Answer: Network access control (NAC)
Explanation:
The solution that should be used is the network access control. Network access control helps in keeping devices and users that are unauthorized out of ones private network.
In this case, since one will like to prevent the laptops from connecting to the network unless anti-virus software and the latest operating system patches are installed, then the network access control can be used. One can only give access to the device that it wants to give access to and prevent others from connecting.
Answer:
Uneven use of resources
Explanation:
Potential problem associated with supporting multi - user operation without hardware support is:
Uneven use of resources: In a situation where we assign a set of resources to user 1 and if a new user comes, then it would be difficult to allocate new resources to him. The processor would get confused between the two users. And the tasks would not be completed. This can affect task processing.
<u>Output:</u>
f1 in A
f2 in A
f1 in B
f2 in A
f1 in A
f2 in A
f1 in B
f2 in B
<u>Explanation:</u>
In this snippet, the code makes use of virtual functions. A virtual function is defined as a function that is defined in the base class and redefined in the derived class. If the derived function accesses the virtual function, the program will get executed with the derived class’s version of the function.
In this code, we define the virtual function f1() in class A and also redefine it in class B which is the derived class of A. While executing the program, the function g which takes the object b (class B’s object) as a parameter. It will print class B’s version of f1() rather than class A’s version. This is working off the virtual function.
The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.
#include<stdio.h>
#include<string.h>
int* uniqueValue(int input1,int input2[])
{
int left, current;
static int arr[4] = {0};
int i = 0;
for(i=0;i<input1;i++)
{
current = input2[i];
left = 0;
if(current > 0)
left = arr[(current-1)];
if(left == 0 && arr[current] == 0)
{
arr[current] = input1-current;
}
else
{
for(int j=(i+1);j<input1;j++)
{
if(arr[j] == 0)
{
left = arr[(j-1)];
arr[j] = left - 1;
}
}
}
}
return arr;
}
