将任何子数组的所有元素与X相乘后，最大化子数组总和

给定一个由N个整数组成的数组arr []和一个整数X。我们可以选择任何子数组，并将其所有元素乘以X。乘法后，找到具有最大和的子数组。任务是使子阵列相乘，以使最终子阵列总和最大化。

例子：

Input: arr[] = { -3, 8, -2, 1, -6 }, X = -1
Output: 15
Choose the sub-array with {-2, 1, -6} and multiply X.
Now the new array is {-3, 8, 2, -1, 6} whose maximum sub-array sum is 15.

Input: arr[] = {1, 2, 4, -10}, X = 2
Output: 14

方法：无法使用贪婪方法解决问题。贪婪方法在许多情况下失败，其中一种是a [] = {-3，8，-2，1，6}和x = -1 。我们将使用动态编程来解决上述问题。令dp [ind] [state] ，其中ind是数组中的当前索引，并且有3种状态。

• 第一种状态定义直到索引ind与X相乘才选择子数组。
• 第二种状态定义索引ind处的当前元素在子数组中，该子数组乘以X。
• 第三状态定义子阵列已经被选择。

Kadane的算法已与DP一起使用，以同时获得最大子阵列和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define N 5
  
// Function to return the maximum sum
int func(int idx, int cur, int a[], int dp[N][3], int n, int x)
{
  
    // Base case
    if (idx == n)
        return 0;
  
    // If already calculated
    if (dp[idx][cur] != -1)
        return dp[idx][cur];
  
    int ans = 0;
  
    // If no elements have been chosen
    if (cur == 0) {
  
        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = max(ans, a[idx] + func(idx + 1, 0, a, dp, n, x));
  
        // Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x));
    }
    else if (cur == 1) {
  
        // Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x));
  
        // End the sub-array multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x));
    }
    else
  
        // No more multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x));
  
    // Memoize and return the answer
    return dp[idx][cur] = ans;
}
  
// Function to get the maximum sum
int getMaximumSum(int a[], int n, int x)
{
  
    // Initialize dp with -1
    int dp[n][3];
    memset(dp, -1, sizeof dp);
  
    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
        maxi = max(maxi, func(i, 0, a, dp, n, x));
  
    return maxi;
}
  
// Driver code
int main()
{
    int a[] = { -3, 8, -2, 1, -6 };
    int n = sizeof(a) / sizeof(a[0]);
    int x = -1;
  
    cout << getMaximumSum(a, n, x);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
    static int N = 5;
  
// Function to return the maximum sum
static int func(int idx, int cur, int a[],
                int dp[][], int n, int x)
{
  
    // Base case
    if (idx == n)
    {
        return 0;
    }
  
    // If already calculated
    if (dp[idx][cur] != -1)
    {
        return dp[idx][cur];
    }
  
    int ans = 0;
  
    // If no elements have been chosen
    if (cur == 0)
    {
  
        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = Math.max(ans, a[idx] + 
                func(idx + 1, 0, a, dp, n, x));
  
        // Choose the sub-array and multiply x
        ans = Math.max(ans, x * a[idx] + 
                func(idx + 1, 1, a, dp, n, x));
    } 
    else if (cur == 1)
    {
  
        // Choose the sub-array and multiply x
        ans = Math.max(ans, x * a[idx] + 
                func(idx + 1, 1, a, dp, n, x));
  
        // End the sub-array multiplication
        ans = Math.max(ans, a[idx] + 
                func(idx + 1, 2, a, dp, n, x));
    } 
    else // No more multiplication
    {
        ans = Math.max(ans, a[idx] + 
                func(idx + 1, 2, a, dp, n, x));
    }
  
    // Memoize and return the answer
    return dp[idx][cur] = ans;
}
  
// Function to get the maximum sum
static int getMaximumSum(int a[], int n, int x)
{
  
    // Initialize dp with -1
    int dp[][] = new int[n][3];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            dp[i][j] = -1;
        }
    }
  
    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
    {
        maxi = Math.max(maxi, func(i, 0, a, dp, n, x));
    }
  
    return maxi;
}
  
// Driver code
public static void main(String[] args) 
{
    int a[] = {-3, 8, -2, 1, -6};
    int n = a.length;
    int x = -1;
    System.out.println(getMaximumSum(a, n, x));
  
}
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
  
N = 5
  
# Function to return the maximum sum
def func(idx, cur, a, dp, n, x) :
   
  
    # Base case
    if (idx == n) :
        return 0 
  
    # If already calculated
    if (dp[idx][cur] != -1):
        return dp[idx][cur] 
  
    ans = 0 
  
    # If no elements have been chosen
    if (cur == 0) :
  
        # Do not choose any element and use
        # Kadane's algorithm by taking max
        ans = max(ans, a[idx] + func(idx + 1, 0, a, dp, n, x)) 
  
        # Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x)) 
       
    elif (cur == 1) :
  
        # Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x)) 
  
        # End the sub-array multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x)) 
       
    else :
  
        # No more multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x)) 
  
    # Memoize and return the answer
    dp[idx][cur] = ans 
      
    return dp[idx][cur]
   
  
# Function to get the maximum sum
def getMaximumSum(a, n, x) :
   
  
    # Initialize dp with -1
    dp = [[-1 for i in range(3)] for j in range(n)]
      
  
    # Iterate from every position and find the
    # maximum sum which is possible
    maxi = 0 
    for i in range (0, n) :
        maxi = max(maxi, func(i, 0, a, dp, n, x)) 
  
    return maxi 
   
  
# Driver code
  
a =  [ -3, 8, -2, 1, -6 ]   
n = len(a) 
x = -1 
  
print(getMaximumSum(a, n, x)) 
  
# This code is contributed by ihritik

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
    static int N = 5;
  
// Function to return the maximum sum
static int func(int idx, int cur, int []a,
                int [,]dp, int n, int x)
{
  
    // Base case
    if (idx == n)
    {
        return 0;
    }
  
    // If already calculated
    if (dp[idx,cur] != -1)
    {
        return dp[idx,cur];
    }
  
    int ans = 0;
  
    // If no elements have been chosen
    if (cur == 0)
    {
  
        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = Math.Max(ans, a[idx] + 
                func(idx + 1, 0, a, dp, n, x));
  
        // Choose the sub-array and multiply x
        ans = Math.Max(ans, x * a[idx] + 
                func(idx + 1, 1, a, dp, n, x));
    } 
    else if (cur == 1)
    {
  
        // Choose the sub-array and multiply x
        ans = Math.Max(ans, x * a[idx] + 
                func(idx + 1, 1, a, dp, n, x));
  
        // End the sub-array multiplication
        ans = Math.Max(ans, a[idx] + 
                func(idx + 1, 2, a, dp, n, x));
    } 
    else // No more multiplication
    {
        ans = Math.Max(ans, a[idx] + 
                func(idx + 1, 2, a, dp, n, x));
    }
  
    // Memoize and return the answer
    return dp[idx,cur] = ans;
}
  
// Function to get the maximum sum
static int getMaximumSum(int []a, int n, int x)
{
  
    // Initialize dp with -1
    int [,]dp = new int[n,3];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            dp[i,j] = -1;
        }
    }
  
    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
    {
        maxi = Math.Max(maxi, func(i, 0, a, dp, n, x));
    }
  
    return maxi;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []a = {-3, 8, -2, 1, -6};
    int n = a.Length;
    int x = -1;
    Console.WriteLine(getMaximumSum(a, n, x));
  
}
}
  
/* This code contributed by PrinciRaj1992 */

输出：

15

时间复杂度： O(N * 3)
辅助空间： O(N * 3)