Answer:
The code is given below in Java with appropriate comments
Explanation:
//Import the input.
import java.util.Scanner;
class CensoredWords
{
//Define the main method.
public static void main(String args[ ])
{
//Define the variables.
String userInput="" ;
//Define the scanner object
Scanner scobj = new Scanner(System.in);
//Accept the userInput.
System.out.print("Enter String: ");
userInput=scobj.nextLine();
//Check if the input contains darn.
//Print censored.
if(userInput.toUpperCase().indexOf("DARN") != -1)
System.out.printf("Censored");
//IF the input does not contains darn
//Print userInput.
else
System.out.printf(userInput)
return;
}
}
I'm guessing 8? But I'm not 100% positive
<u>Explanation:</u>
Hey there! you need not to panic about it ,your program didn't have Driver program i.e main program! the correct & working code is given below:
// C++ program to count even digits in a given number .
#include <iostream>
using namespace std;
// Function to count digits
int countEven(int n)
{
int even_count = 0;
while (n > 0)
{
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
n = n / 10;
}
cout << "Even count : "
<< even_count;
if (even_count % 2 == 0 )
return 1;
else
return 0;
}
// Driver Code
int main()
{
int n;
std::cin >>n;
int t = countEven(n);
return 0;
}
Answer:
Recently, with the new and advanced hacking algorithms and affordable high-performance computers available to adversaries, the 36 character computer suggested passwords can easily be insecure.
Explanation:
The 8 length passwords generated pseudo-randomly by computers are not secure as there are new algorithms like the brute force algorithm that can dynamically obtain the passwords by looping through the password length and comparing all 36 characters to get the right one.
And also, the use of high-performance computers makes these algorithms effective
