Codeforces Round 628 (Div. 2) |
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Finished |
You are given a positive integer $$$x$$$. Find any such $$$2$$$ positive integers $$$a$$$ and $$$b$$$ such that $$$GCD(a,b)+LCM(a,b)=x$$$.
As a reminder, $$$GCD(a,b)$$$ is the greatest integer that divides both $$$a$$$ and $$$b$$$. Similarly, $$$LCM(a,b)$$$ is the smallest integer such that both $$$a$$$ and $$$b$$$ divide it.
It's guaranteed that the solution always exists. If there are several such pairs $$$(a, b)$$$, you can output any of them.
The first line contains a single integer $$$t$$$ $$$(1 \le t \le 100)$$$ — the number of testcases.
Each testcase consists of one line containing a single integer, $$$x$$$ $$$(2 \le x \le 10^9)$$$.
For each testcase, output a pair of positive integers $$$a$$$ and $$$b$$$ ($$$1 \le a, b \le 10^9)$$$ such that $$$GCD(a,b)+LCM(a,b)=x$$$. It's guaranteed that the solution always exists. If there are several such pairs $$$(a, b)$$$, you can output any of them.
2 2 14
1 1 6 4
In the first testcase of the sample, $$$GCD(1,1)+LCM(1,1)=1+1=2$$$.
In the second testcase of the sample, $$$GCD(6,4)+LCM(6,4)=2+12=14$$$.
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