ACPC10A - What’s Next

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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
Dushyant Singh: 2015-12-02 16:15:13

@jinesl - Use if ( b == ( a + c ) / 2.0 ) instead of if ( b == ( a + c ) / 2 )

jinesl: 2015-12-01 15:15:57

This thing is blowing me up... I can't understand why I'm getting wrong answer...

rameshkonatala: 2015-11-17 07:52:04

Should I convert next no. in GP into fraction and display it?

Sid: 2015-11-16 23:15:44

Do take a note of this line above "three integers (−10, 000 < a1 , a2 , a3 < 10, 000) where a1 , a2 , and a3 are distinct." This proves 1, 1, 1 will never be a test case.

One silly mistake that I was doing that gave me many WAs was the statement "while (a1 != 0 && a2 != 0 && a3 != 0)". Its only later that I realized this would negate legitimate test cases like -3 0 3, -8 -4 0... i.e. any case with 0 in it(Oops!)

HEMANTH SAVASERE: 2015-11-16 13:40:37

hey guys according to definition of AP common difference can be any constant 0 or any negative number
so 1 1 1 is an ap and also gp

suresh007: 2015-10-28 14:50:27

i think there is no need to check for 1,1,1 sequence as number are distinct(input section).

sarthak_8: 2015-10-17 13:56:31

should I put a condition to satisfy the constraints?

Last edit: 2015-10-17 13:58:02
cqui: 2015-10-16 10:46:12

a1 , a2 , and a3 are distinct 1 1 1 should not be input.
should we check for wrong entries?

johnnycash: 2015-10-07 22:25:49

check for silly mistake......its a simple one but costed me many WA because of that reason...
(c++)

harshmmodi: 2015-10-01 19:56:38

@ nonushikhar... 111 is a GP and not an AP since the given condition says that c.d. and c.ratio both are non-zero integer. Hence c.d. of AP cannot be Zero but c.ratio can be 1

Last edit: 2015-10-01 19:58:30

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