ACFRAC - Another Continuous Fractions Problem

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The problem description is the same as the problem CFRAC and CFRAC2.

Input & Output

Multiple test cases, a single line with a single uppercase character C indicates the end of the input. The number of test cases will be less than 1000.

For each test case:

The first line of the input contains a single uppercase character A or B.A denotes that the input following character A and output format of this test case is the same as problem CFRAC, otherwise the input following character B and output format of this test case is the same as problem CFRAC2. But please pay attention that: the width and the height of the image after the character B will not appear in the input; the original fraction will not appear in the output of the test case of type A.

The example will make everything clear.

Example

Input:
A
75 34
B
..........1......
2.+.-------------
............1....
....4.+.---------
..............1..
........1.+.-----
................1
............5.+.-
................1
C

Output:
Case 1:
..........1......
2.+.-------------
............1....
....4.+.---------
..............1..
........1.+.-----
................1
............5.+.-
................1
Case 2:
75 34

hide comments
nadstratosfer: 2019-07-11 08:46:01

p !=q for all cases of type A. Still, p > q exists in testdata as opposed to CFRAC, but the statement doesn't specify what to output there. For cases like 30 10, print:
....1
2.+.-
....1

Don't complain about this being hard in C. A pythonist on SPOJ has to deal with idiotic time limits, malformed inputs, implicit casting of results to 32-bit integer etc. every day while you get "AC in 1 go!!!1" without ever noticing the issues. And don't even start with me about having to add "irrelevant code"; half of the C code I've ever written deals with stupid stuff a programmer should never have to worry about. Just learn some Python and experience the fun of handling a problem like this with it.

Bharath Reddy: 2014-04-04 17:01:32

If we solve CFRAC and CFRAC2 in Python, then this is very simple..
We can just combine the two

(Tjandra Satria Gunawan)(曾毅昆): 2013-01-08 09:18:13

@Santiago Palacio: I agree with you, the constraints is not clear and need big int to solve this... I think it's already hard to solve even with constraints p and q < 2^64... now I still can't solve this problem because p and q > 2^64 (overflow)...

Santiago Palacio: 2013-01-08 05:26:07

Why would you have to put those constraints? (p,q can be >= 2^64). That just limits some languages or creates the absolutely innesesary need of using code that isn't relevant to this problem... just my opinion.

Last edit: 2013-01-08 05:26:25
(Tjandra Satria Gunawan)(曾毅昆): 2012-06-02 19:57:27

There are 2 types of tricky test case on this problem:
1) p==q
2) p>2^64 and q>2^64
I'm getting WA, i can't solve this problem, because it's always overflow :(


Added by:Fudan University Problem Setters
Date:2007-11-01
Time limit:4s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: C99 ERL JS-RHINO
Resource:XX Colombian National Programming ACM 2006, test data by Blue Mary