Easy Solution with detailed explanation (100 % accurate)
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is centered around efficiently finding the GCD values for a series of numbe
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is centered around efficiently finding the GCD values for a series of numbers based on divisors. The key insight is to leverage the relationship between divisors and their contributions to the GCD of pairs. Instead of calculating GCDs for every possible pair, which would be inefficient, we can focus on the divisors of the numbers involved.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Divisor Calculation**: First, we need to precompute the divisors for all numbers up to a certain limit (in this case, 50000). For each integer i, we will mark all its multiples as having i as a divisor.\n2. **Counting Divisors**: Next, for the list of numbers provided in nums, we will count how many times each divisor appears using a Counter. This helps in determining how many pairs of numbers can be formed with a particular divisor.\n3. **GCD Pair Contribution**: For each divisor, we will calculate how many pairs contribute to the GCD being that divisor. This involves:\n    1. Starting from the largest divisor down to the smallest.\n    2. For each divisor f, determine how many pairs can be formed and update the count of pairs for its multiples.\n4. **GCD Accumulation**: We will accumulate the counts of pairs for each GCD value in descending order to facilitate quick queries.\n5. **Answer Queries**: For each query, we can efficiently determine how many pairs have a GCD greater than the query value by leveraging a prefix sum or cumulative count of pairs.\n\n# Complexity\n- Time complexity: 0(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 0(n+k)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nN = 50001\ndivs = [[] for _ in range(N)]\nfor i in range(1, N):\n    for j in range(i, N, i):\n        divs[j].append(i)\nclass Solution:\n    def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:\n        c = Counter(chain.from_iterable(divs[ni] for ni in nums))\n        gcds = defaultdict(int)\n        for f in sorted(c.keys(), reverse=True):\n            nf = gcds[f] = gcds[f] + c[f] * (c[f] - 1) // 2\n            if not nf:\n                del gcds[f]\n                continue\n            for ff in divs[f]:\n                gcds[ff] -= nf\n            gcds[f] = nf\n        gcd_i = sorted(gcds.keys())\n        gcd_ac = list(accumulate(gcds[i] for i in gcd_i))\n        return [gcd_i[bisect_right(gcd_ac, q)] for q in queries]\n```
['Math', 'Python', 'Python3']