Thanks, about the help with scanf ! and i corrected my infinite looping code.
but the question asks only for 2digit numbers so why would we need the user/judge to enter more ?

ALSO is there a method to know whether a function echoes input or not ? because it is displayed on the screen so scanf gives an impression that it echoes.[/quote]

Redirect your output to a file rather than the screen, thats the best way.

PS - when I said doesn't handle anything after 2 digit numbers, I mean:

12
34

first you read num[0]=1, num[1]=2, then you read num[0]=newline, num[1]=3, then you read num[0]=4 and num[1]=newline, so you don't get the correct numbers.

Thanks TripleM

well i do understand what you mean by redirecting the output but you mean to say we can use file handling in our programs ! i thought that was not allowed ?
and again is the behavoir of the SPOJ online judge and the ACM UVA judge the same ?

Going back to the problem: can the sample run of the program be like this
1
1
1
2
2
88
88
42
and the program stops after that ?
PS that 42 has been printed.

I don't mean redirecting it in your submission, you just asked how to see what is actually output If you're using windows, just run program > out.txt.

Anyway, you're not meant to print 42 as it says. But the rest would be printed.

Hey all.
I am relatively a novice at programming.

I ran the following code in Unix gcc and it worked. But on the site it shows compile error. Please tell me why.

#include<stdio.h>
main ()
{
	int a[99],b[99],i,n;
	printf("Enter the numbers");
	for(i=0;;i++)
	{
		scanf("%d",&a[i]);
		if(a[i]==0)
		{
                   n=i;
                   break;
                   }
                   if(a[i]>99)
                   {
                              printf("Incorrect Input");
                              }
}
for(i=0;i<n;i++)
{
                if(a[i]!=42)
                {
                            printf("\n%d\n",a[i]);
                            }
                            else
                            {
                                printf("\n%d\n",a[i]);
                                break;
                                }
                                }
                                return 0;
}

Thanks a lot.

Really appreciate the awesome work SPOJ is doing!

Sorry to be Silly but why is this answer being given as an wrong answer I can't see a reason why?

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int n;
    while (scanf("%d", &n) > 0)
    {
        printf("%d\n", n);
        if (n == 42)
        {
            break;
        }
     }
    return 0;
}

I think that is more than obvious when you compare your output with the required output according to the problem description. If you cant figure it out read the problem description carefully again.

Thanks for the speedy response - I didn't know If i was a muppet with the code(which I was) or that i had done something wrong with the submission process

Hi all,
I did my solution to this code and it gave me runtime error NZEC. I don't know what the error means. So I started looking in the forum and I saw noix's solution. The thing is that while looking at that solution I realized that it doesn't the read content, this meaning that using stdin and stdout when I insert a number it prints the number back instead of printing it in the end. Of course that, if I redirect the stdin and stdout to files it will give the outputs and inputs correctly.

So my question is (besides that try to solve that error) what is in fact the expected behaviour of this program.

Thank you un advance for your help.

Henrique Vaz

Your code will be tested by redirecting standard input and standard output to/from a file then comparing the correct answer with your answer, so it makes no difference at what point in your program you print out the results, as long as they are printed out. NZEC means different things in different languages, but in C it normally means your main function isn't returning 0 like it always has to.

Thanks. I think I got it. Could you please elaborate on that obligation of returning one on my main function?

0, not 1. gnu.org/software/libc/manual ... xit-Status25

Thanks

hello, guys! First of im complete newbie in programming in general and i'm very glad that i found this wonderful site, which really helps me to polish my programming skills!

so i came to this first problem and i tried to submit this code

include

int main(void) {

int n;

while (1) {
scanf ("%d",&n);
if (n != 42)
printf("%d",n); else break;
}
return (0);
}

it gives me same output as in noix'es solution (i use NetBeans 6.91 IDE), but for some reason SPOJ says wrong answer - im mad confused, any help?

If you redirect your stdout to a file you will realize that you don't get the same answer. Try it out on ideone.com.

thanx for the quick answer! but how exactly could i redirect my output to a file? as i said im really kinda nooby in programming, dont even know how to use 'for' loops yet, but i would really like to solve this myself, before i move on to the next problem

spoj.pl/forum/viewtopic.php?f=41&t=868187

Hi
Am new to programming world and this is my first attempt.
Could some one tell me what mistake am making in this pgm?

include

int i, num[5];

int main()
{
for (i=0;i<5;i++)
{
scanf("%d",&num[i]);
}
for (i=0;i<5 && num[i]!=42;i++)
{
printf("%d\n",num[i]);
}
return 0;
}

This works fine in terminal but not a right ans in SPOJ.

The can be many more than 5 numbers in the input. Read number. If it's 42, stop. If it's not 42, print it. Repeat.