Narcissistic decimal number: Difference between revisions
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<lang perl6>sub is-narcissic(Int $n) { $n == [+] $n.comb »**» $n.chars } |
<lang perl6>sub is-narcissic(Int $n) { $n == [+] $n.comb »**» $n.chars } |
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for 0 .. * { |
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say (grep &is-narcissic, 0 .. *)[^25];</lang> |
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if .&is-narcissic { |
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.say; |
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last if ++state$ >= 25; |
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} |
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}</lang> |
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{{out}} |
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<pre>0 |
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1 |
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2 |
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3 |
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4 |
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5 |
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6 |
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7 |
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8 |
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9 |
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153 |
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370 |
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371 |
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407 |
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Ctrl-C</pre> |
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Here the program was interrupted but if you're patient enough you'll see all the 25 numbers. |
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=={{header|Python}}== |
=={{header|Python}}== |
Revision as of 23:23, 6 March 2014
A Narcissistic decimal number is a positice 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-narcissic(Int $n) { $n == [+] $n.comb »**» $n.chars }
for 0 .. * {
if .&is-narcissic {
.say; last if ++state$ >= 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