Two reasons:
1. You expect to get the number of inputs at the start of your program. That is correct for lots of SPOJ problems, but not here. Your program has to read an unknown number of inputs until it finds '42'.
2. Everything you print will be take as an output of your program. So the judge "judges" the line "enter the total numbers you want to enter", and that is not an expected result.

[quote="abhishek22, post:68, topic:47"]

scanf("%d",a[i]);

[/quote]

Probably this should read scanf("%d",&a[i]); ?
Did you read the post directly above of yours?

include

int main()
{
int n, mq[1001], ms[1001], current, i, j=0;

scanf("%d", &n);

while (n>0)
{
	for (i = 0; i<n; i++)
	scanf("%d",&mq[i]);
	current = 1;

	for (i = 0; i<n; i++){
		while(j>0 && ms[j]==current){
			current++;
			j--;
		}
		if (mq[i]==current){
			current++;
		}else{
			j++;
			ms[j] = mq[i];
		}
	}
	while(j>0 && ms[j]==current){
		current++;
		j--;
	}
	if (current == n+1){
		printf("%s","yes");
	}else{
		printf("%s","no");
	}

	scanf("%d", &n);
}
return 0;

}
why I receive RUN TIME ERROR? Please T_T

Your code appears to deal with some other problem, not TEST ?

This solution no make sense, the solution don't does what it need do.

The code from Noix do:

input
output
input
output
... until you input 42

The problem ask you do

input
input
....until you input 42

output
output
...when 42 stop output.

someone for the love of GOD can explain me what you want we do in this problem ?

it is correct and efficient to do it the way like noix did. (though other approaches can get accepted, too)

• read a value from input stream
• print that value to output stream

include

using namespace std;

int main(){
int arr[10],i, m = 42;
for (i = 0; i < 10;i++){
cin >> arr[i];

	if (arr[i] == m){
		break;
	}
	cout << arr[i] << " ";
}



return 0;
}

Hi, this is the code I put for resolution for this problem, but every time SPOJ says Time Limit Exceeded:

#include
int main(int i)
{
while(scanf("%d",&i)>0 && printf("%d\n",i) && i != 42 );
return 0;
}

Could you please help me on this ?

This was my first question on spoj and i am highly demotivated towards competitve programming.
this was my code

#include <iostream>
using namespace std;
int main() {
int a[10],i;
 for(i=0;i<10;i++)
 { cin>>a[i];
 }
for(i=0;i<10;i++)
 {
  if(a[i]!=42)
   cout<<a[i]<<endl;
   else
   break;
 }
	return 0;
}

It is running fine on codeblocks but is giving errors here on spoj.
Please someone help me.

I wonder why it tell me time limit exceed??

#include <stdio.h>
int main()
{
int i=0,number[10];
do
{
scanf("%d",&number[i]);
i++;
}
while(number[i-1]!=42);

for(int j=0;j<i-1;j++)
{
    printf("%d\n",number[j]);
}

return 0;

}

I tried making as short as possible for fun and I had this:

for (short x = 0; x != 42; scanf("%hd", &x), (x != 42)&& printf("%hd\n", x));
return 0;

But initially I had this:

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <limits.h>

#define ERROR_RET_NULL(x) fprintf(stderr, "ERROR: %s returned NULL in file '%s' at line: %i.", x, __FILE__, (__LINE__ - 2))


int 
main(void) 
{
        while (true)
        {
        	short temp = 0;

                char userInputString[5];
                char *endPtr;
                
                if (fgets(userInputString, 5, stdin) == NULL)
                {
                        ERROR_RET_NULL("fgets");
                        exit(EXIT_FAILURE);
                }
                else
                {
                        char *checkPtr = strchr(userInputString, '\n');
                        if (checkPtr) *checkPtr = '\0';
                        
                        long numberToCastToShort = strtol(userInputString, &endPtr, 10);
                        if (numberToCastToShort > SHRT_MAX || numberToCastToShort < SHRT_MIN)
                        {
                                /* Given the size restriction on the string this is basically deadcode */
                                fprintf(stderr, "ERROR: number too big");
                                exit(EXIT_FAILURE);
                        }
                        else if (numberToCastToShort == 42) break;
                        else temp = (short) numberToCastToShort;
                        
                }
                printf("%d\n", temp);
        }
        
        return 0;
}

I am not getting the output required… It ask for the number but then page closes after entering the number.

I followed the code of noix.

Can anyone tell me why am i getting an error for this code(life universe and everything)

#include <stdio.h>
#include <stdlib.h>

int main()
{
int i,n,t=3;
for(i=0;i<t;i++)
{
scanf("%d",&n);
if(n==42)
break;
else
continue;
}
return 0;
}

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

return 0;

}
I WROTE THIS AND THEY TOLD ME I’M WRONG, CORRECT ME PLEASE

Your code is very bad formatted or bad pasted or both.

#include <stdio.h>

int main (void) {
	int numb;
	do {
          scanf ("%d", &numb);
	      if (numb == 42) break;
	      printf ("%d\n", numb);
    } while (numb != 42); // or simple: while (1);

	return 0;
}

PS

Please don’t shout, QUIETER PLEASE!!!

My first try on SPOJ.
I can’t use argc+argv on SPOJ, can I?

#include <stdio.h>

int main(int argc, char** argv)
{
    int i = 0;
    for(; i < argc; ++i){
        if(argv[i][0] == '4' && argv[i][1] == '2')
            return 0;
        else
            printf("%s\n",argv[i]);
    }
    return 0;
}

No, You should read from standard input.
Example:
scanf("%i\n", &number);