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?

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.

include

define SIZE 10

int main()
{
int a[SIZE],i,j;
printf("INPUT:\n");
for(i=0;i {
scanf("%d\n",&a[i]);
if(a[i]>99)
{
scanf("%d\n",&a[i]);
}
if(a[i]==42)
{
printf("OUTPUT:\n");
for(j=0;j {
printf("%d\n",a[j]);
}
}
}
return 0;
}

dis is my code:

include

int main()
{
int a[5];
int i;
printf("Enter 5 2digit numbers:\t");
for(i=0;i<5;i++)
{
scanf("%d",&a[i]);
}
printf("Output:\n");
for(i=0;a[i]!=42&&a[i]!=NULL&&i<5;i++)
{
printf("%d\n",a[i]);
}
return 0;
}

still i am getting wrong output as an answer...plz help

The judge considers everything that you output. Therefore you should not be prompting for input.

Also, read the problem statement. Where does it say that there will be exactly 5 numbers in the input?

can any geek tell me why the following code is giving a wrong answer.......

include

int main()
{
int x,k=0;

scanf("%d",&x);
while(x>0)
{
if( x==42)
k=1;
if(k==0 && x<100)
printf("%d\n",x);
scanf("%d",&x);
}
return 0;
}

Are you a geek?

Why do you stop after reading zero?

You are assuming that the file ends in '\n'. In reality you should not care how the file ends. You should ony care about finding "42". Use scanf.

i am running this code and it is getting me a run time error.. why ?> plz help me. i m new to programming

include

int main()
{
int num;
scanf("%d", &num);
while(num!=42)
{
printf("%d\n", num);
scanf("%d", &num);
}
}

Please use code tags when posting code.

You always need to return 0; from main in C.

Just thought I'd share mine since I haven't seen it in this topic yet.

#include <stdio.h>
int main(void) {
    char x[3];
    while(atoi(gets(x))!=42)puts(x);
    return 0;
}

Alternatively with fgets() and fputs():

#include <stdio.h>
int main(void) 
{
    char x[4];
    while (atoi(fgets(x,4,stdin))!=42)
    {
    	fputs(x,stdout);
    }
    return 0;
}

Do you understand why this solution is risky?