Answer:
switchboard
Explanation:
The option that he should implement would be a switchboard interface design. These design structures for graphic user interfaces use a single main page, large icons/buttons, a fixed navigation menu, and all the necessary functionality right in front of the user. This design is made with simplicity in mind in order to make it as easy as possible for a new user to pick up and efficiently and intuitively navigate the user interface. Therefore, since Calvin needs a simple yet professional design, this would be the best implementation.
I think it’s Apple ios
It’s definitely not Microsoft but my guess is Apple
Inaccurate and misleading
Answer:
# include <iostream.h>
# include <stdio.h>
# include <string.h>
using namespace std;
class citizen
{
int i;
public string name[30];
public long int phonenumber[30];
public void addindividual(string name1)
{
If (i<=30)
{ int flag=0;
for(int j=0; j<=i;j++)
{
if (strcmp(name[i], name1)
{
flag=1;
}
else
{
flag=0;
}
}
If (flag)
{
if (i<30)
{
for(j=i+1;j<=30; j++)
{
cout<<"Enter the name:"; getchar(name[j]);
cout<<"Enter the phone number:"; cin>>phonenumber[j];
i++;
}
else
{
cout<<"The person already exists";
exit();
}
}
else
{
cout<<"array is full:";
exit();
}
}
}
Void main()
{
string str;
cout<<" Enter name:";
getline(cin, str); ;
citizen c1=new citizen();
c1.addindividual(name1);
}
Explanation:
With a little more effort you can make the program allow the user to enter any number of details, but less than 30 overall. We have used here flag, and as a programmer we know why we use the Flag. It is used to check whether certain Boolean condition is fulfilled or not. Here, we are checking whether a given name is present in the array of names, and if it is not present, we add that to the list. And if the name is present, we print, it already exist.
Answer:
def prime_generator(s, e):
for number in range(s, e+1):
if number > 1:
for i in range(2, number):
if (number % i) == 0:
break
else:
print(number)
prime_generator(6,17)
Explanation:
I believe you want to ask the prime numbers between s and e.
- Initialize a for loop that iterates from s to e
- Check if the number is greater than 1. If it is, go inside another for loop that iterates from 2 to that number. If the module of that number to any number in range (from 2 to that number) is equal to 0, this means the number is not a prime number. If the module is not equal to zero, then it is a prime number. Print the number
