js \n直接显示字符串_显示N个字符的最短时间

js \n直接显示字符串

Problem statement:

问题陈述：

You need to display N similar characters on a screen. You are allowed to do three types of operation each time.

您需要在屏幕上显示N个相似的字符。每次允许您执行三种类型的操作。

You can insert a character,
您可以插入一个字符，
You can delete the last character
您可以删除最后一个字符
You can copy and paste all displayed characters. After copy operation count of total written character will become twice.
您可以复制并粘贴所有显示的字符。复制操作后，总书写字符数将变为两倍。

All the operations are assigned different times.

所有操作均分配有不同的时间。

Time for insertion is X
插入时间为X
Time for deletion is Y
删除时间为Y
Time for copy, paste is Z
复制时间，粘贴为Z

You need to output minimum time to display N characters on the screen using these operations.

使用这些操作，您需要输出最短时间以在屏幕上显示N个字符 。

Input:

输入：

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer N denoting the number of same characters to write on the screen. The next line contains time for insertion, deletion, and copying respectively.

输入的第一行包含一个整数T，表示测试用例的数量。然后是T测试用例。每个测试用例都包含一个整数N，该整数N表示要在屏幕上写入的相同字符的数量。下一行分别包含插入，删除和复制的时间。

Output:

输出：

Print the minimum time to display N characters on the screen.

打印最短时间以在屏幕上显示N个字符。

Constraints:

限制条件：

 1 <= T <= 100
1 <= N <= 100
1 <= X, Y, Z <= N

Example:

例：

Input:
Test case: 2
First test case,
N, number of characters to be displayed = 9
Value of X, Y, Z respectively,
1 2 1
N, number of characters to be displayed = 10 
Value of X, Y, Z respectively,
2 5 4
Output:
minimum time for first test case: 5
minimum time for second test case: 14

Explanation:

说明：

For the first test case no of character to be displayed is: 9
Time for insertion is 1
Time for deletion is 2
Time for copy paste is 1
Say the similar character is 'a'
So the best way is
Insert two character
Time is 2
Copy paste
Total character so far is 4
Total time 3
Copy paste again
Total character so far is 8
Total time 4
Insert gain
Total character so far is 9
Total time 5
This is the most optimum way we can solve this

Solution Approach:

解决方法：

This is a recursive problem

这是一个递归问题

And we have a few possibilities,

我们有几种可能性，

Simply insert
只需插入
Copy-paste
复制粘贴
Delete
删除

Now we need to understand the optimal option based on the situation

现在我们需要根据情况了解最佳选择

Say a situation where we need to reach character 13 and we are at character 7 displayed already

假设有一种情况，我们需要到达第13个字符，并且已经显示了第7个字符

Also, Cost of Inserting 6 digits is much higher than one-time copy paste and deleting ( just consider this case, it may not happen always if copy paste option is costly)

另外，插入6位数字的成本比一次性复制粘贴和删除要高得多(仅考虑这种情况，如果复制粘贴选项成本很高，则可能不会总是发生)

So, the question is when the "delete" options is useful

因此，问题在于何时使用“删除”选项

It's useful for such case what I mentioned. So if we formulate the recursion

我提到的这种情况对这种情况很有用。所以如果我们制定递归

Say,

说，

N = number of characters to be displayed.

N =要显示的字符数。

So, if n is odd only then we need deletion if n is we even don't need that.

所以，如果n是奇数只有这样，我们需要删除，如果n是我们甚至不需要这样。

Now, we have our recursion, so we can write the recursive function. Since there will be many overlapping sub-problems, we can wither use top-down DP or bottom-up DP. I have used memoization in my implementation. As an exercise, you should be able to convert the above recursion to tabulation DP.

现在，我们有了递归，因此我们可以编写递归函数。由于将存在许多重叠的子问题，因此我们可以使用自顶向下的DP或自底向上的DP。我在实现中使用了记忆。作为练习，您应该能够将上述递归转换为表格DP。

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int dp[101];
int minimum(int a, int b)
{
return a > b ? b : a;
}
int my(int cur, int x, int y, int z)
{
if (cur == 1) {
return x;
}
// meoization, dont compute what's already computed
if (dp[cur] != -1)
return dp[cur];
if (cur % 2 == 0) { //for even n
dp[cur] = minimum(x + my(cur - 1, x, y, z), z + my(cur / 2, x, y, z));
}
else { //for odd n
dp[cur] = minimum(x + my(cur - 1, x, y, z), z + y + my((cur + 1) / 2, x, y, z));
}
return dp[cur];
}
int main()
{
int t, n, item;
cout << "enter number of testcase\n";
cin >> t;
for (int i = 0; i < t; i++) {
cout << "Enter number of characters\n";
cin >> n;
int x, y, z;
cout << "Insert time for insertion, deletion and copy respectively\n";
cin >> x >> y >> z;
for (int i = 0; i <= n; i++)
dp[i] = -1;
cout << "Minimum time is: " << my(n, x, y, z) << endl;
}
return 0;
}

Output:

输出：

enter number of testcase
3 
Enter number of characters
8 
Insert time for insertion, deletion and copy respectively 
3 2 5 
Minimum time is: 16 
Enter number of characters
8 
Insert time for insertion, deletion and copy respectively 
1 1 3 
Minimum time is: 7
Enter number of characters
3 
Insert time for insertion, deletion and copy respectively 
1 4 6 
Minimum time is: 3