Fastest way to generate number like 66666 when the number of digits is given

Question

I have an interesting problem where I want to generate a big number (~30000 digits) but it has to be all identical digits, like 66666666666666.......

So far I have done this by:

def fillWithSixes(digits):
    result = 0
    for i in range(digits):
        result *= 10
        result += 6
    return result

However, this is very inefficient, and was wondering if there is any better way? Answer in cpp or java is okay too.

Edit:

Let's not just solve for 666666..... I want it to be generic for any number. How about 7777777777.... or 44444........ or 55555...?
String operations are worse, the increase from current complexity of O(n) to O(n^2).

what do you mean by unique digits.... also, you can instead make `result` a string and concatenate with `result += '6'` then return `int(result)` but if the number is as big as you said, I'm pretty sure that will reach the `int` limit — R Nar, Nov 16 '15 at 20:35
or, even better, just `return int('6'*digits)` since `'str'*int = 'strstrstr...'#int times` — R Nar, Nov 16 '15 at 20:36
@RNar - python's Ints are arbitrarily big. `int('6' * 10000)` works great. — Seth, Nov 16 '15 at 20:45
[depends on your version of python](http://stackoverflow.com/questions/7604966/maximum-and-minimum-values-for-ints) — R Nar, Nov 16 '15 at 20:46

score 6 · Accepted Answer · edited Nov 17 '15 at 02:07

6

You may use the formula 666...666 = 6/9*(10**n-1), where n is the number of digits.

So, in Python, you would write that as

n = int(input())
a = 6 * (10**n - 1) // 9
print(a)

edited Nov 17 '15 at 02:07

jfs

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answered Nov 16 '15 at 20:36

wythagoras

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1

I found this to be on the order of 10^5 times faster than any other answer. +1 – dpwilson Nov 16 '15 at 21:11
2

@pjs `6 * (10**n - 1) / 9` – Peter Wood Nov 16 '15 at 21:27
@Nick replace `6` with the digit you want. `(10**n - 1) / 9` is the desired number of 1's, and can then be scaled by any digit. – pjs Nov 16 '15 at 22:11

score 1 · Answer 2 · answered Nov 16 '15 at 20:38

You can use ljust or rjust:

number = 6
amount_of_times_to_repeat = 30000
big_number = int("".ljust(amount_of_times_to_repeat, str(number)))
print big_number

In one single line:

print int("".ljust(30000, str(6)))

Or:

new_number = int("".ljust(30000, str(6)))

score 0 · Answer 3 · edited May 23 '17 at 12:30

The fastest method to generate such numbers with 100000+ digits is decimal.Decimal():

from decimal import Decimal as D

d = D('6' * n)

Measurements show that 6 * (10**n - 1) // 9 is O(n*log n) while D('6' * n) is O(n). Though for small n (less than ~10000), the former can be faster.

Decimal internal representation stores decimal digits directly. If you need to print the numbers latter; str(Decimal) is much faster than str(int).

Fastest way to generate number like 66666 when the number of digits is given

3 Answers3