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
2 4 8
3 6 7

Case 1: Yes
Case 2: No

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
2021-04-05 15:34:14
Got AC ?
"Yes"
2021-03-16 17:29:54
@abu_rifat ans is ok ,but what is the logic?
2020-07-22 16:15:13
Output it in cout<<"Case "<<i<<": Yes"<<'\n'; this format. :)
2020-06-16 19:04:34
those Case 1 2 stuff sucks, cost me WA, but m happy finally it got AC :)
2020-06-09 16:21:31
very bad problem, got WA just because wrong display of output,no focus on concept but only focussing on visuals
2020-05-26 18:46:29
Printing the output is difficult compared to the question itself. Cost me a WA as I printed yes/no in uppercase :p
2020-05-21 09:16:07
Hey print "Case 1" too otherwise you get wrong answer
2020-05-16 06:56:43
Problem is easy, but we need to format the output as the 1st output shows "Case 1: Yes", 2nd output shows "Case 2: No" similarly 3rd output will show "Case 3: Yes/No" and so on ..

Last edit: 2020-05-16 06:57:45
2020-05-01 13:28:13
Find GCD(A,B), If(C%GCD) No. Else Yes.
2020-04-24 22:47:25
Use this format otherwise you will unnecessarily waste time thinking that there is some logical error
cout<<"Case "<<i<<":"<<" Yes/No"<<"\n";


Last edit: 2020-04-24 22:48:51
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