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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.
Example
Sample Input |
Sample Output |
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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2018-06-08 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" |
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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. |