Guys... his question was very specific. He wasn't asking for info about the knapsack problem.

How can I get A[n] to increase like a binary counter so that it checks every possible combination of items??

There's two ways to go about this. First off, what you are looking for, is a counting method. I'll assume that A[n] is an array of ints. The following code performs the kind of incremental counting you are looking for.

Code:

#include <stdio.h>
void zeroarr (int * array, size_t size) {
int i;
for (i = 0; i < size; ++i) {
array[i] = 0;
}
return;
}
void printarr (int * array, size_t size) {
int i;
for (i = 0; i < size; ++i) {
printf ("%d", array[i]);
}
printf ("\n");
return;
}
// Return 1 if overflow
// This is what you want.
int incrarr (int * arr, size_t size) {
int i;
for (i = 0; i < size; ++i) {
if (arr[i] == 1) {
// 1 encountered. Set to 0 and increment next value
arr[i] = 0;
}
else {
// 0 encountered. Set to 1, and stop incrementing
arr[i] = 1;
break;
}
}
// Detect overflow
return i == size;
}
int main (void) {
int a[8];
zeroarr(a, 8);
for (;;) {
printarr(a, 8);
if (incrarr(a, 8))
break;
}
}

A much more elegant way to go about solving the knapsack problem would be to, instead of using a counting array, use a recursive function to perform the solution. There is a direct analogue between counting and basic tree recursion.