Codeforces Round 655 (Div. 2) |
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Finished |
In Omkar's last class of math, he learned about the least common multiple, or $$$LCM$$$. $$$LCM(a, b)$$$ is the smallest positive integer $$$x$$$ which is divisible by both $$$a$$$ and $$$b$$$.
Omkar, having a laudably curious mind, immediately thought of a problem involving the $$$LCM$$$ operation: given an integer $$$n$$$, find positive integers $$$a$$$ and $$$b$$$ such that $$$a + b = n$$$ and $$$LCM(a, b)$$$ is the minimum value possible.
Can you help Omkar solve his ludicrously challenging math problem?
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \leq t \leq 10$$$). Description of the test cases follows.
Each test case consists of a single integer $$$n$$$ ($$$2 \leq n \leq 10^{9}$$$).
For each test case, output two positive integers $$$a$$$ and $$$b$$$, such that $$$a + b = n$$$ and $$$LCM(a, b)$$$ is the minimum possible.
3 4 6 9
2 2 3 3 3 6
For the first test case, the numbers we can choose are $$$1, 3$$$ or $$$2, 2$$$. $$$LCM(1, 3) = 3$$$ and $$$LCM(2, 2) = 2$$$, so we output $$$2 \ 2$$$.
For the second test case, the numbers we can choose are $$$1, 5$$$, $$$2, 4$$$, or $$$3, 3$$$. $$$LCM(1, 5) = 5$$$, $$$LCM(2, 4) = 4$$$, and $$$LCM(3, 3) = 3$$$, so we output $$$3 \ 3$$$.
For the third test case, $$$LCM(3, 6) = 6$$$. It can be shown that there are no other pairs of numbers which sum to $$$9$$$ that have a lower $$$LCM$$$.
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