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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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2017-04-29 15:16:02
there is only two cases case 1 and case 2 or more,and space between case and 1 or not. |
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2017-04-29 15:13:27
i didn't understood input/output format. |
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2017-04-11 18:41:20
very easy AC in one go...! |
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2017-03-31 16:36:47
Gcd is enough. |
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2017-02-18 19:33:57
space between : and Yes / No (*it cause me two WA) |
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2017-02-14 22:08:14
simple one for who knows diophantine equation... |
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2017-01-07 14:19:01
easy :) my 20th..got wa due to output format.Use Diophantine Equation...hope this helps:) |
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2016-12-02 15:43:08
Linear Diophantine Equations condition |
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2016-11-26 05:11:31
trivial |
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2016-11-25 20:15:01
You don't have to just print "Yes" and "No". Literally, you gotta print Case 1: ... Case 2: ... |