How many times is c found in n

Question

n = int(input("Input n: "))
c = int(input("Input c: "))

def it_show(n, c):
    counter= 0
    for i in range(1, n+1):
        if i < 10:
            if i == c:
                counter+= 1
        if i > 9:
            digit = i % 10
            if digit == c:
                counter+= 1
            i //= 10
    return counter

print(it_show(n, c))

From the start sory my English is not so good.

In the task it is emphasizes to work with numbers, no strings.

I need to input n and c.Program need to return in how many numbers from 1 to n(including n), c show up.Etc. n=20 and c=5 return is 2.Because from 1 to 20, 5 show up in numbers 5 and 15, that is 2 numbers.In this code for this case it works.But if I input n=14 and c=1 return is 2, but it needs to bee 6 because from 1 to 14(including 14) there are 6 numbers that have c(1) in them (1, 10, 11, 12, 13, 14).

Answer 1

You can use a while to cycle through the digits.

def it_show(n, c):

  counter = 0

  for i in range(1, n+1):

    while i > 0:

      if i % 10 == c: # if the right digit is equal to c
        counter += 1
        break # counter increased, and no need to check the other digits

      i //= 10 # integer division to eliminate the right digit, which was already checked

  return counter

print(it_show(20,5))
print(it_show(14,1))