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
2016-12-14 06:57:33
very easy .....just check any case involving 0 ie 0 2 4
2016-12-12 11:00:10
Finally AC after 4 WA.
Try this test case :
-5 25 -125
Answer should be GP 625
2016-12-04 11:53:30
*No need to stress on data type all ints, costed me 4 WAs :(
2016-11-23 00:04:55
625 125 5 can,t be a test case because it,s common ratio is not an integer
2016-10-23 17:20:45
a,b,c mus be distinct,read the problem :) So,1,1,1 obviously not in the test case .
And corner case may be:
625 125 25
ANs:
GP 5
2016-10-03 13:52:11
can anybody give me some corner cases?
2016-09-29 08:56:50
Why 1 1 1 should be AP while clearly above it stated that difference should be a non zero integer, anyway i guess its not in test case.. :)

Last edit: 2016-09-29 09:04:57
2016-09-25 20:04:29 hacker_sk
enjoy :)

Last edit: 2016-09-25 20:07:33
2016-09-16 16:55:25
common ratio should be integer
2016-08-07 10:28:19
1 1 1 can never be case because all three numbers should be distinct.
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