Question

HALIX PROGRAMMING C++

1. Write Halix machine code to read and add 5 values.

Test cases: 1 2 3 4 5 ==>15 -1 -2 -3 -4 23 ==> 13 0 0 0 0 0 ==> 0

2. Write Halix machine code to read a value: Double it if the nu halve it when it's odd.

Test cases: 14 ==> 28 0 ==>0 25==>12 -2==>-4 -7==>-3

3. Write a program that reads ans echoes (display what was just values that are terminated by the value 777.

Pseudo Code: int val;

cin >> val;

while (val != 777)

{

cout > val;

cin >> val;

} cout << 4444;

Test cases: 1 2 777 ==> 1 2 4444 777 => 4444 25 777 ==> 25 4444 -2 7 9 4 3 84 777 ==> -2 7 9 4 3 84 4444

4. Write a program that reads input values terminated by the value 9999, and computes the product of the negative values, and the sum of the positive value.

Pseudo-code: int val, negProd=1, posSum=0;

cin >> val;

while (val != 9999)

{

if (val < 0)

negProd *= val;

else

posSum += val;

cin >> val;

}

cout << posSum << negProd;

cout << 5555

Test cases: 1 2 9999 ==> 1 3 9999=>1 0 1 -2 2 -5 -5 ==> 40 3

1 -2 2 -5 -5 ==> -40 3 -2 7 9 4 3 -84 9999 ==>168 23

Answer 1

1.

#include <iostream>
using namespace std;
int main(){
int a, b, c, d, e;

cout << "Enter Five numberes ";
cin >> a >> b >> c >> d >> e ;

c = a + b + c + d + e;
cout <<"Sum of the numbers: " << c << endl;

return 0;
}

2.

#include <iostream>
using namespace std;
int main(){
int n,res;
cout << "Enter Number: ";
cin >> n;

if ( n % 2 == 0){
   res = n*2;
cout << res ;
   }else{
   res = n/2;
cout << res;
   }
return 0;
}

3.

#include <iostream>
using namespace std;
int main(){
int n,res;
   cout<<"Enter numbers ";
   while(1){ //While 1 is always Ture
       cin >> n;
       if (n==777){
           break;
       }else{
           cout << n <<" " ;
       }
   }
   cout<<"4444 ";
return 0;
}

4.

#include <iostream>
using namespace std;
int main(){
int n,neg_product=1,pos_sum=0;
   cout<<"Enter numbers ";
   while(1){ //While 1 is always Ture
       cin >> n;
       if (n==9999){
           break;
       }else if (n>0){
           pos_sum +=n;
       }else if (n<0){
           neg_product *=n;
       }
   }
   cout<< neg_product << " "<< pos_sum;
return 0;
}

I give the screenshot of code also.

Thank You :-)