Narcissistic decimal number

Narcissistic decimal number is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A Narcissistic decimal number is a positive decimal number, $n$ in which if there are $m$ digits in the number then the sum of all the individual digits of the number raised to the power $m$ is equal to $n$.

For example, if $n$ is 153 then $m$, the number of digits is 3 and we have $1^{3}+5^{3}+3^{3}=1+125+27=153$ and so 153 is a narcissistic decimal number.

The task is to generate and show here, the first 25 narcissistic numbers.

C

<lang c>#include <stdio.h>

typedef long long xint;

1. define MAX_LEN 16

xint dpow[MAX_LEN + 1][11];

void init(void) { int i, p; for (p = 0; p <= MAX_LEN; p++) for (i = 0; i <= 10; i++) dpow[p][i] = p ? dpow[p-1][i] * i : 1; }

void narc(int power, int pos, xint value, xint dsum) { xint i, ds, v, ten;

if (!pos) { if (value == dsum) printf(" %lld", value); return; }

ten = dpow[pos - 1][10]; for (i = (pos == power); i < 10; i++) { ds = dsum + dpow[power][i]; v = value + i * ten;

if (ds >= v + ten) break;

if (v <= ds + dpow[power][9] * (pos - 1)) narc(power, pos - 1, v, ds); } }

int main(void) { int i; init();

for (i = 1; i <= 16; i++) { printf("length %d:", i); narc(i, i, 0, 0); putchar('\n'); }

return 0; }</lang>

Output:

length 1: 1 2 3 4 5 6 7 8 9
length 2:
length 3: 153 370 371 407
length 4: 1634 8208 9474
length 5: 54748 92727 93084
length 6: 548834
length 7: 1741725 4210818 9800817 9926315
length 8: 24678050 24678051 88593477
length 9: 146511208 472335975 534494836 912985153
^C

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