Proizvolov的身份

给定大小为N的两个数组A和B。数组A按升序排列，而B按降序排列。这两个数组都是1到2N的数字的子序列。任务是找到两个数组的绝对差之和。

Sum = |A₁ – B₁| + |A₂ – B₂| + |A₃ – B₃| + …. + |A_N – B_N|

例子：

Input : A[] = {1, 2, 3, 4, 5}, B[] = {10, 9, 8, 7, 6}
Output : 25

Input : A[] = {1, 5, 6, 8, 10, 12}, B[] = {11, 9, 7, 4, 3, 2}
Output : 36

朴素的方法：朴素的方法是运行一个循环并找到绝对差之和。

高效的方法： Proizvolov的恒等式是一个与正整数差之和有关的恒等式。它指出如果我们采用第一个2N个整数并将它们划分为N个数字的两个子集。

按升序排列一个子集： A ₁ 2 3 <…。 N

以降序排列另一个子集： B ₁ > B ₂ > B ₃ >…。 > B _N

然后，总和| A ₁ – B ₁ | + | A ₂ – B ₂ | + | A ₃ – B ₃ | +…。 + | A _N – B _N |总是等于N ²

下面是上述方法的实现：

C++

// CPP program to implement proizvolov's identity
#include
using namespace std;
  
// Function to implement proizvolov's identity
int proizvolov(int a[], int b[], int n)
{
    // According to proizvolov's identity
    return n*n;
}
  
// Driver code
int main()
{
    int a[] = {1, 5, 6, 8, 10}, b[] = {9, 7, 4, 3, 2};
      
    int n = sizeof(a) / sizeof(a[0]);
      
    // Function call
    cout << proizvolov(a, b, n);
      
    return 0;
}

Java

// Java program to implement proizvolov's identity
class GFG
{
    // Function to implement proizvolov's identity
    static int proizvolov(int a [], int b [], int n)
    {
        // According to proizvolov's identity
        return n * n;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int a [] = {1, 5, 6, 8, 10};
        int b [] = {9, 7, 4, 3, 2};
          
        int n = a.length;
          
        // Function call
        System.out.println(proizvolov(a, b, n));
    }
}
  
// This code is contributed by ihritik

Python3

# Python3 program to implement
# proizvolov's identity
  
# Function to implement 
# proizvolov's identity
def proizvolov(a, b, n): 
    return n * n
  
# Driver code
a = [ 1, 5, 6, 8, 10 ] 
b = [ 9, 7, 4, 3, 2 ] 
n = len(a) 
  
# Function call
print(proizvolov(a, b, n, )) 
  
# This code is contributed by nidhiva

C#

// C# program to implement proizvolov's identity
using System;
  
class GFG
{
    // Function to implement proizvolov's identity
    static int proizvolov(int [] a, 
                          int [] b, int n)
    {
        // According to proizvolov's identity
        return n * n;
    }
      
    // Driver code
    public static void Main () 
    {
        int [] a = {1, 5, 6, 8, 10};
        int [] b = {9, 7, 4, 3, 2};
          
        int n = a.Length;
          
        // Function call
        Console.WriteLine(proizvolov(a, b, n));
    }
}
  
// This code is contributed by ihritik

输出：

25