Answer:
a)
#include <iostream>
using namespace std;
int main() {
bool a,b,c;
cin>>a>>b;
if(a^b)//X-OR operator in C++.
c=true;
else
c=false;
cout<<c;
return 0;
}
b)
#include <iostream>
using namespace std;
int main() {
bool a,b,c,d;
cin>>a>>b>>c;
if((a^b)^c)//X-OR operator in C++.
d=true;
else
d=false;
cout<<d;
return 0;
}
Explanation:
The above written programs are in C++.There is an operator (^) called X-OR operator in C++.It returns true if the number of 1's are odd and returns false if the number of 1's are even.
In the if statement I have user X-OR operator(^) to find the result and storing the result in another boolean variable in both the questions.
Answer:
4. A Denial of Service attack (DDOS attack).
Explanation:
A DDOS attack is a malicious attempt to disrupt the normal traffic to a service.
In essence, it sends an enormous amount of requests to the service, until the server is overwhelmed because it can't handle that much traffic, and collapses in an overflow.
Thus, regular users are not able to access their services.
Usually, attackers use a botnet, a network of "zombie" computers that have been previously infected with a malware that allows the attacker to remotely control them, then the botnet starts to send a flood of traffic from different locations, and make the attacker difficult to detect or track.
Your answer is true your welcome!
<u>Active Directory</u> consist of many users and their information.
Answer:
Each time you insert a new node, call the function to adjust the sum.
This method has to be called each time we insert new node to the tree since the sum at all the
parent nodes from the newly inserted node changes when we insert the node.
// toSumTree method will convert the tree into sum tree.
int toSumTree(struct node *node)
{
if(node == NULL)
return 0;
// Store the old value
int old_val = node->data;
// Recursively call for left and right subtrees and store the sum as new value of this node
node->data = toSumTree(node->left) + toSumTree(node->right);
// Return the sum of values of nodes in left and right subtrees and
// old_value of this node
return node->data + old_val;
}
This has the complexity of O(n).
Explanation:
