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.
Example
Sample Input |
Sample Output |
2 |
Case 1: Yes |
Problem Setter: Md Abdul Alim, Dept. of Computer Science, Bangladesh University of Business & Technology
hide comments
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namitp:
2017-04-11 18:41:20
very easy AC in one go...! |
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lonelybanboo:
2017-03-31 16:36:47
Gcd is enough. |
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stark_attack:
2017-02-18 19:33:57
space between : and Yes / No (*it cause me two WA) |
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code96maker:
2017-02-14 22:08:14
simple one for who knows diophantine equation...
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ayushgupta1997:
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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satya_jha123:
2016-12-02 15:43:08
Linear Diophantine Equations condition |
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spartax:
2016-11-26 05:11:31
trivial |
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cnexans:
2016-11-25 20:15:01
You don't have to just print "Yes" and "No". Literally, you gotta print
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kushalanand:
2016-11-13 17:07:03
a=3 b=7 c=11 AC solution giving Yes. weird. |
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E Naveen Kumar:
2016-10-27 08:55:42
AC in a GO...!! |
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 |