BCEASY - Bottom Coder (Easy)

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Some of you may be familiar with the TopCoder (TM) contest. Its exact rules are not important for this problem, but know that the most important part of it is writing a program according to the given specification. Many times the contestant ends up with a program which would work perfectly – if only he could change a couple of characters (like, replacing "=" by "==" in C, etc.). Unfortunately, even the best programmers sometimes aren't able to spot these tiny but necessary changes until it's too late... and that's why we developed a brand-new BottomCoder training for them!

The idea is very simple – you're given a problem specification, a source code, and a list of permitted modifications. Your task is to find a modification which would cause the program to behave according to the specification.

Specification: "Write a program which outputs EXACTLY 42 asterisks and NOTHING more (e.g. NO end-of-line markers, like "\n", ...)"

The code you are supposed to modify:

int i, n=42;
main() {
  for(i=0; i<n; i--) {
    printf("*");
  }
}

As this is a really, really simple problem, you are only permitted to make exactly ONE of these modifications to the source: 1) Add one character to the source. 2) Delete one character from the source. 3) Replace one character in the source by a different one.

Moreover, it would be definitely too easy if we asked you to find just one solution, so you'll need to find TWO DIFFERENT solutions in order to obtain credit for this problem. (There are exactly three different solutions, so don't worry, it can be done!)

Input

There is no input for given problem.

Output

Your submission should consist of two parts. The first part should contain the first of your solutions. A single line with the letter "Q" follows. (Note that the letter Q is used as a separator. You will have to do without inserting the letter Q in at least one of your solutions :) After this line you should add your second solution.

You don't need to worry much about the exact formatting of your submission. The exact judging procedure will look as follows:

The first occurrence of the letter Q is found, the input is split into two parts. Any whitespace in each of the parts is removed. It is checked whether the two submissions differ and whether each of them was obtained from the original program by an allowed change. Each of your two submissions is compared to each of the three correct solutions.

Example

Output:
int i, n=42; main(
){ for(i=0; i<n; i--)   { printf("?"); } }
Q
int i, n=41; main() { for(i=0; i<n;i--) { printf("*"); } }

(syntactically valid (but incorrect) submission)


hide comments
Vladimir Kirichenkoff: 2011-07-13 22:04:31

nice one

impulse: 2011-07-11 09:16:05

@Roman Sol , i really don't get why i keep getting WA, i've test my solution, please tell me, my ID is 5363658
LOL, it's a stupid mistake, AC!!!!

Last edit: 2011-07-11 09:18:29
Boyke Ferdinandes: 2011-06-15 07:19:11

Hello,

I found 2 solutions with rules number 1(Add) and 3(Modify).
I tried to find another (last) solution, is it according to the last rules (no 2 which is Delete) ?
Or is it still possible for me to use rules number 1 or 3 ?

Jargon: 2011-05-21 23:02:50

@arshak

You made two edits in the first one (you can't change the -- to a ++).

arshak minasyan: 2011-05-18 15:01:48

hey guys,
I have edited the code
int i, n=42; main(
){ for(i=0; i<n; i++) { printf("*"); } }
Q
int i, n=42; main() { for(i=n; i>0;i--) { printf("*"); } }


but it still WA, why can anyone tell me?

Ashutosh: 2011-04-17 07:17:58

will it not take more than two ..... many time i got WA ??
its bad!!

luis alberto carvajal martinez: 2011-03-17 05:44:50

i use the operator ∼ to one of the solution but it gave EXACTLY 43 "*" ...F**K jejeje there it goes my second solution

sudarshan: 2011-03-02 10:54:01

i think the line>> for(i=0;i<n;i--)should be changed as for(i=0;i<n;i++) or for(i=n;i>0;i--)

Azamat Ahmedov: 2011-02-15 05:58:44

I think we must replace "*" to any character.

Micha³ Rec³aw: 2010-10-23 12:40:50

(problem solved)

Last edit: 2010-10-25 09:21:21

Added by:Roman Sol
Date:2005-05-13
Time limit:0.100s
Source limit:10000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:TEXT
Resource:IPSC 2005