Codeforces Round 768 (Div. 2) |
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Finished |
You are given two arrays $$$a$$$ and $$$b$$$ of $$$n$$$ positive integers each. You can apply the following operation to them any number of times:
Find the minimum possible value of $$$\max(a_1, a_2, \ldots, a_n) \cdot \max(b_1, b_2, \ldots, b_n)$$$ you can get after applying such operation any number of times (possibly zero).
The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 100$$$) — the number of test cases. Description of the test cases follows.
The first line of each test case contains an integer $$$n$$$ ($$$1\le n\le 100$$$) — the length of the arrays.
The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10\,000$$$) where $$$a_i$$$ is the $$$i$$$-th element of the array $$$a$$$.
The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \le b_i \le 10\,000$$$) where $$$b_i$$$ is the $$$i$$$-th element of the array $$$b$$$.
For each test case, print a single integer, the minimum possible value of $$$\max(a_1, a_2, \ldots, a_n) \cdot \max(b_1, b_2, \ldots, b_n)$$$ you can get after applying such operation any number of times.
361 2 6 5 1 23 4 3 2 2 533 3 33 3 321 22 1
18 9 2
In the first test, you can apply the operations at indices $$$2$$$ and $$$6$$$, then $$$a = [1, 4, 6, 5, 1, 5]$$$ and $$$b = [3, 2, 3, 2, 2, 2]$$$, $$$\max(1, 4, 6, 5, 1, 5) \cdot \max(3, 2, 3, 2, 2, 2) = 6 \cdot 3 = 18$$$.
In the second test, no matter how you apply the operations, $$$a = [3, 3, 3]$$$ and $$$b = [3, 3, 3]$$$ will always hold, so the answer is $$$\max(3, 3, 3) \cdot \max(3, 3, 3) = 3 \cdot 3 = 9$$$.
In the third test, you can apply the operation at index $$$1$$$, then $$$a = [2, 2]$$$, $$$b = [1, 1]$$$, so the answer is $$$\max(2, 2) \cdot \max(1, 1) = 2 \cdot 1 = 2$$$.
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