Munchausen numbers: Difference between revisions
(→{{header|Perl 6}}: Added Perl 6 solution) |
(Added Algol 68) |
||
Line 7: | Line 7: | ||
;Task requirements |
;Task requirements |
||
Finds all Munchausen numbers between 1 and 5000 |
Finds all Munchausen numbers between 1 and 5000 |
||
=={{header|ALGOL 68}}== |
|||
<lang algol68># Find Munchausen Numbers between 1 and 5000 # |
|||
# note that 6^6 is 46 656 so we only need to cosider numbers consisting of 0 to 5 # |
|||
# table of Nth powers # |
|||
[]INT nth power = ([]INT( 1, 1, 2 * 2, 3 * 3 * 3, 4 * 4 * 4 * 4, 5 * 5 * 5 * 5 * 5 ))[ AT 0 ]; |
|||
FOR d1 FROM 0 TO 5 DO |
|||
INT d1 part = d1 * 1000; |
|||
FOR d2 FROM 0 TO 5 DO |
|||
INT d2 part = d2 * 100; |
|||
FOR d3 FROM 0 TO 5 DO |
|||
INT d3 part = d3 * 10; |
|||
FOR d4 FROM 0 TO 5 DO |
|||
INT digit power sum := 0; |
|||
IF d1 > 0 THEN |
|||
digit power sum := nth power[ d1 ] |
|||
+ nth power[ d2 ] |
|||
+ nth power[ d3 ] |
|||
+ nth power[ d4 ] |
|||
ELIF d2 > 0 THEN |
|||
digit power sum := nth power[ d2 ] |
|||
+ nth power[ d3 ] |
|||
+ nth power[ d4 ] |
|||
ELIF d3 > 0 THEN |
|||
digit power sum := nth power[ d3 ] |
|||
+ nth power[ d4 ] |
|||
ELSE |
|||
digit power sum := nth power[ d4 ] |
|||
FI; |
|||
INT number = d1 part + d2 part + d3 part + d4; |
|||
IF digit power sum = number THEN |
|||
print( ( whole( number, 0 ), newline ) ) |
|||
FI |
|||
OD |
|||
OD |
|||
OD |
|||
OD</lang> |
|||
{{out}} |
|||
<pre> |
|||
1 |
|||
3435 |
|||
</pre> |
|||
=={{header|C}}== |
=={{header|C}}== |
||
Line 36: | Line 80: | ||
<pre>1 |
<pre>1 |
||
3435</pre> |
3435</pre> |
||
=={{header|C sharp|C#}}== |
=={{header|C sharp|C#}}== |
Revision as of 19:18, 22 September 2016
- Definition of Munchausen numbers
A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, is n itself.
- Task requirements
Finds all Munchausen numbers between 1 and 5000
ALGOL 68
<lang algol68># Find Munchausen Numbers between 1 and 5000 #
- note that 6^6 is 46 656 so we only need to cosider numbers consisting of 0 to 5 #
- table of Nth powers #
[]INT nth power = ([]INT( 1, 1, 2 * 2, 3 * 3 * 3, 4 * 4 * 4 * 4, 5 * 5 * 5 * 5 * 5 ))[ AT 0 ];
FOR d1 FROM 0 TO 5 DO
INT d1 part = d1 * 1000; FOR d2 FROM 0 TO 5 DO INT d2 part = d2 * 100; FOR d3 FROM 0 TO 5 DO INT d3 part = d3 * 10; FOR d4 FROM 0 TO 5 DO INT digit power sum := 0; IF d1 > 0 THEN digit power sum := nth power[ d1 ] + nth power[ d2 ] + nth power[ d3 ] + nth power[ d4 ] ELIF d2 > 0 THEN digit power sum := nth power[ d2 ] + nth power[ d3 ] + nth power[ d4 ] ELIF d3 > 0 THEN digit power sum := nth power[ d3 ] + nth power[ d4 ] ELSE digit power sum := nth power[ d4 ] FI; INT number = d1 part + d2 part + d3 part + d4; IF digit power sum = number THEN print( ( whole( number, 0 ), newline ) ) FI OD OD OD
OD</lang>
- Output:
1 3435
C
Adapted from Zack Denton's code posted on Munchausen Numbers and How to Find Them. <lang C>#include <stdio.h>
- include <math.h>
int main() {
for (int i = 1; i < 5000; i++) { // loop through each digit in i // e.g. for 1000 we get 0, 0, 0, 1. int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; // find the sum of the digits // raised to themselves sum += pow(digit, digit); } if (sum == i) { // the sum is equal to the number // itself; thus it is a // munchausen number printf("%i\n", i); } } return 0;
}</lang>
- Output:
1 3435
C#
<lang csharp>Func<char, int> toInt = c => c-'0';
foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);</lang>
- Output:
1 3435
JavaScript
<lang javascript>for (let i of [...Array(5000).keys()] .filter(n => n == n.toString().split() .reduce((a, b) => a+Math.pow(parseInt(b),parseInt(b)), 0))) console.log(i);</lang>
- Output:
1 3435
Perl 6
<lang perl6>sub is_munchausen ( Int $n ) {
constant @powers = 0, |map { $_ ** $_ }, 1..9; $n == @powers[$n.comb].sum;
} .say if .&is_munchausen for 1..5000;</lang>
- Output:
1 3435
zkl
<lang zkl>[1..5000].filter(fcn(n){ n==n.split().reduce(fcn(s,n){ s + n.pow(n) },0) }) .println();</lang>
- Output:
L(1,3435)