带有添加，删除和追加操作的表格N-copy字符串

由N个字母“ G”副本组成的字符串称为N复制字符串。即“ GGGG”是一个4副本字符串，因为它包含4个字母“ G”的副本。最初，我们有一个空字符串，可以对其执行以下三个操作：

1. 添加单个字母“ G”，成本为X。
2. 删除成本为X的单个字母“ G”。
3. 将到目前为止形成的字符串以Y的成本附加到其自身，即我们拥有字符串“ GG”，可以将其附加到其自身以制成“ GGGG”
   给定N，X和Y，可以找到使用上述操作生成N复制字符串的最低成本

例子：

Input : N = 4, X = 2, Y = 1 
Output : 4 
Explanation: 
1) Initially, the string it empty, we perform one Add operation to form string ‘G’. 
2) Next, we perform a operation of type 3, and form the string ‘GG’. 
3) Again we perform the operation 3 and form ‘GGGG’, which is the required N-copy string. 
Cost = X + Y + Y = 2 + 1 + 1 = 4
Input : N = 4, X = 1, Y = 3 
Output : 4 
We can perform 4 consecutive Add operations to form ‘GGGG’. 
Cost = X + X + X + X = 1 + 1 + 1 + 1 = 4 

可以使用动态编程方法解决该问题。
如果我们仔细分析这些操作，很容易看出，仅当存在多余的副本时才执行删除操作，只有在此删除操作之前涉及到类型3的操作时，才会出现这种情况，因为如果前面有一个类型为1的操作，那么它是没有意义的，因为连续的add和remove操作只会增加成本，而不会在我们的字符串引入任何新的副本。
因此，我们可以说我们有以下操作，
1)添加单个字符“ G”，成本为X
2)将字符串附加到其自身，然后删除单个字符“ G”，成本为Y + X
3)将字符串附加到其自身，成本为Y
注意：现在我们将参考这些修改后的操作
现在，可以使用O(n)DP方法解决它。

Let dp[i] represent the minimum cost to form a i-copy string 
then the state transitions can be:
1) If i is odd, 
Case 1: Add a character to the (i-1)-copy string, 
Case 2: Perform an operation of type 2 on the ((i+1)/2)-copy string
i.e dp[i] = min(dp[i – 1] + X, dp[(i + 1) / 2)] + Y + X) 
2) If i is even 
Case 1: Add a character to the (i-1)-copy string, 
Case 2: Perform an operation of type 3 on the (i/2)-copy string
i.e dp[i] = min(dp[i – 1] + X, dp[i / 2] + Y) 

C++

// CPP code to find minimum cost to form a
// N-copy string
#include 
 
using namespace std;
 
// Returns the minimum cost to form a n-copy string
// Here, x -> Cost to add/remove a single character 'G'
// and y-> cost to append the string to itself
int findMinimumCost(int n, int x, int y)
{
    int* dp = new int[n + 1];
 
    // Base Case: to form a 1-copy string we
    // need to perform an operation of type
    // 1(i.e Add)
    dp[1] = x;
 
    for (int i = 2; i <= n; i++) {
        if (i & 1) {
 
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 2 operation
            //        on ((i + 1) / 2)-copy string
            dp[i] = min(dp[i - 1] + x, dp[(i + 1) / 2] + y + x);
        }
        else {
 
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 3 operation on
            //        (i/2)-copy string
            dp[i] = min(dp[i - 1] + x, dp[i / 2] + y);
        }
    }
    return dp[n];
}
 
// Driver Code
int main()
{
    int n = 4, x = 2, y = 1;
    cout << findMinimumCost(n, x, y);
    return 0;
}

Java

// Java code to find minimum cost to form a
// N-copy string
class Solution
{
  
// Returns the minimum cost to form a n-copy string
// Here, x -> Cost to add/remove a single character 'G'
// and y-> cost to append the string to itself
static int findMinimumCost(int n, int x, int y)
{
    int dp[] = new int[n + 1];
  
    // Base Case: to form a 1-copy string we
    // need to perform an operation of type
    // 1(i.e Add)
    dp[1] = x;
  
    for (int i = 2; i <= n; i++) {
        if ((i & 1)!=0) {
  
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 2 operation
            //        on ((i + 1) / 2)-copy string
            dp[i] = Math.min(dp[i - 1] + x, dp[(i + 1) / 2] + y + x);
        }
        else {
  
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 3 operation on
            //        (i/2)-copy string
            dp[i] = Math.min(dp[i - 1] + x, dp[i / 2] + y);
        }
    }
    return dp[n];
}
  
// Driver Code
public static void main(String args[])
{
    int n = 4, x = 2, y = 1;
    System.out.println( findMinimumCost(n, x, y));
     
}
}
//contributed by Arnab Kundu

Python3

# Python3 code to find minimum cost to
# form a N-copy string
 
# function returns the minimum cost to
# form a n-copy string Here, x->Cost to
# add/remove a single character 'G' and
# y->cost to append the string to itself
 
def findMinimumCost(n, x, y):
    dp = [0 for i in range(n + 1)]
     
    # base case: ro form a 1-copy string
    #  we need tp perform an operation
    # of type 1(i,e Add)
    dp[1] = x
     
    for i in range(2, n + 1):
        if i & 1:
            # case1. Perform a Add operation
            #        on (i-1)copy string
            # case2. perform a type 2 operation
            #        on((i+1)/2)-copy string
            dp[i] = min(dp[i - 1] + x,
                        dp[(i + 1) // 2] + y + x)
        else:
 
            # case1. Perform a Add operation
            #        on (i-1)-copy string
            # case2. Perform a type # operation
            #        on (i/2)-copy string
            dp[i] = min(dp[i - 1] + x,
                        dp[i // 2] + y)
     
    return dp[n]
     
# Driver code
n, x, y = 4, 2, 1
 
print(findMinimumCost(n, x, y))
             
# This code is contributed
# by Mohit Kumar

C#

// C# code to find minimum cost to form a
// N-copy string
using System;
 
class GFG
{
 
// Returns the minimum cost to form a n-copy string
// Here, x -> Cost to add/remove a single character 'G'
// and y-> cost to append the string to itself
static int findMinimumCost(int n, int x, int y)
{
    int[] dp = new int[n + 1];
 
    // Base Case: to form a 1-copy string we
    // need to perform an operation of type
    // 1(i.e Add)
    dp[1] = x;
 
    for (int i = 2; i <= n; i++)
    {
        if ((i & 1)!=0)
        {
 
            // Case1. Perform a Add operation on
            //     (i-1)-copy string,
            // Case2. Perform a type 2 operation
            //     on ((i + 1) / 2)-copy string
            dp[i] = Math.Min(dp[i - 1] + x,
                             dp[(i + 1) / 2] + y + x);
        }
        else
        {
 
            // Case1. Perform a Add operation on
            //     (i-1)-copy string,
            // Case2. Perform a type 3 operation on
            //     (i/2)-copy string
            dp[i] = Math.Min(dp[i - 1] + x,
                             dp[i / 2] + y);
        }
    }
    return dp[n];
}
 
// Driver Code
public static void Main()
{
    int n = 4, x = 2, y = 1;
    Console.WriteLine(findMinimumCost(n, x, y));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP

 Cost to add/remove a
// single character 'G' and y-> cost to
// append the string to itself
function findMinimumCost($n, $x, $y)
{
    $dp[$n + 1] = array();
 
    // Base Case: to form a 1-copy string
    // we need to perform an operation of
    // type 1(i.e Add)
    $dp[1] = $x;
 
    for ($i = 2; $i <= $n; $i++)
    {
        if ($i & 1)
        {
 
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 2 operation
            //        on ((i + 1) / 2)-copy string
            $dp[$i] = min($dp[$i - 1] + $x,
                          $dp[($i + 1) / 2] + $y + $x);
        }
        else
        {
 
            // Case1. Perform a Add operation on
            //        (i-1)-copy string,
            // Case2. Perform a type 3 operation on
            //        (i/2)-copy string
            $dp[$i] = min($dp[$i - 1] + $x,
                          $dp[$i / 2] + $y);
        }
    }
    return $dp[$n];
}
 
// Driver Code
$n = 4;
$x = 2;
$y = 1;
echo findMinimumCost($n, $x, $y);
 
// This code is contributed by Sach_Code
?>

Javascript

 
    // Javascript code to find minimum cost to form a 
    // N-copy string
     
    // Returns the minimum cost to form a n-copy string
    // Here, x -> Cost to add/remove a single character 'G'
    // and y-> cost to append the string to itself
    function findMinimumCost(n, x, y)
    {
        let dp = new Array(n + 1);
 
        // Base Case: to form a 1-copy string we
        // need to perform an operation of type 
        // 1(i.e Add)
        dp[1] = x;
 
        for (let i = 2; i <= n; i++) {
            if ((i & 1)!=0) {
 
                // Case1. Perform a Add operation on 
                //        (i-1)-copy string,
                // Case2. Perform a type 2 operation 
                //        on ((i + 1) / 2)-copy string
                dp[i] = Math.min(dp[i - 1] + x, dp[parseInt((i + 1) / 2, 10)] + y + x);
            }
            else {
 
                // Case1. Perform a Add operation on 
                //        (i-1)-copy string,
                // Case2. Perform a type 3 operation on
                //        (i/2)-copy string
                dp[i] = Math.min(dp[i - 1] + x, dp[parseInt(i / 2, 10)] + y);
            }
        }
        return dp[n];
    }
     
    let n = 4, x = 2, y = 1;
    document.write(findMinimumCost(n, x, y));
 

输出：

4

时间复杂度O(n)
辅助空间O(n)