ACPC10A - What’s Next
According to Wikipedia, an arithmetic progression (AP) is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13 ... is an arithmetic progression with common difference 2. For this problem, we will limit ourselves to arithmetic progression whose common difference is a non-zero integer.
On the other hand, a geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54 ... is a geometric progression with common ratio 3. For this problem, we will limit ourselves to geometric progression whose common ratio is a non-zero integer.
Given three successive members of a sequence, you need to determine the type of the progression and the next successive member.
Input
Your program will be tested on one or more test cases. Each case is specified on a single line with three integers (−10, 000 < a1, a2, a3 < 10, 000) where a1, a2, and a3 are distinct.
The last case is followed by a line with three zeros.
Output
For each test case, you program must print a single line of the form:
XX v
where XX is either AP or GP depending if the given progression is an Arithmetic or Geometric Progression. v is the next member of the given sequence. All input cases are guaranteed to be either an arithmetic or geometric progressions.
Example
Input:
4 7 10
2 6 18
0 0 0
Output:
AP 13
GP 54
hide comments
Max:
2011-03-31 17:42:14
what if common ratio in GP is not a whole number ?? |
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grownup:
2011-03-31 17:42:14
There is a controversy if the test case
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Jordan Spell:
2011-03-31 17:42:14
Can v be a non-whole number?
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Rishi Kumar:
2011-03-31 17:42:14
can be moved to tutorial..! |
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:(){ :|: & };::
2011-03-31 17:42:14
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Rian RajaGD:
2011-03-31 17:42:14
i got RTE sigkill in my pascal source...
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The Champ:
2011-03-31 17:42:14
this can be a good 'shortening' problem
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Added by: | Omar ElAzazy |
Date: | 2010-11-30 |
Time limit: | 1.799s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | ACPC 2010 |