hello everyone i am fairly new to programming in c++ and i quite dont understand what the problem is asking you to do ?
also what is meant by brute force ?
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Aug '12last reply
Apr '21- 40
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Welcome!
spoj.pl/forum/viewtopic.php?f=41&t=22430
spoj.pl/forum/viewtopic.php?f=41&t=21244
Post back if you still have questions.
Not another new programmer falling into the C++ trap! There have been worse languages, but not many.
Advantages of C++:
Good libraries
Disadvantages of C++:
A type system that is too wimpy to express anything interesting, so that typical code is riddled with type casts, but that nevertheless has undecidable type checking. This is in stark contrast to O'Caml, Haskell, and SML. Even Java does this better.
No type inference at all, so all variable types must be declared.
Horrid syntax
To go with the horrid syntax, an ancient, primitive macro system that is 1) hard to understand 2) non-hygienic, and 3) incapable of complex computations. For a proper macro system, see Scheme's syntax-case macro system.
thnx for the replies guys and those links really helped i thought about it again and the answer has been accepted 
5 months later
whats the problem with this code:
include
int main()
{
int n;
while(scanf("%d", &n) && n!=42)
printf("%d\n", n);
printf("%d", n);
return 0;
}
1 year later
Hi, i am not able to find the error in my program. My code is as follows:
public class Life {
int inputArray[];
Life(int a[]){
inputArray=a;
}
void printResult(){
System.out.println("Started");
//System.out.println(inputArray[0]);
for(int i=0;i<=inputArray.length-1;i++){
if(inputArray[i]==42 &&(inputArray[i]>99)&&(inputArray[i]<-99))
break;
System.out.println(inputArray[i]);
}
System.out.println("FINISHED");
}
public static void main(String[] args) {
int c[]={1,3,4,5,1,2,3,7,8,9,0,1,2,3,42,4,5,6,7,8,4};
Life ob=new Life(c);
ob.printResult();
}
}Welcome! Please use code tags when posting code. What language is this? What error is the judge returning?
[quote="leppy"]Hello leppy... This is Java.. I am new to programming world and I don't know about code tags.. Can you figure out the issue? Thanks for your response.
Your code should look like this.
click on code tag and paste your code between them.
and on SPOJ you should not use public before class if you are using public you class name should be main.
Your algorithm is completely wrong, what problem is asking is you have to print all number until you encounter 42 in the input. Input can be any integer value it is not only between 1 and 100,and no need to print Started or finished in the output you have to print only those things which problem is asking for.
Hope this helps.
class Life
{
int inputArray[];
Life(int a[])
{
inputArray=a;
}
void printResult()
{
System.out.println("Started");
//System.out.println(inputArray[0]);
for(int i=0;i<=inputArray.length-1;i++)
{
if(inputArray[i]==42 &&(inputArray[i]>99)&&(inputArray[i]<-99))
break;
System.out.println(inputArray[i]);
}
System.out.println("FINISHED");
}
public static void main(String[] args)
{
int c[]={1,3,4,5,1,2,3,7,8,9,0,1,2,3,42,4,5,6,7,8,4};
Life ob=new Life(c);
ob.printResult();
}
}class Main {
int inputArray[];
Main(int a[]){
inputArray=a;
}
void printResult(){
for(int i=0;i<=inputArray.length-1;i++){
if(inputArray[i]==42)
break;
System.out.println(inputArray[i]);
}
}
public static void main(String[] args) {
int c[]={42,3,4,5,1,2,3,7,8,9,0,1,2,3,4,5,6,7,8,4};
Main ob=new Main(c);
ob.printResult();
}
}what is this "int c[]={42,3,4,5,1,2,3,7,8,9,0,1,2,3,4,5,6,7,8,4};"
your code prints nothing.
how you are reading the input? what are you doing with "c[]".
Read the problem statement once again.
Still you don't understand what problem is asking for.
answer my question.-->How you are reading the input?
Thanks for your help.. But,I think that they should be more worried about logic instead of the source of input...Anyways I will take care of all the things that you have told me and try to solve some good problems.
How come the judge understand your logic until you process the given input and print the desired output..
I don't understand what actually you are trying to do.
anyway..good luck. 
If you are looking for a maximum on the number of integers in the input, there is none. There could be 5, there could be one billion.
2 years later
Hi everyone, i don't understand what is wrong in my code... if someone can help me...
#include <stdio.h>
int main(){
int num;
do {
scanf("%d", &num);
printf("%d", num);
} while(num != 42);
return 0;
}1 month later
Hey guys,
I am particularly new to coding I tried with C while giving a solution to this problem but I'm always getting an SIGSEGV error
Pleas Help!!
Here is my code
#include <stdio.h>
int main() {
int a[20], x, count=0;
do{
scanf("%d\n", &x);
a[count]=x;
count++;
}while(x!=42);
x=0;
for(x; x<count-1; x++){
printf("%d\n", a[x]);
}
return 0;
}
Read the forum, there are lots of questions asking this. And answers 
If I'm allowed to quote myself:
9 months later
#include <stdio.h>
int main(void) {
// your code here
int a;
while(a!=42)
{
scanf("%d",&a);
if(a==42)
break;
printf("%d",a);
}
return 0;
}
Can u find problem in this code ?
is there any problem with this code.??it gives the output but still its showing error.
#include<stdio.h>
int main()
{
int i,n,j;
int ch[5];
scanf("%d",&n);
for(j=0;j<n;j++)
{
scanf("%d",&ch[j]);
}
for(i=0;i<n;i++)
{
if(ch[i]<ch[i+1])
printf("\n%d",ch[i]);
else
break;
}
if(ch[i]>ch[i+1])
printf("\n%d",ch[i]);
return 0;
}
8 months later
I Have corrected your problem in my code.
here you go.
#include <stdio.h>
int main(void)
{
// your code here
int n;
while(scanf("%d", &n)&&n<=99)
{
if (n==42)
break;
else
printf("%d\n", n);
}
return 0;
}
10 months later
#include
using namespace std;
int main(){
int batas;
cin >> batas;
int input[batas];
for(int i=0; i<batas; i++){
cin >> input[i];
}
for(int i=0; i<batas; i++){
if(input[i]==42){
exit(0);
} else {
cout << input[i] << endl;
}
}
cin.get();
return 0;
}
what’s the problem?
1 month later
hey guys! tried to submit my first problem but failed wonder what’s wrong with this one, any help?
int i,size=1,ar[100],ans;
for(i=1;ans!=42;)
{
if(ans==42)
break;
cin>>ans;
ar[i]=ans;
size++;
i++;
}
for(i=1;i<size;i++)
cout<<ar[i]<<"\n";2 months later
here is the simple solution in c++…
#include
using namespace std;
int main(){
int x ;
while( x !=42){
cin >> x;
if(x !=42){
cout << x << endl;
}
}
}
Hello atifasr,
A few remarks about your code:
What do you think happens to your program if the input contains more than 100 numbers ?
Your variables size and i are effectively the same thing. They’re both initialized to 1, and incremented at the same time.
The variable ans is used uninitialized on the first iteration of the loop.
You’re checking if ans == 42 at the start of a loop that will be entered only if ans != 42. That condition in your if can never be true.
You are THEN storing the input into ans, and not checking if it is now == 42. When 42 is encountered, it is going to be put in your array, supposing less than 100 numbers have been read so far.
I suggest you think about what you want to do exactly and what the constraints are, sort everything out so that it makes sense, and then start typing code.
5 months later
what is the problem in my code-:
#include<stdio.h>
#include<stdlib.h>
int main(){
int input, outputr[10],i=0,j=0;
while(i<=5){
scanf("%d",&input);
outputr[i]=input;
i++;
}
while(j<i){
if(outputr[j]==42)
{break;}
else{
printf("%d\n",outputr[j]);}
j++;
}
return 0;
}
3 months later
