i'm having the same "wrong answer". I don't know what's wrong! i get the right output! D:

#include <iostream>
using namespace std;
int main() {
int n, n2, n3, n4, n5;
cin>>n;
cin>>n2;
cin>>n3;
cin>>n4;
	switch(n4==42)
	{
		break;
		case 1:
			cin>>n5;
			cout<<n<<endl;
			cout<<n2<<endl;
			cout<<n3<<endl;
		break;
}
}

The "sample input" is not the only test case that your code will be tested on.

Some other test cases:

42

1
2
3
4
4
5
5
6
7
8
9
42
12
12
13

Here's a good description / explanation on how to solve this problem Life universe and everything problem in C++, C, and Java:

swageroo.com/wordpress/solvi ... verything/138

include

include

void main()
{ int n;
while(1)
{ cin>>n;
if(n==42)
break;
cout<<"\n"< }
getch();
}

this is the code which i wrote but it shows a compilation error.can someone please tell me what is wrong with this code?

Conio is not in the c standard library.

If you click on compilation error it should give you more info. cin does not exist in the namespaces that you are using. Either use std::cin or at the top, after your includes,put:

using namespace std;

Please use code tags when posting code.

#include<iostream>
CODE: SELECT ALL
using namespace std;
int main()
{ int n;
while(1)
{ cin>>n;
if(n==42)
break;
cout<<"\n"<<n<<"\n";
}
return 0;
}

these are the errors it is showing after the correction
prog.cpp:2: error: function definition does not declare parameters
prog.cpp: In function 'int main()':
prog.cpp:7: error: 'cin' was not declared in this scope
prog.cpp:10: error: 'cout' was not declared in this scope

CODE: SELECT ALL

int number;

while(true)
{
cin >> number;
if(number == 40)
break;
cout << number;
}

}
dainik bhasker5

sir i am now working in dev c++ 4.9.9.2 and in which language i should summit my solution there is no option for this language

which compiler i should choose here i know coding in c++4.9.9.2

gcc.gnu.org/26

include

include

int main()
{
int s;
for(;
{
cin>>s;
if (s!=42)
cout<else
break;
}
return 0;
}

The SPOJ judge shows compilation error and I'm not able to figure it out.

If you click on where it says compilation error it will show what the compilation error is.

include

using namespace std;
int main()
{
int i,n,a[20];
cout<<"How Many Numbers Do u want to use as input :- ";
cin>>n;cout< for( i=0;i {
cin>>a[i];
}
cout<<"\nHere your output:- ";
for( i=0;i {
if(a[i]!=42)
cout<<" " < else if(a[i]==42)
break;
}
}
i hv written the prgrm using array 1st u hv to give all the no that u want to use as input & then it gives u the output as mentioned in the qstn, so where am i wrng??

include

int main(void)
{
int n,c=0,i;
int a[c];
printf(" input \n");
while(1)
{
scanf("%d",&n);
a[c]=n;
if(a[c]==42)
break;
c++;
}
printf("OUTPUT \n");
for(i=0;i printf("%d \n",a[i]);
return 0;

i ran this code but its giving wrong answer .please help me to correct this code.I am not getting my mistake
}

include

int main(void)
{
int i[5],k;

for(k=0;k<=4;k++)
{
scanf("%d",&i[k]);
}

for(k=0;k<=4;k++){
if(i[k]!=42)
{printf("%d\n",i[k]);}
else break;	}return 0;

}

Hi Folks,

For the users that have most recently posted in this topic. Welcome!

Everything that you output will be considered for the answer, so this will cause WA.

cout << "enter the numbers" << endl;

Please use code tags when posting code (see the nice box around the code above?)

I see all of you assuming that there will be a certain number of inputs. djs1234 assumes that it will be less than or equal to 20. This is not true. The problem states that you should continue to take input until it reads 42, it does not say if there will be 5, 100, 10000000 numbers in the input.

while not done
  read input
  if number = 42 then done
  else print number
end while

include

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

what the hell is wrong !!!
why it has not been accepted ... it is succesfully running in my compiler