Narcissistic decimal number: Difference between revisions
m (→{{header|Perl 6}}: proper spelling) |
m (sp.) |
||
Line 1: | Line 1: | ||
{{draft task}} |
{{draft task}} |
||
A [http://mathworld.wolfram.com/NarcissisticNumber.html Narcissistic decimal number] is a |
A [http://mathworld.wolfram.com/NarcissisticNumber.html Narcissistic decimal number] is a positive decimal number, <math>n</math> in which if there are <math>m</math> digits in the number then the sum of all the individual digits of the number raised to the power <math>m</math> is equal to <math>n</math>. |
||
For example, if <math>n</math> is 153 then <math>m</math>, the number of digits is 3 and we have <math>1^3+5^3+3^3 = 1+125+27 = 153</math> and so 153 is a narcissistic decimal number. |
For example, if <math>n</math> is 153 then <math>m</math>, the number of digits is 3 and we have <math>1^3+5^3+3^3 = 1+125+27 = 153</math> and so 153 is a narcissistic decimal number. |
Revision as of 23:42, 6 March 2014
A Narcissistic decimal number is a positive decimal number, in which if there are digits in the number then the sum of all the individual digits of the number raised to the power is equal to .
For example, if is 153 then , the number of digits is 3 and we have and so 153 is a narcissistic decimal number.
The task is to generate and show here, the first 25 narcissistic numbers.
Perl 6
Here is a straightforward, naive implementation. It works but takes ages. <lang perl6>sub is-narcissistic(Int $n) { $n == [+] $n.comb »**» $n.chars }
for 0 .. * {
if .&is-narcissistic {
.say; last if ++state$ >= 25;
}
}</lang>
- Output:
0 1 2 3 4 5 6 7 8 9 153 370 371 407 Ctrl-C
Here the program was interrupted but if you're patient enough you'll see all the 25 numbers.
Python
This solution pre-computes the powers once.
<lang python>from __future__ import print_function from itertools import count, islice
def narcissists():
for digits in count(0): digitpowers = [i**digits for i in range(10)] for n in range(int(10**(digits-1)), 10**digits): div, digitpsum = n, 0 while div: div, mod = divmod(div, 10) digitpsum += digitpowers[mod] if n == digitpsum: yield n
for i, n in enumerate(islice(narcissists(), 25), 1):
print(n, end=' ') if i % 5 == 0: print()
print()</lang>
- Output:
0 1 2 3 4 5 6 7 8 9 153 370 371 407 1634 8208 9474 54748 92727 93084 548834 1741725 4210818 9800817 9926315