Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i], k <= 100

Solution: Brute Force

Time complexity: O(n²)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int countPairs(vector<int>& nums, int k) {
    const int n = nums.size();
    int ans = 0;
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j)
        ans += (nums[i] == nums[j]) && (i * j % k == 0);
    return ans;
  }
};

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