For loop and eliminating repetition - c++

Question

Geeting!
I'm working on a console application which read in numbers put them in an array and the output tells you those numbers position in the array where the number is larger then 1000.
so if you have an array like this :
int be[] = { 1001, 1001, 1001, 101, 1001 };
The output should tell you that from the first to the third position you have 1000+ numbers and from the fifth to the last position you have 1000+ numbers. So the outcome would be like this:
1 3 5 6
I solved it with for and if like this:
for (int i=1;i<=5;i++)
{
if (be[i]>=1000)
{
cout<<i<<" ";
for (int j=1;j<5-1;j++)
{
if (be[i+j]<=1000)
{
cout<<i+j-1<<" ";
break;
}
}
}
}
And the main problem is that the outcome will be this:
1 3 2 3 3 3 5 6
So after the break i want the first for loop to continue where the secont one ended but I don't exactly know how to do it. If there's anyone who got the basic idea from this vague explanation and can give me some hint or tips I would be most grateful!

Answer 1

You only need one loop and a bunch of checks.
Here's some pseudo code:
loop through elements
if number > 1000 and not started
print index
else if number <=1000
started = false
print index-1
if started = true
print index - 1 // last element of array

Answer 2

The only comment: you have no 6th position in your array, that is why I output 5 5 lik from 5th element to 5th element
http://ideone.com/k3gF03
int main(void) {
int be[] = { 1001, 1001, 1001, 101, 1001 };
int beSize = (sizeof(be)/sizeof(*be));
bool fl = false;
int i=0;
for (i;i<beSize;i++)
{
if (be[i]>=1000)
{
if (!fl) {
fl=true;
cout<<i+1<<" ";
}
} else {
if (fl) {
fl=false;
cout<<i<<" ";
}
}
}
if (fl) { cout<<i<<" "; }
return 0;
}

Answer 3

If I have understood you correctly you need something like the following
#include <iostream>
int main()
{
const size_t low_bound = 1000;
int be[] = { 1001, 1001, 1001, 101, 1001 };
const size_t N = sizeof( be ) / sizeof( *be );
for ( int x : be ) std::cout << x << ' ';
std::cout << std::endl;
for ( size_t i = 0; i < N; i++ )
{
if ( be[i] >= low_bound )
{
size_t j = i + 1;
while ( j < N && be[j] >= low_bound ) j++;
--j;
std::cout << i + 1 << "-" << j + 1 << " ";
i = j;
}
}
std::cout << std::endl;
}
The program output is
1001 1001 1001 101 1001
1-3 5-5
Take into account that array indices start at 0.

Answer 4

To have the first loop continue where the second ended just do that
int j = 1;
for (int i=j;i<=5;i++)
{
if (be[i]>=1000)
{
cout<<i<<" ";
for (;j<4 and be[i+j]>1000 ;j++);
if(j < 5 )
{
cout<<i+j-1<<" ";
}
}
}
have not really checked if it does what you want?

Answer 5

Very simple and not polite code. Sorry. Array in c++ begin with index 0. So there is a first dummy element.
int be[] = { 99, 1001, 1001, 1001, 101, 1001 }; // 99 is dummy
bool in=false;
int j;
int i;
for (i=1;i<=5;i++)
{
if (be[i]>=1000 && in==false)
{
in=true;
cout << i << " ";
}
if (be[i]<1000 && in==true)
{
in=false;
cout << i-1 << " ";
}
}
if (be[i]<1000 && in==true)
cout << i << " "; // or i-1 if you prefer the last one