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Allisa [31]
4 years ago
10

NOTE: in mathematics, division by zero is undefined. So, in C , division by zero is always an error. Given a int variable named

callsReceived and another int variable named operatorsOnCall, write the necessary code to read values into callsReceived and operatorsOnCall and print out the number of calls received per operator (integer division with truncation will do). HOWEVER: if any value read in is not valid input, just print the message "INVALID". Valid input for operatorsOnCall is any value greater than 0. Valid input for callsReceived is any value greater than or equal to 0.
Computers and Technology
1 answer:
Sergio [31]4 years ago
8 0

Answer:

#include <stdio.h>

int main()

{

   int callsReceived, operatorsOnCall;

   scanf("%d", &callsReceived);

   scanf("%d", &operatorsOnCall);

if(callsReceived>=0 && operatorsOnCall>0)

{

   printf("%d",callsReceived/operatorsOnCall);

}

else

{

   printf("INVALID") ;

} }

Explanation:

The programming language to use here is C.

There is an int type variable named callsReceived    

Another int type variable is operatorsOnCall

scanf is used to read values into callsReceived and operatorsOnCall

printf is used to print the number of calls received per operator which is found by dividing the calls received with the operators on call i.e. callsReceived / operatorsOnCall

Here the callsReceived is the numerator and should be greater than or equal 0 and operatorsOnCall is the denominator and should be less than 0. So an IF condition is used to check if the value stored in operatorsOnCall is greater than 0 and if the value stored in callsReceived is greater than or equal to 0 in order to precede with the division. AND (&&) logical operator is used to confirm that both these conditions should  hold for the IF statement to evaluate to TRUE.

In case the input is not valid, the else part is displayed which means that the message INVALID will be displayed.

In C++ cin is used to read from callsReceived and operatorsOnCall and cout is used to display the results of division or the INVALID message in case the IF condition is false. Rest of the code is the same.

cin >> callsReceived;

cin >> operationsOnCall;

if(callsReceived>=0 && operatorsOnCall>0)

   cout << (callsReceived/operationsOnCall);

else

   cout << "INVALID";

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g 1. Write a program that asks the user for a number greater than 5 and prints all values between 1 and n that are multiples of
a_sh-v [17]

PROGRAM 1

#include <iostream>  

using namespace std;

int main() {

   int n;

   // user input taken for n

   cout<<"Enter any number greater than five: ";

   cin>>n;

   

   cout<<"The multiples of 5 between 1 to "<<n<<" are shown below."<<endl;

   

   // displaying multiples of 5 between 1 to n

   for(int h=1; h<=n; h++)

   {

       // if number is completely divisible by 5, remainder will be 0

       if( (h%5)==0)

           cout<<h<<endl;

   }

 

   return 0;

}

OUTPUT

Enter any number greater than five: 22

The multiples of 5 between 1 to 22 are shown below.

5

10

15

20

The program does not implements any input validation since this is not mentioned in the question.

User input is taken. All the multiples of 5 computed inside a for loop, and displayed to the console.

PROGRAM 2

#include <iostream>

using namespace std;

int main() {

   // variables to hold respective values

   int n;

   int sum=0;

   double avg=0;

   

   do

   {

       // user input taken for n

       cout<<"Enter any number between 1 and 100: ";

       cin>>n;

       if(n<1 || n>100)

           cout<<"Invalid number. Enter valid number."<<endl;

       cout<<""<<endl;

       

   }while(n<1 || n>100);

   

   // computing sum and average of numbers from 1 to n

   for(int h=1; h<=n; h++)

   {

       sum = sum+h;

   }

   

   avg=avg+(sum/n);

   

   cout<<"Sum of all the numbers from 1 to "<<n<<" is "<<sum<<endl;

   

   cout<<""<<endl;

   cout<<"Average of all the numbers from 1 to "<<n<<" is "<<avg<<endl;

   

   return 0;

}

OUTPUT

Enter any number between 1 and 100: 111

Invalid number. Enter valid number.

Enter any number between 1 and 100: 0

Invalid number. Enter valid number.

Enter any number between 1 and 100: 23

Sum of all the numbers from 1 to 23 is 276

Average of all the numbers from 1 to 23 is 12

The program not implements any input validation since this is mentioned in the question.

User input is taken inside do-while loop till valid input is obtained.

The sum of all numbers from 1 to n computed inside a for loop, average computed outside for loop.

Both values are displayed to the console.

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3 years ago
A cashier distributes change using the maximum number of five dollar bills, followed by one dollar bills. For example, 19 yields
posledela

Answer:

The correct program to this question as follows:

Program:

//header file

#include <stdio.h> //include header file for using basic function

int main() //defining main method

{

int amountToChange=19,numFives,numOnes; //defining variable

numFives = amountToChange/5; //holding Quotient  

numOnes = amountToChange%5; //holding Remainder

printf("numFives: %d\n", numFives); //print value

printf("numOnes: %d\n", numOnes); //print value

return 0;

}

Output:

numFives: 3  

numOnes: 4  

Explanation:

In the above program first, a header file is included then the main method is declared inside the main method three integer variable is defined that are "amountToChange, numFives, and numOnes", in which amountToChange variable a value that is "19" is assigned.

  • Then we use the numFives and the numOnes variable that is used to calculate the number of 5 and 1 , that is available in the amountToChange variable.
  • To check this condition we use (/ and %) operators the / operator is used to hold Quotient value and the % is used to hold Remainder values and after calculation prints its value.

6 0
3 years ago
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