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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.
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Sample Input |
Output for Sample Input |
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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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2021-04-05 15:34:14
Got AC ? "Yes" |
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2021-03-16 17:29:54
@abu_rifat ans is ok ,but what is the logic? |
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2020-07-22 16:15:13
Output it in cout<<"Case "<<i<<": Yes"<<'\n'; this format. :) |
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2020-06-16 19:04:34
those Case 1 2 stuff sucks, cost me WA, but m happy finally it got AC :) |
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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 |
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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 |
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2020-05-21 09:16:07
Hey print "Case 1" too otherwise you get wrong answer |
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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 |
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2020-05-01 13:28:13
Find GCD(A,B), If(C%GCD) No. Else Yes. |
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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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