contestada

Write a program that allows users to enter a series of numbers and displays frequency of each number. Implement the following: 1. The user can enter a maximum of 20 numbers or enter-1 to end the series 2. Only numbers between 0..9 is allowed (or -1 ); if the user enters a number outside the valid input, inform the user and ask them to enter a valid number 3. Display the frequency of each number in the output. If the user input is 0,0,1,2,2,4,5,8,8,8,9,9, -1 the output will be:

Respuesta :

frknkrtrn

Answer:

zero = one = two = three = four = five = six = seven = eight = nine = 0

while True:

   number = int(input("Enter a number: "))

   if number == -1:

       break

   elif 0 <= number <= 9:

       if number == 0:

         zero += 1

       elif number == 1:

         one += 1

       elif number == 2:

         two += 1

       elif number == 3:

         three += 1

       elif number == 4:

         four += 1

       elif number == 5:

         five += 1

       elif number == 6:

         six += 1

       elif number == 7:

         seven += 1

       elif number == 8:

         eight += 1

       elif number == 9:

         nine += 1

   else:

       print("Invalid input!")

print("0 -> " + str(zero) + "," + " 1 -> " + str(one) + "," + " 2 -> " + str(two) + "," + " 3 -> " + str(three))

print("4 -> " + str(four) + "," + " 5 -> " + str(five) + "," + " 6 -> " + str(six) + "," + " 7 -> " + str(seven))

print("8 -> " + str(eight) + "," + " 9 -> " + str(nine))

Explanation:

Initialize the variables to hold the count for each digit

Initialize a while loop that iterates until a specified condition is reached

Inside the loop:

Get a number from the user and check

If the number is -1, stop the loop

If it is between 0 to 9, increase corresponding count value by 1

If it is out of range, print an error message

When the loop is done, print each count variable

Otras preguntas