大小为K的子集，乘积等于两个完美平方的差

给定一个大小为N的整数数组arr []和一个整数K ，任务是计算数组大小为K的子集的数量，其元素乘积可以表示为² – b ² 。

例子：

Input: arr[] = {1, 2, 3} K = 2
Output: 2
Explaination:
All possible subsets of length 2 with their products are given below:
{1, 2} = 2
{2, 3} = 6
{1, 3} = 3
Since, only 3 can be expressed as (2² – 1², therefore only one such subset exists.

Input: arr[] = {2, 5, 6} K = 2
Output: 2
Explaination:
All possible contiguous sub-sequences with their products given below:
{2, 5} = 10
{2, 6} = 12
{5, 6} = 30
Since, only 12 can be expressed as (4² – 2²), only one such subset exists.

方法：

1. 生成大小为K的所有子集。
2. 计算所有子集的乘积。
3. 仅当数字为奇数或被4整除时，数字才能表示为两个数字的平方差。
4. 因此，用满足此条件的乘积计算所有子集。

下面是上述方法的实现：

Python3

# Python3 implementation of the approach
  
import itertools
  
# Function to return the
# Count of required sub-sequences
def count_seq(arr, n, k):
  
    # ans is Count variable
    ans = 0
  
    for seq in itertools.combinations(arr, k):
  
        # product of seq
        pro = 1 
      
        for ele in seq:
            pro *= ele
      
        # checking form of a2-b2
        if ((pro % 4) != 2): 
            ans += 1
    return ans
  
# Driver code
if __name__ == "__main__":
    arr = [2, 5, 6]
    n = len(arr)
    k = 2
    print(count_seq(arr, n, k))

输出：

1