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ENIGMATH - PLAY WITH MATH |
You would have been fed up with competitive programming questions so far, now it is time to solve little math.
Assume you have a equation A × x - B × y = 0
For a given value of A and B, find the minimum positive integer value of x and y that satisfies this equation.
Input
First line contains T, number of test cases 0 ≤ T ≤1000 followed by T lines.
First line of each test case contains two space separated integers A and B. 1 ≤ A, B ≤1 000 000 000.
Output
For each test case, output a single line containing two integers x and y (separated by a single space).
Example
Input: 1 2 3 Output: 3 2
Note:
- Brute force won't pass the given constraint.
- Negative number cases are avoided to make the problem easy.
Added by: | B.R.ARVIND |
Date: | 2013-09-12 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
hide comments
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2018-09-14 03:56:52
Cari KPK nya habistu , a sama b nya di bagi sama KPK nya |
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2018-05-30 15:53:35
its my 50th |
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2018-01-23 15:40:03
just find the lcm then divide lcm by a and b...x and y will be respective values....AC in one go... |
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2018-01-03 11:24:07
AC in one Go...!! |
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2017-12-09 18:50:13
dont read comments.... |
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2017-12-02 03:23:07
Print minimum values of (x,y) not value of x&y, and not minimum value of x followed by matching y. Print each result on a separate line, don't insert additional newlines like comments suggest. Crap statement and crap judge: if a < b print the bigger value first else WA. WTF! |
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2017-09-05 05:15:37
Easiest question... Don't forget to print new line. Costed me one WA |
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2017-08-20 16:39:14
i used a/=gcd(a,b); b/=gcd(a,b); n resulted 1 WA..:( |
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2017-07-30 03:45:42
AC in one go :) |
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2017-04-08 09:28:06
easy !! accepted in one go!!!use gcd |