Custom iteration through array in C printing every nth value - c

Question

I'm currently working on a homework assignment and what I have to do is ask for the size of the array and what I would like to loop through it (how many things I should skip through). Example (I bolded what would be the users response)
How large is the array?
10
How would you like to iterate through it?
3
Conditions:
Then it should loop through every third item and then start over looping through the items that haven't been picked until the last one. I must also start on the first item (0) which you can see in my code that I provided.
This is what I thought I could do...
int amount=0, often=0, counter=0, x;
printf("Size of Array? ");
scanf("%d", &amount);
printf("iteration size? ");
scanf("%d", &often);
// sets the array (1 means it exists)
int[] array[amount];
for(x=0; x<amount; x++)
array[x] = 1;
// the part that I'm confused on
// supposed to iterate through everything
for(x=0; x<amount; x++){
if(array[0] == 1){
printf("#0\n");
array[0]=0;
}
if(array[x]==1)
counter++;
if(counter == often){
printf("#%d\n", x);
array[x]=0;
counter=0;
}
return 0;
}
However, I need to continuously loop through until the entire array is done except for the last one. Mine stops after I loop through the array once. This is the output I have verse the output I want.
My output:
0
3
6
9
Wanted output:
0
3
6
9
4
8
5
2
7
Notice how the the wanted output loops through it again picking the third number that hasn't been picked yet. That is where I'm confused and how to fix this problem. Any guidance or information would be great, thanks.

Answer 1

You need to keep looping until all numbers have been printed,
for instance like this:
int total = amount;
counter = often - 1;
while (total > 0) {
for(x=0; x<amount; x++){
if(array[x]==1)
counter++;
if(counter == often){
printf("#%d\n", x);
array[x]=0;
counter=0;
total--;
}
}
}
Edit: The entire program:
#include <stdio.h>
#include <stdlib.h>
int main (void) {
int amount=0, often=0, counter=0, x;
printf("Size of Array? ");
scanf("%d", &amount);
printf("iteration size? ");
scanf("%d", &often);
// sets the array (1 means it exists)
int array[amount]; // Fixed typo in declaration
for(x=0; x<amount; x++)
array[x] = 1;
// the part that I'm confused on
// supposed to iterate through everything
int total = amount;
counter = often - 1;
while (total > 0) {
for(x=0; x<amount; x++){
if(array[x]==1)
counter++;
if(counter == often){
printf("#%d\n", x);
array[x]=0;
counter=0;
total--;
}
}
}
return 0;
}
Compilation:
gcc -std=c99 -Wall a.c -o a
Running the program:
Size of Array? 10
iteration size? 3
#0
#3
#6
#9
#4
#8
#5
#2
#7
#1

Answer 2

Let's start out with a loop which skips every three numbers:
for(int i = 0; i < amount; i += 3)
{
print(i);
array[i] = 0;
}
Now, this only seems to work for 0, 3, 6, 9.
We want to be able to start at the first existing value, and repeat this process until nothing remains.
So let's put another loop on the outside, to count how many items remain:
int total = amount;
while(total > 0)
{
for(int i = 0; i < amount; i += 3)
{
print(i);
array[i] = 0;
total--;
}
}
Good so far, but we still start at 0 every time. We want to start at a particular value, and skip until the 3rd number is reached.
int total = amount;
while(total > 0)
{
for(int i = getStartingIndex(array, 3); i < amount; i += 3)
{
while(array[i] == 0 && i < amount)
i++;
print(i);
array[i] = 0;
total--;
}
}
Let's define our getStartingIndex function:
int getStartingIndex(int array[], int skipCount)
{
int index = 0;
for(int i = 0; i < skipCount; i++)
{
while(array[index] == 0)
index++;
index++;
}
return index - 1;
}
Now, we start at the value specified, and print numbers until we reach amount. And then we go back to the beginning, but we start at the third unused number.

Answer 3

We need to keep the current step size, it will run from zero to the desired step size and each time we visit an index and mark it as visited we will reset that counter to zero. Our main loop will iterate over the array one by one, until there are no unvisited indices left. We make sure the index does not run out of bounds of the array, so when index hits the length of the array, we reset it to zero.
If you don't want the last element to be printed, just change the while loop's conditional to while (visited_count < total_length - 1)
int main() {
int array[10] = { 0 };
int total_length = 10;
int visited_count = 0;
int index = 0;
int step_size = 3;
int current_step = 3;
while (visited_count != total_length) /* while there are unvisited elements */
{
if (!array[index] && current_step == step_size) /* array[index] has value zero, and we are at the desired step count, so print that element and mark it as visited */
{
array[index] = 1;
visited_count++;
printf("%d\n", index);
current_step = 0; /* resetting the step counter */
}
index++; /* this is how we iterate over the array, incrementing it at each iteration */
if (index == total_length) /* checking for out-of-bounds of the array, if we hit the total length, the index should be reset. */
{
index = 0;
}
if (!array[index]) /* now that we have stepped through the array, we increment the step counter if that element is not visited. */
{
current_step++;
}
}
return 0;
}