给定大小为N的数组a 。任务是合并数组中的所有元素并找到最大可能值。如下所述，可以合并数组中的两个元素。
If i and j are two indexes of the array(i ≠ j). Merging jth element into ith element makes a[i] as a[i] – a[j] and remove a[j] from the array.
例子：
Input : a[] = {2 1 2 1} (n == 4)
Output : 4
Merge 3rd element into 2nd element then the array becomes {2, -1, 1}
Merge 3rd element into 2nd element then the array becomes {2, -2}
Merge 2rd element into 1nd element then the array becomes {4}
Input: a[] = {1, 3, 5, -2, -6}
Output: 17
Merge 4th element into 3rd element then the array becomes {1, 3, -7, -6}
Merge 2rd element into 3rd element then the array becomes {1, -10, -6}
Merge 2nd element into 1st element then the array becomes {11, -6}
Merge 2rd element into 1st element then the array becomes {17}
方法：
- 如果数组同时包含正负元素，则将数组中所有元素的绝对值相加
- 如果数组仅包含肯定元素。然后从所有其他元素的总和中减去最小元素
- 如果数组仅包含否定元素。首先，将所有元素替换为其绝对值。然后从所有其他元素的总和中减去最小元素
下面是上述方法的实现：
C++
// CPP program to maximum value after
// merging all elements in the array
#include
using namespace std;
// Function to maximum value after
// merging all elements in the array
int Max_sum(int a[], int n)
{
// To check if positive and negative
// elements present or not
int pos = 0, neg = 0;
for(int i = 0; i < n; i++)
{
// Check for positive integer
if(a[i] > 0)
pos = 1;
// Check for negative integer
else if(a[i] < 0)
neg = 1;
// If both positive and negative
// elements are present
if(pos == 1 and neg == 1)
break;
}
// To store maximum value possible
int sum = 0;
if(pos==1 and neg==1)
{
for(int i=0; i < n ; i++)
sum += abs(a[i]);
}
else if(pos == 1)
{
// To find minimum value
int mini = a[0];
sum = a[0];
for(int i=1; i < n; i++)
{
mini = min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2*mini;
}
else if(neg == 1)
{
// Replace with absolute values
for(int i = 0; i < n; i++)
a[i] = abs(a[i]);
// To find minimum value
int mini = a[0];
sum = a[0];
for(int i=1; i < n; i++)
{
mini = min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2*mini;
}
// Return the required sum
return sum;
}
// Driver code
int main()
{
int a[] = {1, 3, 5, -2, -6};
int n = sizeof(a) / sizeof(a[0]);
// Function call
cout << Max_sum(a, n);
return 0;
} Java
// Java program to maximum value after
// merging all elements in the array
import java.io.*;
class GFG
{
// Function to maximum value after
// merging all elements in the array
static int Max_sum(int a[], int n)
{
// To check if positive and negative
// elements present or not
int pos = 0, neg = 0;
for(int i = 0; i < n; i++)
{
// Check for positive integer
if(a[i] > 0)
pos = 1;
// Check for negative integer
else if(a[i] < 0)
neg = 1;
// If both positive and negative
// elements are present
if((pos == 1) && (neg == 1))
break;
}
// To store maximum value possible
int sum = 0;
if((pos == 1) && (neg == 1))
{
for(int i = 0; i < n ; i++)
sum += Math.abs(a[i]);
}
else if(pos == 1)
{
// To find minimum value
int mini = a[0];
sum = a[0];
for(int i = 1; i < n; i++)
{
mini = Math.min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2*mini;
}
else if(neg == 1)
{
// Replace with absolute values
for(int i = 0; i < n; i++)
a[i] = Math.abs(a[i]);
// To find minimum value
int mini = a[0];
sum = a[0];
for(int i = 1; i < n; i++)
{
mini = Math.min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2*mini;
}
// Return the required sum
return sum;
}
// Driver code
public static void main (String[] args)
{
int []a = {1, 3, 5, -2, -6};
int n = a.length;
// Function call
System.out.println (Max_sum(a, n));
}
}
// This code is contributed by ajit.Python3
# Python 3 program to maximum value after
# merging all elements in the array
# Function to maximum value after
# merging all elements in the array
def Max_sum(a, n):
# To check if positive and negative
# elements present or not
pos = 0
neg = 0
for i in range(n):
# Check for positive integer
if(a[i] > 0):
pos = 1
# Check for negative integer
elif(a[i] < 0):
neg = 1
# If both positive and negative
# elements are present
if(pos == 1 and neg == 1):
break
# To store maximum value possible
sum = 0
if(pos==1 and neg==1):
for i in range(n):
sum += abs(a[i])
elif(pos == 1):
# To find minimum value
mini = a[0]
sum = a[0]
for i in range(1,n,1):
mini = min(mini, a[i])
sum += a[i]
# Remove minimum element
sum -= 2*mini
elif(neg == 1):
# Replace with absolute values
for i in range(n):
a[i] = abs(a[i])
# To find minimum value
mini = a[0]
sum = a[0]
for i in range(1,n):
mini = min(mini, a[i])
sum += a[i]
# Remove minimum element
sum -= 2*mini
# Return the required sum
return sum
# Driver code
if __name__ == '__main__':
a = [1, 3, 5, -2, -6]
n = len(a)
# Function call
print(Max_sum(a, n))
# This code is contributed by
# Surendra_GangwarC#
// C# program to maximum value after
// merging all elements in the array
using System;
class GFG
{
// Function to maximum value after
// merging all elements in the array
static int Max_sum(int[] a, int n)
{
// To check if positive and negative
// elements present or not
int pos = 0, neg = 0;
for (int i = 0; i < n; i++)
{
// Check for positive integer
if (a[i] > 0)
pos = 1;
// Check for negative integer
else if (a[i] < 0)
neg = 1;
// If both positive and negative
// elements are present
if ((pos == 1) && (neg == 1))
break;
}
// To store maximum value possible
int sum = 0;
if ((pos == 1) && (neg == 1))
{
for (int i = 0; i < n; i++)
sum += Math.Abs(a[i]);
}
else if (pos == 1)
{
// To find minimum value
int mini = a[0];
sum = a[0];
for (int i = 1; i < n; i++)
{
mini = Math.Min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2 * mini;
}
else if (neg == 1)
{
// Replace with absolute values
for (int i = 0; i < n; i++)
a[i] = Math.Abs(a[i]);
// To find minimum value
int mini = a[0];
sum = a[0];
for (int i = 1; i < n; i++)
{
mini = Math.Min(mini, a[i]);
sum += a[i];
}
// Remove minimum element
sum -= 2 * mini;
}
// Return the required sum
return sum;
}
// Driver code
public static void Main(String[] args)
{
int[] a = { 1, 3, 5, -2, -6 };
int n = a.Length;
// Function call
Console.WriteLine(Max_sum(a, n));
}
}
// This code is contributed by
// sanjeev2552输出：
17