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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
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 |
hide comments
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2011-03-31 17:42:14 Max
what if common ratio in GP is not a whole number ?? |
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2011-03-31 17:42:14 grownup
There is a controversy if the test case 1 1 1 arrives because 1 1 1 can be an AP as well as a GP.And the other thing is the precision in case of a GP. |
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2011-03-31 17:42:14 Jordan Spell
Can v be a non-whole number? For example... 500 250 125....ratio (1/2) Would v be 62.5? 62 1/2? |
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2011-03-31 17:42:14 Rishi Kumar
can be moved to tutorial..! |
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2011-03-31 17:42:14 :(){ :|: & };:
This problem can be made way harder by including some random series and test for HP. |
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2011-03-31 17:42:14 Rian RajaGD
i got RTE sigkill in my pascal source... can you help me? how to solve this RTE problem? |
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2011-03-31 17:42:14 The Champ
this can be a good 'shortening' problem Re:(Debanjan)Very true,this better suited in either challenge or tutorial. Last edit: 2010-12-15 16:29:35 |