i have erased the cout <<"the output is:-" and cin>>n;
but still i m not understanding if the no of input integers have not been specified ... how the compiler will understand when he had to stop taking inputs
it is still showing wrong answer

The problem statement tells you to stop reading input after reading 42.

Please post your updated code.

while not done
   Read a number
   If number is 42 
      Done
   Else
      Print number

include

using namespace std;
main()
{
int a[10], n=5, c;
for(c=0;c {
cin >>a[c];
}
for(c=0;c {
if(a[c]!=42)
{
cout << a[c]<< '\n';
}
else
{
return 0;
}
};
return 0;
}

even if i made it to stop after reading 42 ... i'm unable to remove 'n' or the limit of for loop ... it needs an upper limit to be defined. for(c=0; ???;c++)

Look up the "while" loop instead of the "for" loop.

Or you could use:
for(scanf("%d",&c); c != 42; scanf("%d",&c))

While Delshire's code does work and gets AC, it has one interesting quirk.

while(a <= 100)

It is given in the problem statement that the numbers "are integers of one or two digits" so this works fine. It would be more acceptable to see:

while(true)

or, since it is given that there is always a 42 in the input statement:

while(a != 42)

This last segment also removes the need for the extra break, giving the loop a single exit point.

#include <iostream>
using namespace std;
int main() {
	int a = 0;
	while(a != 42) {
		cin >> a;
		if(a != 42)
			cout << a << endl;
	}
}

Hello! Thanks for the quick reply, i brought another piece of code, a little more 'formal' i think:

#include <iostream>
using namespace std;
void displayOutput(int myArray[], int qOfNumbers);
int main() 
{
int inputValues[100] = {};
for(int i = 0; inputValues[i] < 100; i++)
	{
		int holder;
		cin >> holder;
		inputValues[i] = holder; 
		if (inputValues[i] == 42){
		displayOutput(inputValues, i);
		break;
		}
    }
}
void displayOutput(int myArray[], int qOfNumbers)
{
	for(int a = 0; a < qOfNumbers; a++)
	{
		cout << myArray[a] << "\n";
	}
}

int inputValues[100] = {};

This is scary. There is nothing in the problem statement that tells you a specific constraint on the number of inputs in the test case. While it does work in this case, in a normal setting you would run the risk of buffer overrun.

So i should go with:

int inputValues[100] = {0};

Thanks for the reply.

No. You have declared an array of 100 elements. If there are 105 elements in the input then you have a serious problem.

Ok so i should limit the input entry so 'they' can not enter more than 99 values, right?

No. Nowhere in the problem statement does it declare what the maximum amount of numbers to be input is. Because there is no constraint you cannot safely declare an array.

If a constraint were given then it would not be your responsibility to enforce it.

Cool! Thanks!

#include<iostream>
using namespace std;
int main()
{
   int a[10],i;
   for(i=0;i<5;i++)
     cin>>a[i];
   for(i=0;i<5;i++)
   {
       if(a[i]==42)
        break;
      cout<<"\n"<<a[i];
   }
}

I dont know why my code is not working..its giving the input and output perfectly..Kindly help....

Przepraszam test 3 9 jest niepoprawny (Chodzi o to ze test 3 9 nie wystepuje.

Tak innego przykładu ale nawet patrząc na twój widać że np. ulica o numerze 2 łączy się z ulicami o numerach : 15, 5, 1, 16, 4, 3. Czyli aż sześcioma ulicami.

No ale co z tego wynika? Czy nie da się wyznaczyć trasy jakiej szuka Franek?
Uwierz, to zadanie nie bez powodu jest w kategorii "łatwe".

18 1
1 2
2 3
3 3
3 2
2 4
4 5
5 6
6 5
5 7
7 8
8 9
9 10
10 11
11 12
12 2
2 1
1 2
2 1

Odpowiedź brzmi NIE ??

#include <stdio.h>
int main()
{
    int p, t, k;
    long long x, y;
    bool test = true;
    scanf("%d %d", &t, &p);    
    long long int tab[t][2];
    for(int i =0; i<t; i++)
    {
                  scanf("%lld %lld", &x, &y);
                  tab[i][0]=x;
                  tab[i][1]=y;
                  if(test == true)
                  {
                  if(i == 0)
                  {
                   if(x != p)
                   {
                        test = false;                        
                   }
                  }
                  if(i != 0) 
                  {
                   if(x != k)
                    {
                       test = false;                  
                    }
                  } 
                  k=y;
                  }
    }   
    if( y != p)
      test = false;             
    if(test==true)
    {
    for(int i =0; i<t; i++)
    {
     for(int j=0; j<t; j++)
     {
             if(i!=j)
             {
                     if((tab[i][0] == tab[j][0]) && (tab[i][1] == tab[j][1]))
                     {
                     test=false;
                     break;
                     }
             }
     }
     if(test==false)
      break;
    }
    }
    if(test == true)
      printf("TAK");
    else
      printf("NIE");
    return 0;
}

Mam coś takiego. Nie wiem co może być jeszcze źle

Co do testu - odpowiedzią jest "TAK".
Co do programu - masz błędny algorytm.
Zastanów się nad czymś takim: Jeśli, którąś ulicą dojdzie się do danego skrzyżowania, to jakąś inną trzeba z niego wyjść. Czyż nie?
A może próbowałeś kiedyś rysować pewne figury "jednym pociągnięciem ołówka"? Które figury da się tak narysować?
A tak w ogóle, to przeczytaj cały ten wątek, bo znajdziesz tu dużą podpowiedź.