计算所有子数组，它们的总和可以被拆分为两个整数的平方差

给定一个数组arr [] ，任务是对所有可以求和的子数组计数为两个整数的平方之差。
例子：

Input: arr[] = {1, 3, 5} 
Output: 6 
Explanation: 
There are six subarrays which can be formed from the array whose sum can be split as the difference of squares of two integers. They are: 
Sum of the subarray {1} = 1 = 1² – 0² 
Sum of the subarray {3} = 3 = 2² – 1² 
Sum of the subarray {5} = 5 = 3² – 2² 
Sum of the subarray {1, 3} = 4 = 2² – 0² 
Sum of the subarray {3, 5} = 8 = 3² – 1² 
Sum of the subarray {1, 3, 5} = 9 = 5² – 4²
Input: arr[] = {1, 2, 3, 4, 5} 
Output: 11 
 

方法：为了解决此问题，需要进行观察。除了可以以((4 * N)+ 2)形式表示的那些平方以外，任何数字都可以表示为两个平方的差，其中N是整数。因此，可以按照以下步骤计算答案：

• 遍历数组以找到给定数组的所有可能的子数组。
• 如果数字K可以表示为((4 * N)+ 2) ，其中N是某个整数，则K + 2始终是4的倍数。
• 因此，对于子数组K的每个和，检查(K + 2)是否为4的倍数。
• 如果是，则该特定值不能表示为平方差。

下面是上述方法的实现：

C++

// C++ program to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
#include 
using namespace std;
 
// Function to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
int Solve(int arr[], int n)
{
    int temp = 0, count = 0;
 
    // Loop to iterate over all the possible
    // subsequences of the array
    for (int i = 0; i < n; i++) {
        temp = 0;
        for (int j = i; j < n; j++) {
 
            // Finding the sum of all the
            // possible subsequences
            temp += arr[j];
 
            // Condition to check whether
            // the number can be split
            // as difference of squares
            if ((temp + 2) % 4 != 0)
                count++;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 };
    int N = sizeof(arr) / sizeof(int);
 
    cout << Solve(arr, N);
    return 0;
}

Java

// Java program to count all the
// non-contiguous subarrays whose
// sum can be split as the
// difference of the squares
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
static int Solve(int arr[], int n)
{
    int temp = 0, count = 0;
 
    // Loop to iterate over all the possible
    // subsequences of the array
    for(int i = 0; i < n; i++)
    {
       temp = 0;
       for(int j = i; j < n; j++)
       {
            
           // Finding the sum of all the
           // possible subsequences
           temp += arr[j];
            
           // Condition to check whether
           // the number can be split
           // as difference of squares
           if ((temp + 2) % 4 != 0)
               count++;
       }
    }
     
    return count;
}
 
// Driver Code
public static void main(String [] args)
{
    int arr[] = { 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 };
    int N = arr.length;
 
    System.out.println(Solve(arr, N));
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program to count all the non-contiguous
# subarrays whose sum can be split
# as the difference of the squares
 
# Function to count all the non-contiguous
# subarrays whose sum can be split
# as the difference of the squares
def Solve(arr, n):
 
    temp = 0; count = 0;
 
    # Loop to iterate over all the possible
    # subsequences of the array
    for i in range(0, n):
        temp = 0;
        for j in range(i, n):
 
            # Finding the sum of all the
            # possible subsequences
            temp = temp + arr[j];
 
            # Condition to check whether
            # the number can be split
            # as difference of squares
            if ((temp + 2) % 4 != 0):
                count += 1;
         
    return count;
 
# Driver code
arr = [ 1, 2, 3, 4, 5,
        6, 7, 8, 9, 10 ];
N = len(arr);
print(Solve(arr, N));
 
# This code is contributed by Code_Mech

C#

// C# program to count all the
// non-contiguous subarrays whose
// sum can be split as the
// difference of the squares
using System;
 
class GFG{
 
// Function to count all the
// non-contiguous subarrays whose
// sum can be split as the
// difference of the squares
static int Solve(int []arr, int n)
{
    int temp = 0, count = 0;
 
    // Loop to iterate over all
    // the possible subsequences
    // of the array
    for(int i = 0; i < n; i++)
    {
       temp = 0;
       for(int j = i; j < n; j++)
       {
            
          // Finding the sum of all the
          // possible subsequences
          temp += arr[j];
           
          // Condition to check whether
          // the number can be split
          // as difference of squares
          if ((temp + 2) % 4 != 0)
              count++;
       }
    }
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 };
    int N = arr.Length;
 
    Console.Write(Solve(arr, N));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

// Javascript program to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
 
// Function to count all the non-contiguous
// subarrays whose sum can be split
// as the difference of the squares
function Solve(arr, n)
{
    let temp = 0, count = 0;
 
    // Loop to iterate over all the possible
    // subsequences of the array
    for (let i = 0; i < n; i++) {
        temp = 0;
        for (let j = i; j < n; j++) {
 
            // Finding the sum of all the
            // possible subsequences
            temp += arr[j];
 
            // Condition to check whether
            // the number can be split
            // as difference of squares
            if ((temp + 2) % 4 != 0)
                count++;
        }
    }
 
    return count;
}
 
// Driver code
    let arr = [ 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 ];
    let N = arr.length;
    document.write(Solve(arr, N));
 
// This code is contributed by souravmahato348.

输出：

40

时间复杂度： O(N) ² ，其中N是数组的大小