To form counter of N element

Hi,

Can anyone let me know how can i form a counter of say 30 elements so that to meet following requirement,

I need to access 5 elements out of 30 elements and out of these 5 elements once the one value is printed (i.e now access for remaining 4 element will be left) other one should get added immediately to this 5 element set, from remaining total 25 elements. In this way 5 element count will be there always.

I am trying this as follows, by forming vector of 30 elements accessing its 5 elements as,
for (iter = vector1.begin(); it < vector1.end();it++)
{
for(int count = 5; count >= 1 ; count--)
{
cout << "Print value<" <<count<<"> : "<< *it <<endl;
it++;
}
}

This displays 5 elements at a time but counter "count" waits for to finish off its remaining 4 elements however i want to maintain this counter to 5 elements always.

Appreciate any help in this.

Thanks.

1) Please use code tags when posting code.

2)

Code:

---

for (iter = vector1.begin(); it < vector1.end();it++)
{
  for(int count = 5; count >= 1 ; count--)
  {
      cout << "Print value<" <<count<<"> : "<< *it <<endl;
      it++;
  }
}

---

What is "it"? It isn't declared anywhere.

If you're typing your code into the message, don't do it. Copy and paste your code from your code editor into the message. I don't know if "it" is an iterator that was declared before and you're erroneously using it in the loop or if you made a typing error.

Quote:

---

I need to access 5 elements out of 30 elements and out of these 5 elements once the one value is printed (i.e now access for remaining 4 element will be left) other one should get added immediately to this 5 element set, from remaining total 25 elements. In this way 5 element count will be there always.

---

I barely understood any of your requirements. How do you determine which element to choose from the 30 elements? Every fifth one? A random element?

Maybe you should post an illustration using just 10 numbers, and what you're trying to do.

Regards,

Paul McKenzie