Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
import turtle
window = turtle.Screen()
tr = turtle.Turtle()
tr.forward(100)
tr.left(90)
tr.forward(100)
tr.left(90)
tr.forward(100)
tr.left(90)
tr.forward(100)
tr.back(100)
tr.left(120)
tr.forward(75)
tr.right(78)
tr.forward(60)
window.mainloop()
In my code, we use the turtle module for the graphics to draw the house with a roof.
Answer:
1 14 3
Explanation:
If that code fragment is put under a main() function and then it will yield the above output. According to the function something(), only changes in value second parameter will be reflected and first will be unchanged because second is passed by reference and the first one is passed as value so only address of 's' is passed, so, only its value is changed and the rest are same.
Answer:
//The Employee Class
public class Employee {
char name;
long ID;
//The constructor
public Employee(char name, long ID) {
this.name = name;
this.ID = ID;
}
//Method Get Person
public void getPerson (char newName, long newId){
this.ID = newName;
this.ID = newId;
}
//Method Print
public void print(){
System.out.println("The class attributes are: EmpName "+name+" EmpId "+ID);
}
}
The working of the class is shown below in another class EmployeeTest
Explanation:
public class EmployeeTest {
public static void main(String[] args) {
Employee employee1 = new Employee('a', 121);
Employee employee2 = new Employee('b', 122);
Employee employee3 = new Employee('c', 123);
employee1.print();
employee2.print();
employee3.print();
}
}
In the EmployeeTest class, Three objects of the Employee class are created.
The method print() is then called on each instance of the class.
