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CEQU - Crucial Equation |
Let us see the following equation,
ax+by=c
Given three positive integers a, b and c. You have to determine whether there exists at least one solution for some integers value of x and y where x, y may be negative or non-negative integers.
For example if a=2, b=4 and c=8 then the equation will be 2x+4y=8, and hence, for x=2 and y=1, there exists a solution.
Let us see another example for a=3, b=6 and c=7, so the equation will become 3x+6y=7 and there exists no solution satisfying this equation.
Input
Input starts with an integer T (1<=T<=105) denoting the number of test cases. Each test case contains three integers a, b, and c. (1<=a, b, c<=106).
Output
For each test case of input print the case number and “Yes” if there exists at least one solution, print “No” otherwise.
Sample Input |
Output for Sample Input |
2 |
Case 1: Yes |
Problem Setter: Md Abdul Alim, Dept. of Computer Science, Bangladesh University of Business & Technology
Added by: | Alim |
Date: | 2014-10-15 |
Time limit: | 3s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 GOSU |
Resource: | Own Problem |
hide comments
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2018-06-24 13:28:37
the basic-most question of diophantine equation. |
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2018-06-20 19:46:39
follow the oupt frmt!! |
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2018-06-08 07:51:06
Take care of how to print the output! |
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2017-08-12 07:50:25
AC in one go!........ |
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2017-07-14 14:25:41
Bézout's identity :) |
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2017-07-11 07:51:49
AC in one go. Easy one |
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2017-07-02 10:21:11
My nth.. AC in one go :D |
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2017-06-29 14:23:19
how to print output |
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2017-05-29 21:22:50
AC in one go :) My 50th !! :D |
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2017-05-23 07:19:29
Don't forget to add space between : and "yes" or "no" |