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@freeloki
Created March 29, 2020 17:54
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def print_formatted(number):
# your code goes here
#print("Number is: {}".format(number))
if number >= 1 and number <= 99:
num_str = str(bin(number)).replace("0b", "")
padding_count = len(num_str)
#print("Padding count is: {} ".format(padding_count))
for x in range(1, number + 1, 1):
format_data = "{}{}{}{}"
# input boyu 1 ise soldan 4 2 ise soldan 3 boşluk bırak
# default 2 üç boşluk koyalım. boyu 1 ise bir boşluk daha ekleyelim.
bin_x = str(bin(x)).replace("0b", "")
if len(bin_x) == 1:
bin_x = padding_count * " " + bin_x
elif len(bin_x) == 2:
bin_x = (padding_count-1) * " " + bin_x
elif len(bin_x) == 3:
bin_x = (padding_count-2) * " " + bin_x
elif len(bin_x) == 4:
bin_x = (padding_count-3) * " " + bin_x
elif len(bin_x) == 5:
bin_x = (padding_count-4) * " " + bin_x
elif len(bin_x) == 6:
bin_x = (padding_count-5) * " " + bin_x
elif len(bin_x) == 7:
bin_x = (padding_count-6) * " " + bin_x
str_x = str(x)
if len(str(x)) == 1:
str_x = (padding_count-1)*" " + str_x
if len(str(x)) == 2:
str_x = (padding_count-2)*" " + str_x
hex_x = str(hex(x)).replace("0x", "")
hex_x = hex_x.upper()
if len(hex_x) == 1:
hex_x = padding_count*" " + hex_x
elif len(hex_x) == 2:
hex_x = (padding_count-1)*" " + hex_x
oct_x = str(oct(x))
oct_x = oct_x.replace("0o","")
if len(oct_x) == 1:
oct_x = padding_count*" " + oct_x
elif len(oct_x) == 2:
oct_x = (padding_count-1)*" " + oct_x
elif len(oct_x) == 3:
oct_x = (padding_count-2)*" " + oct_x
print(format_data.format(str_x, oct_x, hex_x, bin_x))
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