Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Answer:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, int> numbers;
cout << "Enter numbers, 0 to finish" << endl;
int number;
while (true) {
cin >> number;
if (number == 0) break;
numbers[number]++;
}
for (pair<int, int> element : numbers) {
std::cout << element.first << ": occurs " << element.second << " times" << std::endl;
}
}
Explanation:
One trick used here is not to keep track of the numbers themselves (since that is not a requirement), but start counting their occurrances right away. An STL map< > is a more suitable construct than a vector< >.
Answer:
The correct answer for the given question is Integer.parseInt( string variable );
Explanation:
Integer.parseInt( string variable ); is the method in a java programming language that convert the string into the integer value. It takes a string variable and converted into the integer.
Following are the program in java which convert the string value into an integer value.
class Main
{
public static void main(String []args) // main function
{
String str1 = "10009";
// variable declaration
int k = Integer.parseInt(str1);
// convert the string into integer.
System.out.println("Converted into Int:" + k);
}
}
Output:
Converted into Int:10009
Convert.toInt( stringVariable );
Convert.parseInt( stringVariable,Integer.toInt( stringVariable ); are not any method to convert the string into integer .
Therefore the correct answer is :Integer.parseInt( stringVariable );
Answer:
the change that computers go through
Explanation:
like 6 generation
