def minTimeForWritingChars(N, insrt,
remov, cpy):
# method returns minimum time
# to write 'N' characters
if N == 0:
return 0
if N == 1:
return insrt
# declare dp array and initialize
# with zero
dp = [0] * (N + 1)
# first char will always take insertion time
dp[1] = insrt
# loop for 'N' number of times
for i in range(2, N + 1):
# if current char count is even then
# choose minimum from result for (i-1)
# chars and time for insertion and
# result for half of chars and time
# for copy
if i % 2 == 0:
dp[i] = min(dp[i - 1] + insrt,
dp[i // 2] + cpy)
# if current char count is odd then
# choose minimum from
# result for (i-1) chars and time for
# insertion and
# result for half of chars and time for
# copy and one extra character deletion
else:
dp[i] = min(dp[i - 1] + insrt,
dp[(i + 1) // 2] +
cpy + remov)
return dp[N]
# Driver Code
if __name__ == "__main__":
N = 9
insrt = 1
remov = 2
cpy = 1
print(minTimeForWritingChars(N, insrt,
remov, cpy))
# This code is contributed
# by vibhu4agarwalC#
// C# program to write characters in
// minimum time by inserting, removing
// and copying operation
using System;
class GFG
{
// method returns minimum time to write
// 'N' characters
static int minTimeForWritingChars(int N, int insert,
int remove, int copy)
{
if (N == 0)
return 0;
if (N == 1)
return insert;
// declare dp array and initialize with zero
int[] dp = new int[N + 1];
for(int i = 0; i < N + 1; i++)
dp[i] = 0;
// first char will always take insertion time
dp[1] = insert;
// loop for 'N' number of times
for (int i = 2; i <= N; i++)
{
/* if current char count is even then
choose minimum from result for (i-1)
chars and time for insertion and
result for half of chars and time
for copy */
if (i % 2 == 0)
dp[i] = Math.Min(dp[i - 1] + insert,
dp[i / 2] + copy);
/* if current char count is odd then
choose minimum from
result for (i-1) chars and time for
insertion and
result for half of chars and time for
copy and one extra character deletion*/
else
dp[i] = Math.Min(dp[i - 1] + insert,
dp[(i + 1) / 2] + copy + remove);
}
return dp[N];
}
// Driver code
static void Main()
{
int N = 9;
int insert = 1, remove = 2, copy = 1;
Console.Write(minTimeForWritingChars(N, insert,
remove, copy));
}
}
//This code is contributed by DrRoot_输出
5时间复杂度：O(N)
辅助空间： O(N)。