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Inessa05 [86]
3 years ago
13

In python please!! Write the definition of a function named countPos that needs integer values from standard input until there a

re none left and returns the number that are positive. The function must not use a loop of any kind.
Computers and Technology
1 answer:
jek_recluse [69]3 years ago
8 0

Answer:

Explanation:

The following code is written in Python it doesn't use any loops, instead it uses a recursive function in order to continue asking the user for the inputs and count the number of positive values. If anything other than a number is passed it automatically ends the program.

def countPos(number=input("Enter number: "), counter=0):

   try:

       number = int(number)

       if number > 0:

           counter += 1

           newNumber = input("Enter number: ")

           return countPos(newNumber, counter)

       else:

           newNumber = input("Enter number: ")

           return countPos(newNumber, counter)

   except:

       print(counter)

       print("Program Finished")

countPos()

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Akimi4 [234]

Answer:

sorry can´t understand langues

Explanation:

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7 0
3 years ago
Jasmine plays a game on her computer screen. A moving balloon appears on the screen, and she has to pop the balloon by clicking
Elena-2011 [213]
Please find the answer as attachment.

4 0
2 years ago
Write a do-while loop that continues to prompt a user to enter a number less than 100, until the entered number is actually less
Ad libitum [116K]

Answer:

#include<iostream>//library inclusion

using namespace std;

int main()

{

int userInput;

do//start of do while loop

{

 cout << "Enter a number less than a 100" << endl;

 cin >> userInput;

 if (userInput < 100) //condition

 {

  cout << "YOu entered less than a hundred: " << userInput << endl;

 }

 else

 {

  cout << "your number is greater than 100" << endl;

 }

} while (userInput > 100);//condition for do while

return 0;//termination of int main

}

Explanation:

The program has been commented for you. The do-while loop enters the first loop regardless of the condition. Then after the first iteration, it checks for the condition. If the condition is being met, it will iterate through, again. Otherwise it will break out of the loop and land on the "return 0;" line. Which also happens to be the termination of the program in this case. The if-else condition is used for the user to see when prompted.

6 0
3 years ago
Write an algorithm that prints out all the subsets of 3 elements of a set of n elements. The elements of the set are stored in a
Ksenya-84 [330]

void Print_3_Element_Subsets(int n,const element_type S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

print("No subset found");

}

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

print(S[i],S[j],S[k]);

}

}

}

}

C++ Implementation;

#include<cstdlib>

#include<time.h>

using namespace std;

int S1[5]= {1,2,3,4,5};

int n1 = 5;

int S2[2]={1,2};

int n2 = 2;

void Print_3_Element_Subsets(int n,const int S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

printf("No subset found\n");

}

printf("3 Subsets are\n");

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

cout<<S[i]<<" "<<S[j]<<" "<<S[k]<<endl;

}

}

}

}

int main()

{

cout<<"Case 1"<<endl;

Print_3_Element_Subsets(n1,S1);

cout<<endl<<"Case 2"<<endl;

Print_3_Element_Subsets(n2,S2);

}

OUTPUT

Case 1

3 Subsets are

1 2 3

1 2 4

1 2 5

1 3 4

1 3 5

1 4 5

2 3 4

2 3 5

2 4 5

3 4 5

Case 2

No subset found

3 Subsets are

--------------------------------

Process exited after 0.0137 seconds with return value 0

Press any key to continue . . .

7 0
3 years ago
Define a void function that calculates the sum (+), difference (-), product (*), quotient (/), and modulus (%) of two integer nu
BabaBlast [244]

Answer:

#include <iostream>

using namespace std;

void multipurpose(int &num1,int &num2,int &add,int &subt,int &multi,int &divi,int &modulo )

{

   add=num1+num2;//adding two numbers...

   subt=abs(num1-num2);//subtracting two numbers...

   multi=num1*num2;//multiplying two numbers...

   divi=num1/num2;//Finding quotient of two numbers...

   modulo=num1%num2;//Finding modulo of two numbers...

}

void print(int add,int sub,int divi,int multi,int modulo) //function to print the values.

{

   cout<<"The addition is "<<add<<endl<<"The subtraction is "<<sub<<endl

   <<"The quotient is "<<divi<<endl<<"The multiplication is "<<multi<<endl

   <<"The modulus is "<<modulo<<endl;

}

int main() {

   int a,b,sum=0,mult=0,divi=0,sub=0,modulo=0;

   cin>>a>>b;

   multipurpose(a,b,sum,sub,mult,divi,modulo);

   print(sum,sub,divi,mult,modulo);

return 0;

}

Enter the numbers

12 12

The addition is 24

The subtraction is 0

The quotient is 1

The multiplication is 144

The modulus is 0

Explanation:

I have created a function multipurpose that has a argument list of two numbers ,and variables for addition,subtraction,multiplication,Division,Modulus and these all are passed by reference.Since the function is of type void so it cannot return anything so the only way to store the result is by passing the variables to store the results by  reference.

The other function print has the arguments of all the results here you can pass them by value or by reference because you only need to print the results.

7 0
3 years ago
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