If the distribution of the random order can be skewed, you can use the fact that P(x) = (a * x + b) mod n will visit all integers from 0 to n-1 in a "random-looking" order as x iterates from 0 to n-1, if a is coprime with n. To make finding the coprime less complicated, you can first expand the domain to 2^k, then any 3^m within the domain is suitable. Here is an example program:

```c
#include <iostream>
#include <cstdlib>

using namespace std;

// find the expanded domain
int minPowerOfTwo(int x) {
    int ans = 1;
    while (ans < x) {
        ans = ans << 1;
    }
    return ans;
}

// find the random coprime
int randomCoprime(int x) {
    int ans = 0;
    int y = 3
    while (y <= x) {
        y *= 3;
        ans++;
    }
    return rand() % ans + 1;
}

int main() {

    int arr[] = {1, 2, 3, 4, 5, 6};
    int N = sizeof(arr) / sizeof(arr[0]);
    srand(time(NULL));

    expandN = minPowerOfTwo(N);
    coprimeN = randomCoprime(expandN);

    // P(x) = b, a random offset in [0, expandN]
    int index = rand() % (expandN);
    
    // iterate through the array according to P(x) = (a * x + b) mod n
    for (int i = 0; i < expandN; i++) {
        // skip the expanded part
        if (index < N)
            cout << arr[index] << endl;

        index += coprimeN;
        if (index > expandN) index -= expandN;
    }

    return 0;
}
```
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