can the input have any no.of numbers

is same input can be reapted

#include<stdio.h>

int main()

{

while (1)

{

int i;

scanf("%d", &i);

if i == 42 break;

else

printf("%d/n",i);

}

}

oops.. didnt intent to post that. sorry :)

#include<stdio.h>

int main()

{

int i;

while (1)

{

scanf("%d", &i);

if (i!= 42)

{

printf("%d/n",i);

}

}

#include<stdio.h>
int main()
{
int num;
while(scanf("%d",&num)>0 && num!=42)
printf("%dn",num);
return 0;
}

My code is not uploading .. Run time eroor is coming .. its working very fine GCC compiler..

When stops the input?

IF 2 TIMES 42 REPEAT THEN WHATWE DO?

lol why choose an easy problem like life & the universe? How do you terminate input on TurboSort (Easy) ?

I'm entering my data into vectors but I don't know how to terminate it when the input is done. For example, I know max input is 1000000 integers, what do I do if the user wants to stop before that?

That problem tells you exactly how many numbers you should read before stopping, on the first line.

hi please tell me

what is the problem in

#include <stdio.h>

int main()

{

int i;

do{

scanf("%d",&i);

if(i<42)

{

printf("%dn",i);

}

}while(i<42);

}

Describe what you are doing in words. Then read the problem. Your error is obvious.

#include<stdio.h>

int main()

{

int i;

scanf("%d",&i);

if(i!=42)

{

printf("%d",i);

}

}

#include<stdio.h>

int main()

{

int i;

scanf("%d",&i);

do

{

printf("%d",&i);

}while(i!=42)

}

FAQ

@Gaurav

it straight forward that the number read should not be 42. we dont have to compare it with lesser than or greater than 42. This is what u r doing.

Also the horrible thing i found in your code is checking number in while() as well as inside while. Why cant you just check once, then continue loop and print it every time u are able to read coz of condn met.

like this

cin>>n;

while(n!=42)

{

cout<<"n"<<n;

cin>>n;

}

Its pretty simple. Isn't it????

continued from prev post...

logic is ok but

cin and cout wouldnt work. try stream reading using fread(), fgets()

#include<stdio.h>

void main()

{

int i;

scanf("%d",&i);

for(i=1;i<42;i++)

printf("%d",i);

getch();

}

why we can not use conio.h

Because it is not a standard header file.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Life_the_Universe_and_Everything {
public static void main(String str []) throws NumberFormatException, IOException
{     System.out.println("Enter no of digit u wann enter ");
BufferedReader ob1= new BufferedReader(new InputStreamReader(System.in));
int user_choice= Integer.parseInt(ob1.readLine());
// Enter number in array

int num_hold[]=new int[user_choice];
System.out.println("Entre "+user_choice+" numbers ");
for(int i=0;i<num_hold.length;i++)
{
BufferedReader ob2= new BufferedReader(new InputStreamReader(System.in));
int num= Integer.parseInt(ob2.readLine());
num_hold[i]=num;
}



System.out.println("------------------------");         
for(int m=0;m<num_hold.length;m++)
{
int temp=num_hold[m];

if(temp<=88)
{   System.out.println(temp);
}
else{
break; 
}

}




}
}

whats wrong with this?

I am new here, can you please tell where I am going wrong :-

"

The problem statement:

Your program is to use the brute-force approach in order to find the Answer to Life, the Universe, and Everything. More precisely... rewrite small numbers from input to output. Stop processing input after reading in the number 42. All numbers at input are integers of one or two digits."

i think the given example is wrong

No it isn't.

please buddy tell me how this is working... how the input is being given.... i jus cant understand plz tell me....

FAQ

Hi can someone tell how to take input with Python?

ALSO TELL ME TOOO THAT HOW TO TAKE INPUT IN PYTHON AND PHP ???????????

dont we need an array for storng?

Hello, I am new in this website.Where is the problem at this code.I compiled it and run it in my Dev C++ 4.9.9.2.It worked but Codechef is showing it is wrong.I am getting upset to see that...You know this is a simple problem...:(

//Solution of Life, the Universe, and Everything

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

using namespace std;

int main()

{

system("cls");

int n,*a=NULL,i=0;

cout<<"How many inputs do you have?"<<endl;

cin>>n;

a=new int[n];

cout<<"Input:"<<endl;

do

{

cin>>a[i];

i++;

}while(i<n);

cout<<"Output:"<<endl;

i=0;

while(i<n && a[i]!=42)

{

cout<<a[i]<<endl;

i++;

}

delete []a;

a=NULL;

system("pause");

return 0;

}

See the FAQ.

I want to know that.. how is it supposed to work? like.. first user enter some inputs serially and then the outputs are shown ...if that is so and we cannot ask the user "enter input"..then how do I know when to stop taking input and show output??p

or is it like...enter one input, process it if it is not 42.. ?

which one ??

I agree with Israt.I am also confused how do the we make the compiler stop taking input and start to process????!!!!

There is also a problem that I saw successful submissions:

How could it be a successful submissions.It input output shouldn't match with the I/O given there...this simple program makes me crazy and stop me to go ahead...:(((

CodeChef submission 1526 (C) plaintext list. Status: AC, problem TEST, contest . By crashbird (Elijah), 2009-02-23 18:06:46.
#include<stdio.h>
int main()
{
 int t;
 while(1)
 {
   scanf("%d",&t);
   if(t==42)break;
   printf("%dn",t);
 }
 return 0;
}

That program is perfectly correct. Why do you think it gives the wrong output?

plz help.....

my codes are:

#include<stdio.h>
int main()
{
int a[5],i;
for (i=0;i<5;i++)
{
scanf("%d",&a[i]);
if (a[i]!=42)
{

printf("%dn",a[i]);
}
else
{
break;
}
}
return 0;
}

i am getting correct output.

but here when i submit it shows wrong answer.

why..??

Your code will give the wrong answer if there are more than 5 numbers in the input.

#include<stdio.h>
int main()
{
int c,a[10000],i;
for(i=0;i<10000;i++)
{
scanf("%d",&c);
if(c>=0 && c<=99)
a[i]=c;
else
i--;
}
i=0;
while(a[i]!=42&&i<20000)
{
printf("%dn",a[i]);
i++;
}
return 0;
}
Is this code optimal?

There is nothing 'ill organised' about the problem at all.

All of the sample codes I have read allow for 99 as an input; why do you think they don't?

The problem statement doesn't say you can't enter 42 twice, so of course you aren't meant to assume you can't.

The sample input shows numbers appearing after 42 so it is clear that the input doesn't have to stop there.

As for 'mixing' input and output, I don't think you understand how output works. The input does not appear in the output stream whatsoever, and storing all numbers in a collection is a bad idea. Read the FAQ which is very clear about that.

i hav a code which runs otherwise but give a run-time error after submission

the code is in python

def f():

x=raw_input("Enter a number")

l1=[]

l1.append(x)

while(x!=""):

x1=int(x)

l1.append(x1)

x=raw_input("Enter a number")

l=len(l1)

i=1

while(i<l):

if l1[i]!=42:

print l1[i]

else:

break

i=i+1

f()

I m New 2 codechef

JAVA error ===

Can any1 specify more on NZEC runtime error . i read the details given on the site about this error .

So, i placed the code ::::

System.exit(0);

and i also did handled the error with try , catch but the code still gives the error while uploading .

this code is compiling fine on my end .

i am really stuck with this error with past 3 days .!!!

I'm also using HashSet ,  TreeSet stuctures . Do i need to replace these data structures with ARRAY ...??

Help needed.,....

Is there any limit on the no. of inputs???

time limit exceeded error for following code......plz help me reduce time consumption

#include<stdio.h>

int main(void)

{

unsigned int num;

for(;printf("%d",scanf("%d",&num)!=42););

return(0);

}

I CAN NOT UNDERSTAND  WHAT SHOULD I DO OR NOT

#include<iostream>

using namespace std;

int main( ) {

int a[5] ;

for( int i  = 0; i <= 4; i++ ) {

cin>> a[i];

}

for( int j = 0; j <= 4; j++ ) {

cout<<endl;

if( a[j] == 4 )

break;

cout<< a[j]<<endl;

}

return 0;

}

Are you sure you are trying to solve the right problem? Your code doesn't seem to do anything that you are asked to do. For example, you are meant to keep reading until you read a 42, but you stop after 5 numbers, or stop after you read the number 4..

@Mohit U've compared the no wit 4 instead of 42!

WHAT IS MY PLOBLEM TELL ME PLESE

#include<stdio.h>
void main()

{
int i=0,a[100],j,d=0;

printf("n enter number(negetive to end.): ");

while(d>=0)
{ scanf("%d",&a[i]);
i++;
d=a[i];

}

printf("output: ");

for(j=0;j<i;j++)

{ if(a[j]==42)
break;
else
printf("%d ",a[j]);

}

}

#include <stdio.h>
#include <string.h>
int main()
{
int num[99];
int i,j,temp,a[99];
i=0;
while(scanf("%d",&num[i]))
{
if(num[i]!=42)
{
a[i]=num[i];
d=i;
i++;
}
else
{
break;
}
}
for(i=0;i<d;i++)
{
for(j=0;j<d;j++)
{
if(a[j]>a[i])
{
temp=a[j];
a[j]=a[i];
a[i]=temp;
}
}
}
return 0;
}

what wrong in this program Please anyone help me

while submitting it giving the error wrong answer

#include<iostream>

using namespace std;
int main()
{
int n;
for(int i=0; i<100;i++)
cin>>n;
for(int i=0; i<100;i++)
{
if(n==42)
break;
cout<<n<<"n";
}

return 0;
}

Smbdy plz tell me wts wrong in dis program...

Does the problem say to stop after reading 100 numbers?

#include<stdio.h>
int main()
{
int a;
while(1)
{
scanf("%d",&a);
if(a==42||a>=99)
break;
printf("%d",a);
}
return 0;
}

I want to submit it.... compile it using trubo c.....

then which language i select for it.

#include<iostream>
#include<conio.h>
int main()
{
int n;
for(int i=0; i<100;i++)
cin>>n;
for(int i=0; i<100;i++)
{
if(n==42)
break;
cout<<n<<"n";
}

return 0;
}

I'm new to this site and just testing how things work.

what does M represents, because while doing Life universe problem, the first one in easy, it takes 177M while an almost same solution in the successful submission has 0.0M ?

Also, the input and output shown in the example, has to be in that order or we can take input then show output, then again run loop and take input then output. Because, I submitted the solution as shown in the example, but other submitted in other order and still got successful submission?

1. #include<stdio.h>
2. #include<conio.h>
3. void main()
4. {
5. int x;
6. while(x!=42)
7. {
8. scanf("%d",&x);
9. printf("n");
10. printf("n%d",x);
11. }
12. getch();
13. }
    why the compiler did not accept this program in this site...

1. #include<stdio.h>
2. #include<conio.h>
3. void main()
4. {
5. int x;
6. while(x!=42)
7. {
8. scanf("%d",&x);
9. printf("n");
10. printf("n%d",x);
11. }
12. getch();
13. }
    why the compiler did not accept this program in this site...

#include<stdio.h>
#include<conio.h>
void main()
{
int n;
clrscr();
printf("PLEASE ENTER ANY INTEGER NUMBER OF ONE OR TWO DISITS ");
do
scanf("%d",&n);
while(n!=42);
printf("nn42 found!!!nnPLEASE PRESS ANY KEY TO EXIT");
getch();
}

Here, the constraint on the input we have is that any input number n will be such that (0 <= n < 100).

What shall i do if the number inputed is not within the range? Shall i exit the programme or ask user to input again?

If the problem statement tells you that n will always be between 0 and 100, then n will always be between 0 and 100.

thnks Stephen