Disarium numbers: Difference between revisions
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<br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">
F is_disarium(n)
V digitos = String(n).len
V suma = 0
V x = n
L x != 0
suma += (x % 10) ^ digitos
digitos--
x I/= 10
I suma == n
R 1B
E
R 0B
V limite = 19
V cont = 0
V n = 0
print(‘The first ’limite‘ Disarium numbers are:’)
L cont < limite
I is_disarium(n)
print(n, end' ‘ ’)
cont++
n++
</syntaxhighlight>
{{out}}
<pre>
The first 19 Disarium numbers are:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Action!}}==
{{Trans|PL/M}} which is {{Trans|ALGOL 68}}
<syntaxhighlight lang="action!">
;;; find some Disarium Numbers - numbers whose digit-position power sume
;;; are equal to the number, e.g.: 135 = 1^1 + 3^2 + 5^3
PROC Main()
DEFINE MAX_DISARIUM = "9999"
CARD ARRAY power( 40 ) ; table of powers up to the fourth power ( 1:4, 0:9 )
CARD n, d, powerOfTen, count, length, v, p, dps, nsub, nprev
; compute the n-th powers of 0-9
FOR d = 0 TO 9 DO power( d ) = D OD
nsub = 10
nprev = 0
FOR n = 2 TO 4 DO
power( nsub ) = 0
FOR d = 1 TO 9 DO
power( nsub + d ) = power( nprev + d ) * d
OD
nprev = nsub
nsub ==+ 10
OD
; print the Disarium numbers up to 9999 or the 18th, whichever is sooner
powerOfTen = 10
length = 1
count = 0 n = 0
WHILE n < MAX_DISARIUM AND count < 18 DO
IF n = powerOfTen THEN
; the number of digits just increased
powerOfTen ==* 10
length ==+ 1
FI
; form the digit power sum
v = n
p = length * 10;
dps = 0;
FOR d = 1 TO length DO
p ==- 10
dps ==+ power( p + ( v MOD 10 ) )
v ==/ 10
OD
IF dps = N THEN
; n is Disarium
count ==+ 1;
Put( ' )
PrintC( n )
FI
n ==+ 1
OD
RETURN
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
</pre>
=={{header|Ada}}==
<
procedure Disarium_Numbers is
Line 65 ⟶ 160:
exit when Count = Disarium_Count;
end loop;
end Disarium_Numbers;</
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|ALGOL 60}}==
{{works with|GNU Marst|Any - tested with release 2.7}}
{{Trans|ALGOL W}}
<syntaxhighlight lang="algol60">
begin comment find some Disarium numbers
- numbers whose digit position-power sums are equal to the
number, e.g. 135 = 1^1 + 3^2 + 5^3;
integer array power [ 1 : 9, 0 : 9 ];
integer count, powerOfTen, length, n, d;
comment compute the nth powers of 0-9;
for d := 0 step 1 until 9 do power[ 1, d ] := d;
for n := 2 step 1 until 9 do begin
power[ n, 0 ] := 0;
for d := 1 step 1 until 9 do power[ n, d ] := power[ n - 1, d ] * d
end n;
comment print the first few Disarium numbers;
count := 0;
powerOfTen := 10;
length := 1;
n := -1;
for n := n + 1 while count < 19 do begin
integer v, dps, p;
if n = powerOfTen then begin
comment the number of digfits just increased;
powerOfTen := powerOfTen * 10;
length := length + 1
end;
comment form the digit power sum;
v := n;
dps := 0;
for p := length step -1 until 1 do begin
dps := dps + power[ p, v - ( ( v % 10 ) * 10 ) ];
v := v % 10
end p;
if dps = n then begin
comment n is Disarium;
count := count + 1;
outinteger( 1, n )
end
end n
end
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|ALGOL 68}}==
Finds the first 19 Disarium numbers - to find the 20th would require a lot more time and also the table of digit powers would need to be increased to at least 20th powers (and 64 bit integers would be required).
<
# are equal to the number, e.g. 135 = 1^1 + 3^2 + 5^3 #
# compute the nth powers of 0-9 #
Line 108 ⟶ 250:
FI
OD
END</
{{out}}
<pre>
Line 116 ⟶ 258:
=={{header|ALGOL W}}==
{{Trans|ALGOL 68}}
<
% are equal to the number, e.g. 135 = 1^1 + 3^2 + 5^3 %
integer array power ( 1 :: 9, 0 :: 9 );
Line 154 ⟶ 296:
n := n + 1
end
end.</
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Amazing Hopper}}==
<syntaxhighlight lang="c">
#include <basico.h>
#proto encontrarunDisariumpara(_X_)
#synon _encontrarunDisariumpara siencontréunDisarium
algoritmo
decimales '0'
iterar para ( n=3000000, n, --n )
si encontré un Disarium 'n', entonces{
imprimir( #(utf8("El número ")),n," es Disarium\n")
}
siguiente
terminar
subrutinas
encontrar un Disarium para (n)
i=0
n, obtener tamaño parte entera, mover a 'i'
m=0, tn=n, d=0
iterar mientras ( tn )
último dígito de 'tn', mover a 'd,tn'
d, elevado a 'i', más 'm'
mover a 'm'
--i
reiterar
retornar ' #(m==n) '
</syntaxhighlight>
{{out}}
<pre>
El número 2646798 es Disarium
El número 2427 es Disarium
El número 1676 es Disarium
El número 1306 es Disarium
El número 598 es Disarium
El número 518 es Disarium
El número 175 es Disarium
El número 135 es Disarium
El número 89 es Disarium
El número 9 es Disarium
El número 8 es Disarium
El número 7 es Disarium
El número 6 es Disarium
El número 5 es Disarium
El número 4 es Disarium
El número 3 es Disarium
El número 2 es Disarium
El número 1 es Disarium
</pre>
=={{header|APL}}==
<syntaxhighlight lang="APL">(⊢(/⍨)(⊢=(⍎¨(+/*)⍳∘⍴)∘⍕)¨)0,⍳3000000</syntaxhighlight>
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|AppleScript}}==
This returns the first 19 Disarium numbers.
<syntaxhighlight lang="applescript">on isDisarium(n)
set temp to n
set digitCount to 1
repeat while (temp > 9)
set temp to temp div 10
set digitCount to digitCount + 1
end repeat
set temp to n
set sum to 0
repeat with position from digitCount to 2 by -1
set sum to sum + (temp mod 10) ^ position
set temp to temp div 10
end repeat
return (sum + temp = n)
end isDisarium
local Disaria, n
set Disaria to {}
set n to 0
repeat until ((count Disaria) = 19)
if (isDisarium(n)) then set end of Disaria to n
set n to n + 1
end repeat
return Disaria</syntaxhighlight>
{{output}}
<syntaxhighlight lang="applescript">{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798}</syntaxhighlight>
=={{header|Arturo}}==
<
j: 0
psum: sum map digits x 'dig [
Line 179 ⟶ 413:
]
i: i + 1
]</
{{out}}
Line 203 ⟶ 437:
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f DISARIUM_NUMBERS.AWK
BEGIN {
Line 228 ⟶ 462:
return((sum == n) ? 1 : 0)
}
</syntaxhighlight>
{{out}}
<pre>
The first 19 Disarium numbers:
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|BASIC}}==
==={{header|BASIC256}}===
<
digitos = length(string(n))
suma = 0
Line 307 ⟶ 493:
n += 1
end while
end</
{{out}}
<pre>Same as FreeBASIC entry.</pre>
==={{header|Chipmunk Basic}}===
{{works with|Chipmunk Basic|3.6.4}}
<syntaxhighlight lang="vbnet">100 cls
110 sub isdisarium(n)
120 digitos = len(str$(n))
130 suma = 0
140 x = n
150 while x <> 0
160 r = (x mod 10)
170 suma = suma+(r^digitos)
180 digitos = digitos-1
190 x = int(x/10)
200 wend
210 if suma = n then isdisarium = true else isdisarium = false
220 end sub
230 '
240 limite = 19
250 cnt = 0
260 n = 0
270 print "The first ";limite;" Disarium numbers are:"
280 while cnt < limite
290 if isdisarium(n) then
300 print n;" ";
310 cnt = cnt+1
320 endif
330 n = n+1
340 wend
350 end</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
==={{header|FreeBASIC}}===
<
Function isDisarium(n As Integer) As Boolean
Line 336 ⟶ 551:
n += 1
Loop
Sleep</
{{out}}
<pre>Same as Python entry.</pre>
==={{header|Run BASIC}}===
<
digitos = len(str$(n))
suma = 0 : x = n
Line 363 ⟶ 576:
end if
n = n + 1
wend</
{{out}}
<pre> 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427</pre>
==={{header|True BASIC}}===
<
LET digitos = LEN(str$(n))
LET suma = 0
Line 392 ⟶ 605:
LET n = n + 1
LOOP
END</
{{out}}
<pre> 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427</pre>
==={{header|PureBasic}}===
<
digitos.i = Len(Str(n))
suma.i = 0
Line 427 ⟶ 640:
Wend
Input()
CloseConsole()</
{{out}}
<pre>Same as FreeBASIC entry.</pre>e>
==={{header|Yabasic}}===
<
print "The first", limite, " Disarium numbers are:"
while cont < limite
Line 455 ⟶ 666:
wend
if suma = n then return True else return False : fi
end sub</
{{out}}
<pre> 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427</pre>
=={{header|bc}}==
<syntaxhighlight lang="freebasic">
define is_disarium (num) {
n = num
sum = 0
len = length(n)
while (n > 0) {
sum += (n % 10) ^ len
n = n/10
len -= 1
}
return (sum == num)
}
count = 0
i = 0
while (count < 19) {
if (is_disarium(i)) {
print i, "\n"
count += 1
}
i += 1
}
quit
</syntaxhighlight>
{{out}}
<pre>
0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798
</pre>
=={{header|BCPL}}==
<syntaxhighlight lang="BCPL">get "libhdr"
let length(n) = n < 10 -> 1,
length(n/10) + 1
let pow(b, e) = e = 0 -> 1,
b * pow(b, e-1)
let dps(n) = dpsl(n, length(n))
and dpsl(n, p) = n = 0 -> 0,
pow(n rem 10, p) + dpsl(n/10, p-1)
let disarium(n) = dps(n) = n
let start() be
for n=0 to 2500 if disarium(n)
do writef("%N*N", n)</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427</pre>
=={{header|BQN}}==
<syntaxhighlight lang="BQN">Digits ← {𝕊 0: ⟨⟩; (𝕊⌊𝕩÷10)∾10|𝕩}
DigitPowerSum ← (+´⊢⋆1+↕∘≠)∘Digits
Disarium ← ⊢=DigitPowerSum
Disarium¨⊸/ ↕2500</syntaxhighlight>
{{out}}
<pre>⟨ 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 ⟩</pre>
=={{header|C}}==
<syntaxhighlight lang="c">#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int power (int base, int exponent) {
int result = 1;
for (int i = 1; i <= exponent; i++) {
result *= base;
}
return result;
}
int is_disarium (int num) {
int n = num;
int sum = 0;
int len = n <= 9 ? 1 : floor(log10(n)) + 1;
while (n > 0) {
sum += power(n % 10, len);
n /= 10;
len--;
}
return num == sum;
}
int main() {
int count = 0;
int i = 0;
while (count < 19) {
if (is_disarium(i)) {
printf("%d ", i);
count++;
}
i++;
}
printf("%s\n", "\n");
}
</syntaxhighlight>
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|C#}}==
{{trans|Java}}
<syntaxhighlight lang="C#">
using System;
class DisariumNumbers {
// Method to check if a number is a Disarium number
public static bool IsDisarium(int num) {
int n = num;
int len = num.ToString().Length;
int sum = 0;
int i = 1;
while (n > 0) {
// C# does not support implicit conversion from double to int, so we explicitly convert the result of Math.Pow to int
sum += (int)Math.Pow(n % 10, len - i + 1);
n /= 10;
i++;
}
return sum == num;
}
static void Main(string[] args) {
int i = 0;
int count = 0;
// Find and print the first 19 Disarium numbers
while (count <= 18) {
if (IsDisarium(i)) {
Console.Write($"{i} ");
count++;
}
i++;
}
Console.WriteLine();
}
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|C++}}==
<
#include <iostream>
#include <cmath>
Line 498 ⟶ 890:
std::cout << std::endl ;
return 0 ;
}</
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427</pre>
=={{header|CLU}}==
<syntaxhighlight lang="clu">is_disarium = proc (n: int) returns (bool)
digits: array[int] := array[int]$[]
number: int := n
while n > 0 do
array[int]$addl(digits, n//10)
n := n / 10
end
array[int]$set_low(digits, 1)
digit_power_sum: int := 0
for i: int in array[int]$indexes(digits) do
digit_power_sum := digit_power_sum + digits[i] ** i
end
return(digit_power_sum = number)
end is_disarium
disaria = iter (amount: int) yields (int)
n: int := 0
while amount > 0 do
if is_disarium(n) then
amount := amount - 1
yield(n)
end
n := n + 1
end
end disaria
start_up = proc ()
po: stream := stream$primary_output()
for n: int in disaria(19) do
stream$putl(po, int$unparse(n))
end
end start_up</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798</pre>
=={{header|COBOL}}==
<syntaxhighlight lang="cobol"> IDENTIFICATION DIVISION.
PROGRAM-ID. DISARIUM.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 VARIABLES.
03 CANDIDATE PIC 9(9).
03 DIGITS PIC 9 OCCURS 9 TIMES, REDEFINES CANDIDATE.
03 IDX PIC 99.
03 EXPONENT PIC 99.
03 DGT-POWER PIC 9(9).
03 DGT-POWER-SUM PIC 9(9).
03 CAND-OUT PIC Z(8)9.
03 AMOUNT PIC 99 VALUE 18.
PROCEDURE DIVISION.
BEGIN.
PERFORM DISARIUM-TEST VARYING CANDIDATE FROM ZERO BY 1
UNTIL AMOUNT IS ZERO.
STOP RUN.
DISARIUM-TEST.
MOVE ZERO TO DGT-POWER-SUM.
MOVE 1 TO EXPONENT, IDX.
INSPECT CANDIDATE TALLYING IDX FOR LEADING ZEROES.
PERFORM ADD-DIGIT-POWER UNTIL IDX IS GREATER THAN 9.
IF DGT-POWER-SUM IS EQUAL TO CANDIDATE,
MOVE CANDIDATE TO CAND-OUT,
DISPLAY CAND-OUT,
SUBTRACT 1 FROM AMOUNT.
ADD-DIGIT-POWER.
COMPUTE DGT-POWER = DIGITS(IDX) ** EXPONENT.
ADD DGT-POWER TO DGT-POWER-SUM.
ADD 1 TO EXPONENT.
ADD 1 TO IDX.</syntaxhighlight>
{{out}}
<pre> 0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427</pre>
=={{header|Comal}}==
<syntaxhighlight lang="comal">0010 FUNC dps#(n#) CLOSED
0020 DIM digits#(10)
0030 length#:=0
0040 rest#:=n#
0050 WHILE rest#>0 DO
0060 length#:+1
0070 digits#(length#):=rest# MOD 10
0080 rest#:=rest# DIV 10
0090 ENDWHILE
0100 sum#:=0
0110 FOR i#:=1 TO length# DO
0120 sum#:+digits#(i#)^(length#-i#+1)
0130 ENDFOR i#
0140 RETURN sum#
0150 ENDFUNC dps#
0160 //
0170 amount#:=18
0180 num#:=0
0190 WHILE amount#>0 DO
0200 IF dps#(num#)=num# THEN
0210 amount#:-1
0220 PRINT num#
0230 ENDIF
0240 num#:+1
0250 ENDWHILE
0260 PRINT
0270 END
</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427</pre>
=={{header|Cowgol}}==
<syntaxhighlight lang="cowgol">include "cowgol.coh";
sub pow(base: uint8, exp: uint8): (power: uint32) is
power := 1;
while exp > 0 loop
power := power * base as uint32;
exp := exp - 1;
end loop;
end sub;
sub digit_power_sum(n: uint32): (dps: uint32) is
var digits: uint8[10]; # 2**32 has 10 digits
var digit := &digits[0];
var length: uint8 := 0;
while n > 0 loop
[digit] := (n % 10) as uint8;
digit := @next digit;
length := length + 1;
n := n / 10;
end loop;
dps := 0;
var power: uint8 := 1;
while power <= length loop
digit := @prev digit;
dps := dps + pow([digit], power);
power := power + 1;
end loop;
end sub;
var amount: uint8 := 19;
var candidate: uint32 := 0;
while amount > 0 loop
if digit_power_sum(candidate) == candidate then
amount := amount - 1;
print_i32(candidate);
print_nl();
end if;
candidate := candidate + 1;
end loop;</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798</pre>
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
Finds the first 19 numbers in 425 miliseconds. It uses a look up table for the powers and tests about 5 million numbers per second. However, this is not fast enough to find the 20th number. By my calculation, at this speed, it would only take 77,000 years. In other words, the brute force method can't be used to find the 20th number.
<syntaxhighlight lang="Delphi">
{Table to speed up calculating powers. Contains all the powers
of the digits 0..9 raised to the 0..21 power}
const PowersTable: array [0..21,0..9] of int64 = (
($01,$01,$01,$01,$01,$01,$01,$01,$01,$01),
($00,$01,$02,$03,$04,$05,$06,$07,$08,$09),
($00,$01,$04,$09,$10,$19,$24,$31,$40,$51),
($00,$01,$08,$1B,$40,$7D,$D8,$157,$200,$2D9),
($00,$01,$10,$51,$100,$271,$510,$961,$1000,$19A1),
($00,$01,$20,$F3,$400,$C35,$1E60,$41A7,$8000,$E6A9),
($00,$01,$40,$2D9,$1000,$3D09,$B640,$1CB91,$40000,$81BF1),
($00,$01,$80,$88B,$4000,$1312D,$44580,$C90F7,$200000,$48FB79),
($00,$01,$100,$19A1,$10000,$5F5E1,$19A100,$57F6C1,$1000000,$290D741),
($00,$01,$200,$4CE3,$40000,$1DCD65,$99C600,$267BF47,$8000000,$17179149),
($00,$01,$400,$E6A9,$100000,$9502F9,$39AA400,$10D63AF1,$40000000,$CFD41B91),
($00,$01,$800,$2B3FB,$400000,$2E90EDD,$159FD800,$75DB9C97,$200000000,$74E74F819),
($00,$01,$1000,$81BF1,$1000000,$E8D4A51,$81BF1000,$339014821,$1000000000,$41C21CB8E1),
($00,$01,$2000,$1853D3,$4000000,$48C27395,$30A7A6000,$168F08F8E7,$8000000000,$24FD3027FE9),
($00,$01,$4000,$48FB79,$10000000,$16BCC41E9,$123EDE4000,$9DE93ECE51,$40000000000,$14CE6B167F31),
($00,$01,$8000,$DAF26B,$40000000,$71AFD498D,$6D79358000,$45160B7A437,$200000000000,$BB41C3CA78B9),
($00,$01,$10000,$290D741,$100000000,$2386F26FC1,$290D7410000,$1E39A5057D81,$1000000000000,$6954FE21E3E81),
($00,$01,$20000,$7B285C3,$400000000,$B1A2BC2EC5,$F650B860000,$D39383266E87,$8000000000000,$3B3FCEF3103289),
($00,$01,$40000,$17179149,$1000000000,$3782DACE9D9,$5C5E45240000,$5C908960D05B1,$40000000000000,$2153E468B91C6D1),
($00,$01,$80000,$4546B3DB,$4000000000,$1158E460913D,$22A359ED80000,$287F3C1A5B27D7,$200000000000000,$12BF307AE81FFD59),
($00,$01,$100000,$CFD41B91,$10000000000,$56BC75E2D631,$CFD41B9100000,$11B7AA4B87E16E1,$1000000000000000,$A8B8B452291FE821),
($00,$01,$200000,$26F7C52B3,$40000000000,$1B1AE4D6E2EF5,$4DEF8A56600000,$7C05A810B72A027,$8000000000000000,$EE7E56E3721F2929));
function GetPower(X,Y: integer): int64;
{Extract power from table}
begin
Result:=PowersTable[Y,X];
end;
function IsDisarium(N: integer): boolean;
{Sum all powers of the digits raised to position power}
var S: string;
var I,J: integer;
var Sum: int64;
begin
Sum:=0;
S:=IntToStr(N);
for I:=1 to Length(S) do
begin
Sum:=Sum+GetPower(byte(S[I])-$30,I);
end;
Result:=Sum=N;
end;
procedure ShowDisariumNumbers(Memo: TMemo);
{Show Disarium numbers up to specified limit}
{Processes about 5 million numbers per second}
var I,Cnt: int64;
begin
Cnt:=0;
I:=0;
while I<High(int64) do
begin
if IsDisarium(I) then
begin
Inc(Cnt);
Memo.Lines.Add(IntToStr(Cnt)+': '+IntToStr(I));
if Cnt>=19 then break;
end;
Inc(I);
end;
end;
</syntaxhighlight>
{{out}}
<pre>
1: 0
2: 1
3: 2
4: 3
5: 4
6: 5
7: 6
8: 7
9: 8
10: 9
11: 89
12: 135
13: 175
14: 518
15: 598
16: 1306
17: 1676
18: 2427
19: 2646798
</pre>
=={{header|D}}==
<syntaxhighlight lang="d">import std.stdio;
import std.math;
import std.conv;
bool is_disarium(int num) {
int n = num;
int sum = 0;
ulong len = to!string(num, 10).length;
while (n > 0) {
sum += pow(n % 10, len);
n /= 10;
len--;
}
return num == sum;
}
void main() {
int i = 0;
int count = 0;
while (count < 19) {
if (is_disarium(i)) {
printf("%d ", i);
count++;
}
i++;
}
writeln(" ");
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Dart}}==
<syntaxhighlight lang="dart">import "dart:math";
import "dart:io";
void main() {
var count = 0;
var i = 0;
while (count < 19) {
if (is_disarium(i)) {
stdout.write("$i ");
count++;
}
i++;
}
}
bool is_disarium(numb) {
var n = numb;
var len = n.toString().length;
var sum = 0;
while (n > 0) {
sum += (pow(n % 10, len)).toInt();
n = (n / 10).toInt();
len--;
}
return numb == sum;
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|dc}}==
<syntaxhighlight lang="dc">
[10/ll1+sld0<Lx] sL [d10%ll^ls+ss10/ll1-sld0<Dx] sD[lc1+sc
lnp]sP[Osslisnln0sllLx0ssclnlDxlsln=Pli1+silc18>Ix] sI0si0sclIx
</syntaxhighlight>
{{out}}
<pre>
$ dc -e '[10/ll1+sld0<Lx] sL [d10%ll^ls+ss10/ll1-sld0<Dx] sD[lc1+sc
lnp]sP[Osslisnln0sllLx0ssclnlDxlsln=Pli1+silc18>Ix] sI0si0sclIx'
0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
</pre>
Note that I printed out only 18 disarium numbers because it is getting very slow.
Explained version of the above:
<syntaxhighlight lang="dc">
# Macro for computing the input number length
[10 # pushes 10 to stack
/ # divides input by 10 and stores result on stack
ll # push length on stack
1+ # add one to stack (length)
# p # prints intermediate length (for debugging)
sl # saves length to register l
d # duplicates value (number) on top of stack
0 # pushes 0 to stack
<Lx # executes length macro (L) if number > 0
] sL # end of length macro, store it in L
# is Disarium macro
[d # duplicates value (number) on top of stack
10 # pushes 10 to stack
% # pushes (number % 10) to stack
ll # pushes length to stack
^ # computes (n % 10) ^ len
ls # pushes sum to stack
+ss # computes new sum and stores it in s
10/ # integer division number / 10
ll # pushes length on stack
1- # subtract 1 froml length
sl # stores new length in l
d # duplicates value (number) on top of stack
0 # pushes 0 to stack
<Dx # executes recursively disarium macro (D) if number > 0
] sD # stores disarium macro in D
# Printing and counting macro
[lc1+sc # increments disarium number counter
lnp # print number
]sP # Stores printing macro in P
# Iteration macro
[Oss # stores 0 in register s (sum)
li sn # Stores iteration variable in number register
ln # pushes number to stack
0sl # stores 0 in register l (length)
lLx # runs the length macro
0ss # inititialize sum to 0
cln # clear stack and pushes number onto it
# llp # print the length
lDx # runs the Disarium macro once
lsln # pushes sum and number
=P # runs the printing macro if numbers are equal
li # loads iteration variable
1+si # increments iteration variable
lc18 # pushes counter and 18 on stack
>Ix # runs recursively iteration macro if counter < 18
] sI # end of iteration macro, stores it in I
# Main
0si # Initiate iteration variable
0sc # Initiate disarium numbers counter
lIx # running iteration macro the first time
</syntaxhighlight>
=={{header|Draco}}==
<syntaxhighlight lang="draco">proc nonrec pow(byte base, exp) word:
word p;
p := 1;
while exp>0 do
p := p*base;
exp := exp-1
od;
p
corp
proc nonrec disarium(word n) bool:
[5]byte digits;
short i, len;
word input_n, dps;
dps := 0;
i := 0;
input_n := n;
while n > 0 do
digits[i] := n % 10;
n := n / 10;
i := i + 1
od;
len := i;
for i from 0 upto len-1 do
dps := dps + pow(digits[i], len-i)
od;
dps = input_n
corp
proc nonrec main() void:
word n;
for n from 0 upto 2500 do
if disarium(n) then writeln(n:5) fi
od
corp</syntaxhighlight>
{{out}}
<pre> 0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427</pre>
=={{header|EasyLang}}==
<syntaxhighlight lang=easylang>
func disarium x .
h = x
while h > 0
d[] &= h mod 10
h = h div 10
.
for i = 1 to len d[]
h += pow d[i] (len d[] - i + 1)
.
return if h = x
.
while count < 19
if disarium n = 1
count += 1
print n
.
n += 1
.
</syntaxhighlight>
=={{header|Factor}}==
{{works with|Factor|0.99 2021-06-02}}
<
math.vectors prettyprint sequences ;
Line 512 ⟶ 1,466:
: disarium ( -- list ) 0 lfrom [ disarium? ] lfilter ;
19 disarium ltake [ pprint bl ] leach nl</
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|FOCAL}}==
<syntaxhighlight lang="focal">01.10 F N=0,2500;D 2
01.20 Q
02.10 D 3
02.20 I (N-S)2.4,2.3,2.4
02.30 T %5,N,!
02.40 R
03.10 S Z=N;S L=0
03.20 G 3.7
03.30 S K=FITR(Z/10)
03.40 S L=L+1
03.50 S D(L)=Z-K*10
03.60 S Z=K
03.70 I (-Z)3.3
03.80 S S=0
03.90 F I=1,L;S S=S+D(L-I+1)^I</syntaxhighlight>
{{out}}
<pre>= 0
= 1
= 2
= 3
= 4
= 5
= 6
= 7
= 8
= 9
= 89
= 135
= 175
= 518
= 598
= 1306
= 1676
= 2427</pre>
=={{header|Forth}}==
<syntaxhighlight lang="forth">: pow 1 swap 0 ?do over * loop nip ;
: len 1 swap begin dup 10 >= while 10 / swap 1+ swap repeat drop ;
: dps 0 swap dup len
begin dup while
swap 10 /mod swap
2 pick pow
3 roll +
rot 1- rot
swap
repeat
2drop
;
: disarium dup dps = ;
: disaria 2700000 0 ?do i disarium if i . cr then loop ;
disaria
bye</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798 </pre>
=={{header|FutureBasic}}==
<syntaxhighlight lang="futurebasic">
long c = 0, n = 0, t, i
CFStringRef s
while ( c < 18 )
s = fn StringWithFormat(@"%ld",n)
t = 0
for i = 0 to len(s) - 1
t += intVal(mid(s,i,1))^(i+1)
next
if ( t == n )
print n
c++
end if
n++
wend
HandleEvents
</syntaxhighlight>
=={{header|Go}}==
Line 523 ⟶ 1,577:
Although Go has native unsigned 64 bit arithmetic, much quicker than I was expecting at a little under a minute.
<
import (
Line 680 ⟶ 1,734:
fmt.Println()
}
}</
{{out}}
Line 752 ⟶ 1,806:
=={{header|Haskell}}==
<
where
import Data.Char ( digitToInt)
Line 762 ⟶ 1,816:
solution :: [Int]
solution = take 18 $ filter isDisarium [0, 1 ..]
</
<pre>[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427]</pre>
=={{header|J}}==
<
disarium=:
I. disarium i. 27e5
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</syntaxhighlight>
=={{header|Java}}==
<syntaxhighlight lang="java">
import java.lang.Math;
public class DisariumNumbers {
public static boolean is_disarium(int num) {
int n = num;
int len = Integer.toString(n).length();
int sum = 0;
int i = 1;
while (n > 0) {
sum += Math.pow(n % 10, len - i + 1);
n /= 10;
i ++;
}
return sum == num;
}
public static void main(String[] args) {
int i = 0;
int count = 0;
while (count <= 18) {
if (is_disarium(i)) {
System.out.printf("%d ", i);
count++;
}
i++;
}
System.out.printf("%s", "\n");
}
}
</syntaxhighlight>{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|JavaScript}}==
<syntaxhighlight lang="javascript">
function is_disarium (num) {
let n = num
let len = n.toString().length
let sum = 0
while (n > 0) {
sum += (n % 10) ** len
n = parseInt(n / 10, 10)
len--
}
return num == sum
}
let count = 0
let i = 1
while (count < 18) {
if (is_disarium(i)) {
process.stdout.write(i + " ")
count++
}
i++
}
</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|jq}}==
{{Works with|jq}}
'''Also works with gojq, the Go implementation of jq'''
The naive algorithm is good enough to find the first 19 disarium numbers, ergo `limit(19; ...)` below.
<syntaxhighlight lang=jq>
# To take advantage of gojq's arbitrary-precision integer arithmetic:
def power($in;$b): reduce range(0;$b) as $i (1; . * $in);
# $n is assumed to be a non-negative integer
def is_disarium:
. as $n
| {$n, sum: 0, len: (tostring|length) }
| until (.n == 0;
.sum += power(.n % 10; .len)
| .n = (.n/10 | floor)
| .len -= 1 )
| .sum == $n ;
# Emit a stream ...
def disariums:
range(0; infinite) | select(is_disarium);
limit(19; disariums)
</syntaxhighlight>
{{output}}
<pre>
0
...
2646798
</pre>
=={{header|Julia}}==
<
function disariums(numberwanted)
Line 784 ⟶ 1,936:
println(disariums(19))
@time disariums(19)
</
<pre>
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]
0.555962 seconds (5.29 M allocations: 562.335 MiB, 10.79% gc time)
</pre>
=={{header|Kotlin}}==
<syntaxhighlight lang="kotlin">fun power(n: Int, exp: Int): Int {
return when {
exp > 1 -> n * power(n, exp-1)
exp == 1 -> n
else -> 1
}
}
fun is_disarium(num: Int): Boolean {
val n = num.toString()
var sum = 0
for (i in 1..n.length) {
sum += power (n[i-1] - '0', i)
}
return sum == num
}
fun main() {
var i = 0
var count = 0
while (count < 19) {
if (is_disarium(i)) {
print("$i ")
count++
}
i++
}
println("")
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Lua}}==
Like most other solutions, this stops at 19. Computation time aside, the 20th Disarium number is greater than 2^53, above which double-precision floating-point format (which is Lua's in-built number type) has insufficient precision to distinguish between one integer and the next.
<syntaxhighlight lang="lua">function isDisarium (x)
local str, sum, digit = tostring(x), 0
for pos = 1, #str do
digit = tonumber(str:sub(pos, pos))
sum = sum + (digit ^ pos)
end
return sum == x
end
local count, n = 0, 0
while count < 19 do
if isDisarium(n) then
count = count + 1
io.write(n .. " ")
end
n = n + 1
end</syntaxhighlight>
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|MAD}}==
<syntaxhighlight lang="MAD"> NORMAL MODE IS INTEGER
VECTOR VALUES FMT = $ I7*$
AMOUNT = 19
THROUGH TEST, FOR CAND=0, 1, AMOUNT.E.0
WHENEVER DISARI.(CAND)
PRINT FORMAT FMT,CAND
AMOUNT = AMOUNT-1
END OF CONDITIONAL
TEST CONTINUE
INTERNAL FUNCTION(N)
ENTRY TO LENGTH.
L = 0
THROUGH COUNT, FOR NN=N, 0, NN.E.0
L = L+1
COUNT NN = NN/10
FUNCTION RETURN L
END OF FUNCTION
INTERNAL FUNCTION(BASE,EXP)
ENTRY TO RAISE.
R = 1
THROUGH MUL, FOR E=EXP, -1, E.E.0
MUL R = R*BASE
FUNCTION RETURN R
END OF FUNCTION
INTERNAL FUNCTION(N)
ENTRY TO DISARI.
L = LENGTH.(N)
POWSUM = 0
THROUGH DGTLP, FOR NN=N, 0, NN.E.0
NX = NN/10
DG = NN-NX*10
POWSUM = POWSUM+RAISE.(DG,L)
L = L-1
DGTLP NN = NX
FUNCTION RETURN POWSUM.E.N
END OF FUNCTION
END OF PROGRAM</syntaxhighlight>
{{out}}
<pre> 0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<
DisariumQ[n_Integer] := Module[{digs},
digs = IntegerDigits[n];
Line 806 ⟶ 2,080:
,
{n, 0, \[Infinity]}
]][[2, 1]]</
{{out}}
<pre>{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798}</pre>
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">
/* Function that returns a list of digits given a nonnegative integer */
decompose(num) := block([digits, remainder],
digits: [],
while num > 0 do
(remainder: mod(num, 10),
digits: cons(remainder, digits),
num: floor(num/10)),
digits
)$
disariump(n):=block(
decompose(n),
makelist(%%[i]^i,i,length(%%)),
apply("+",%%),
if n=%% then true)$
disarium_count(len):=block([i:0,count:0,result:[]],
while count<len do (if disariump(i) then (result:endcons(i,result),count:count+1),i:i+1),
result)$
/*Test cases */
disarium_count(18);
</syntaxhighlight>
{{out}}
<pre>
[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427]
</pre>
=={{header|MiniScript}}==
<syntaxhighlight lang="miniscript">
isDisarium = function(n)
num = n
sum = 0
if num == 0 then return true
for i in range(ceil(log(n)), 1)
sum += (n % 10) ^ i
n = floor(n / 10)
end for
return num == sum
end function
foundCnt = 0
cnt = 0
while foundCnt < 19
if isDisarium(cnt) then
foundCnt += 1
print cnt
end if
cnt +=1
end while</syntaxhighlight>
{{out}}
<pre>
0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798
</pre>
=={{header|Miranda}}==
<syntaxhighlight lang="miranda">main :: [sys_message]
main = [Stdout (show (take 18 disaria)), Stdout "\n"]
disaria :: [num]
disaria = filter disarium [0..]
disarium :: num->bool
disarium n = n = sum (zipWith (^) (digits n) [1..])
digits :: num->[num]
digits 0 = [0]
digits n = reverse (digits' n)
where digits' 0 = []
digits' n = (n mod 10) : digits' (n div 10)
zipWith :: (* -> ** -> ***) -> [*] -> [**] -> [***]
zipWith f x y = map f' (zip2 x y)
where f' (x,y) = f x y </syntaxhighlight>
{{out}}
<pre>[0,1,2,3,4,5,6,7,8,9,89,135,175,518,598,1306,1676,2427]</pre>
=={{header|Modula-2}}==
<syntaxhighlight lang="modula2">MODULE DisariumNumbers;
FROM InOut IMPORT WriteLn, WriteCard;
CONST Max = 2500;
VAR n: CARDINAL;
PROCEDURE cpow(base, power: CARDINAL): CARDINAL;
VAR i, result: CARDINAL;
BEGIN
result := 1;
FOR i := 1 TO power DO
result := result * base
END;
RETURN result
END cpow;
PROCEDURE length(n: CARDINAL): CARDINAL;
VAR len: CARDINAL;
BEGIN
len := 1;
WHILE n > 10 DO
INC(len);
n := n DIV 10
END;
RETURN len
END length;
PROCEDURE digitpowersum(n: CARDINAL): CARDINAL;
VAR powsum, exp: CARDINAL;
BEGIN
powsum := 0;
FOR exp := length(n) TO 1 BY -1 DO
powsum := powsum + cpow(n MOD 10, exp);
n := n DIV 10
END;
RETURN powsum
END digitpowersum;
PROCEDURE disarium(n: CARDINAL): BOOLEAN;
BEGIN
RETURN digitpowersum(n) = n
END disarium;
BEGIN
FOR n := 0 TO Max DO
IF disarium(n) THEN
WriteCard(n, 5);
WriteLn
END
END
END DisariumNumbers.</syntaxhighlight>
{{out}}
<pre> 0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427</pre>
=={{header|Nim}}==
<syntaxhighlight lang="nim">import strutils
import math
proc is_disarium(num: int): bool =
let n = intToStr(num)
var sum = 0
for i in 0..len(n)-1:
sum += int((int(n[i])-48) ^ (i+1))
return sum == num
var i = 0
var count = 0
while count < 19:
if is_disarium(i):
stdout.write i, " "
count += 1
i += 1
echo ""
</syntaxhighlight>
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|OCaml}}==
<syntaxhighlight lang="ocaml">(* speed-optimized exponentiation; doesn't support exponents < 2 *)
let rec pow b n =
if n land 1 = 0
then if n = 2 then b * b else pow (b * b) (n lsr 1)
else if n = 3 then b * b * b else b * pow (b * b) (n lsr 1)
let is_disarium n =
let rec aux x f =
if x < 10
then f 2 x
else aux (x / 10) (fun l y -> f (succ l) (y + pow (x mod 10) l))
in
n = aux n Fun.(const id)
let () =
Seq.(ints 0 |> filter is_disarium |> take 19 |> iter (Printf.printf " %u%!"))
|> print_newline</syntaxhighlight>
{{out}}
<pre> 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|Odin}}==
<syntaxhighlight lang="Go">
package disarium
import "core:fmt"
import "core:math"
/* main block start */
main :: proc() {
fmt.print("\nThe first 18 Disarium numbers are:")
count, i: int
for count < 19 {
if is_disarium(i) {
fmt.print(" ", i)
count += 1
}
i += 1
}
fmt.println("")
} /* main block end */
/* proc definitions */
power :: proc(base, exponent: int) -> int {
result := 1
for _ in 1 ..= exponent {
result *= base
}
return result
}
is_disarium :: proc(num: int) -> bool {
n := num
sum := 0
len := n <= 9 ? 1 : cast(int)math.floor_f64(math.log10_f64(auto_cast n) + 1)
for n > 0 {
sum += power(n % 10, len)
n /= 10
len -= 1
}
return num == sum
}
</syntaxhighlight>
{{out}}
<pre>
The first 18 Disarium numbers are: 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Pascal}}==
==={{header|Free Pascal}}===
simply adding one by one and keep track of sums.
<syntaxhighlight lang="pascal">
program disarium;
//compile with fpc -O3 -Xs
{$IFDEF WINDOWS}
{$APPTYPE CONSOLE}
{$ENDIF}
{$IFDEF FPC}
{$Mode Delphi}
uses
sysutils;
{$ELSE}
uses
system.SysUtils;
{$ENDIF}
const
MAX_BASE = 16;
cDigits : array[0..MAX_BASE-1] of char =
('0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F');
MAX_DIGIT_CNT = 31;
type
tDgt_cnt= 0..MAX_DIGIT_CNT-1;
tdgtPows = array[tDgt_cnt,0..MAX_BASE] of Uint64;
tdgtMaxSumPot = array[tDgt_cnt] of Uint64;
tmyDigits = record
dgtPot : array[tDgt_cnt] of Uint64;
dgtSumPot : array[tDgt_cnt] of Uint64;
dgtNumber : UInt64;
digit : array[0..31] of byte;
dgtMaxLen : tDgt_cnt;
end;
const
UPPER_LIMIT = 100*1000*1002;
var
{$Align 32}
dgtPows :tdgtPows;
procedure InitMyPots(var mp :tdgtPows;base:int32);
var
pot,dgt:Uint32;
p : Uint64;
begin
fillchar(mp,SizeOf(mp),#0);
For dgt := 0 to BASE do
begin
p := dgt;
For pot in tDgt_cnt do
begin
mp[pot,dgt] := p;
p := p*dgt;
end;
end;
p := 0;
end;
procedure Out_Digits(var md:tmyDigits);
var
i : Int32;
Begin
with md do
begin
write('dgtNumber ',dgtNumber,' = ',dgtSumPot[0],' in Base ');
For i := dgtMaxLen-1 downto 0 do
write(cDigits[digit[i]]);
writeln;
end;
end;
procedure IncByOne(var md:tmyDigits;Base: Int32);inline;
var
PotSum : Uint64;
potBase: nativeInt;
dg,pot,idx : Int32;
Begin
with md do
begin
//first digit seperate
pot := dgtMaxLen-1;
dg := digit[0]+1;
if dg < BASE then
begin
inc(dgtNumber);
digit[0]:= dg;
dgtPot[0] := dgtPows[pot,dg];
dgtSumPot[0] := dgtSumPot[1] + dgtPot[0];
EXIT;
end;
dec(dgtNumber,Base-1);
digit[0]:= 0;
dgtPot[0]:= 0;
dgtSumPot[0] := dgtSumPot[1];
potbase := Base;
idx := 1;
dec(pot);
while pot >= 0 do
Begin
dg := digit[idx]+1;
if dg < BASE then
begin
inc(dgtNumber,potbase);
digit[idx]:= dg;
dgtPot[idx]:= dgtPows[pot,dg];
PotSum := dgtSumPot[idx+1];
//update sum
while idx>=0 do
begin
inc(PotSum,dgtPot[idx]);
dgtSumPot[idx] := PotSum;
dec(idx);
end;
EXIT;
end;
dec(dgtNumber,(dg-1)*PotBase);
potbase *= Base;
digit[idx]:= 0;
dgtPot[idx] := 0;
dec(pot);
inc(idx);
end;
For pot := idx downto 0 do
Begin
dgtPot[idx] :=0;
dgtSumPot[pot] := 1;
end;
digit[idx] := 1;
dgtPot[idx] :=1;
dgtMaxLen := idx+1;
dgtNumber := potbase;
end;
end;
procedure OneRun(var s: tmyDigits;base:UInt32;Limit:Int64);
var
i : int64;
cnt : Int32;
begin
Writeln('Base = ',base);
InitMyPots(dgtPows,base);
fillchar(s,SizeOf(s),#0);
s.dgtMaxLen := 1;
i := 0;
cnt := 0;
repeat
if s.dgtSumPot[0] = s.dgtNumber then
Begin
Out_Digits(s);
inc(cnt);
end;
IncByOne(s,base);
inc(i);
until (i>=Limit);
writeln ( i,' increments and found ',cnt);
end;
var
{$Align 32}
s : tmyDigits;
T0: TDateTime;
base: nativeInt;
Begin
base := 10;
T0 := time;
OneRun(s,base,2646799);
T0 := (time-T0)*86400;
writeln(T0:8:3,' s');
writeln;
base := 11;
T0 := time;
OneRun(s,base,100173172);
T0 := (time-T0)*86400;
writeln(T0:8:3,' s');
writeln;
{$IFDEF WINDOWS}
readln;
{$ENDIF}
end.
</syntaxhighlight>
{{out|@TIO.RUN}}
<pre>
Base = 10
dgtNumber 0 = 0 in Base 0
dgtNumber 1 = 1 in Base 1
dgtNumber 2 = 2 in Base 2
dgtNumber 3 = 3 in Base 3
dgtNumber 4 = 4 in Base 4
dgtNumber 5 = 5 in Base 5
dgtNumber 6 = 6 in Base 6
dgtNumber 7 = 7 in Base 7
dgtNumber 8 = 8 in Base 8
dgtNumber 9 = 9 in Base 9
dgtNumber 89 = 89 in Base 89
dgtNumber 135 = 135 in Base 135
dgtNumber 175 = 175 in Base 175
dgtNumber 518 = 518 in Base 518
dgtNumber 598 = 598 in Base 598
dgtNumber 1306 = 1306 in Base 1306
dgtNumber 1676 = 1676 in Base 1676
dgtNumber 2427 = 2427 in Base 2427
dgtNumber 2646798 = 2646798 in Base 2646798
2646799 increments and found 19
0.008 s
Base = 11
dgtNumber 0 = 0 in Base 0
dgtNumber 1 = 1 in Base 1
dgtNumber 2 = 2 in Base 2
dgtNumber 3 = 3 in Base 3
dgtNumber 4 = 4 in Base 4
dgtNumber 5 = 5 in Base 5
dgtNumber 6 = 6 in Base 6
dgtNumber 7 = 7 in Base 7
dgtNumber 8 = 8 in Base 8
dgtNumber 9 = 9 in Base 9
dgtNumber 10 = 10 in Base A
dgtNumber 27 = 27 in Base 25
dgtNumber 39 = 39 in Base 36
dgtNumber 109 = 109 in Base 9A
dgtNumber 126 = 126 in Base 105
dgtNumber 525 = 525 in Base 438
dgtNumber 580 = 580 in Base 488
dgtNumber 735 = 735 in Base 609
dgtNumber 1033 = 1033 in Base 85A
dgtNumber 1044 = 1044 in Base 86A
dgtNumber 2746 = 2746 in Base 2077
dgtNumber 59178 = 59178 in Base 40509
dgtNumber 63501 = 63501 in Base 43789
dgtNumber 100173171 = 100173171 in Base 515AA64A
100173172 increments and found 24
0.294 s
</pre>
=={{header|Perl}}==
<
use warnings;
Line 821 ⟶ 2,597:
last if 19 == @D;
}
print "@D\n";</
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|Phix}}==
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">limit</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">19</span>
Line 843 ⟶ 2,619:
<span style="color: #000000;">n</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>
<!--</
{{out}}
<pre>
Line 851 ⟶ 2,627:
===stretch===
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #000080;font-style:italic;">-- translation of https://github.com/rgxgr/Disarium-Numbers/blob/master/Disarium.c</span>
Line 981 ⟶ 2,757:
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%d disarium numbers found (%s)\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">count</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">elapsed</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">time</span><span style="color: #0000FF;">()-</span><span style="color: #000000;">t0</span><span style="color: #0000FF;">)})</span>
<!--</
{{out}}
<pre>
Line 1,033 ⟶ 2,809:
===Iterative approach===
{{trans|Python}}
<
Limit = 19,
D = [],
Line 1,058 ⟶ 2,834:
X := X div 10
end,
Sum == N.</
{{out}}
Line 1,073 ⟶ 2,849:
The cp solver and sat solvers takes about the same time for finding the 7-digits number 2646798, but the sat solver is much faster for checking longer numbers; it took almost 9 minutes to prove that there are no Disarium numbers of length 8..17.
<
% import cp.
Line 1,108 ⟶ 2,884:
to_num(List, Base, Num) =>
Len = length(List),
Num #= sum([List[I]*Base**(Len-I) : I in 1..Len]).</
{{out}}
Line 1,125 ⟶ 2,901:
(And finding the first 36 Disarium numbers in base 36 is even easier: 0..Z.)
=={{header|PicoLisp}}==
<pre>
(de disarium (N)
(let S 0
(for (I . N) (mapcar format (chop N))
(inc 'S (** N I)) )
(= N S) ) )
(let (N 0 C 0)
(until (= C 19)
(and
(disarium N)
(printsp N)
(inc 'C) )
(inc 'N) )
(prinl) )
</pre>
=={{header|Python}}==
<
def isDisarium(n):
Line 1,151 ⟶ 2,943:
print(n, end = " ")
cont += 1
n += 1</
{{out}}
<pre>The first 19 Disarium numbers are:
Line 1,159 ⟶ 2,951:
Based on...{{Trans|ALGOL 68}}... but as the original PL/M compiler only supports 8 and 16-bit unsigned integers, this stops after trying up to 9999 or finding 18 Disarium numbers. Also, PL/M only supports 1-dimensional arrays.<br>
{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)
<
/* SUMS ARE EQUAL TO THE NUMBER, E.G. 135 = 1^1 + 3^2 + 5^3 */
Line 1,228 ⟶ 3,020:
END;
EOF</
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
</pre>
=={{header|Quackery}}==
<syntaxhighlight lang="Quackery"> [ [ [] swap
[ 10 /mod
rot join swap
dup 0 = until ]
drop ] ] is digits ( n --> [ )
[ 0 over digits
witheach
[ i^ 1+ ** + ] = ] is disarium ( n --> b )
[ temp put [] 0
[ dup disarium if
[ dup dip join ]
1+
over size
temp share = until ]
drop ] is disariums ( n --> [ )
19 disariums echo</syntaxhighlight>
{{out}}
<pre>[ 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798 ]</pre>
=={{header|Raku}}==
Not an efficient algorithm. First 18 in less than 1/4 second. 19th in around 45 seconds. Pretty much unusable for the 20th.
<syntaxhighlight lang="raku"
put $disarium[^18];
put $disarium[18];</
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
2646798</pre>
=={{header|Refal}}==
<syntaxhighlight lang="refal">$ENTRY Go {
= <FindDisarium 19 0>;
};
Digits {
0 = ;
s.N, <Divmod s.N 10>: (s.R) s.D = <Digits s.R> s.D;
};
Pow {
s.N 0 = 1;
s.N s.P = <* s.N <Pow s.N <- s.P 1>>>;
};
PowSum {
() e.P = 0;
(s.I e.X) e.P = <+ <Pow s.I e.P> <PowSum (e.X) <+ e.P 1>>>;
e.X = <PowSum (e.X) 1>;
};
Disarium {
e.N, <PowSum <Digits e.N>>: e.N = True;
e.N = False;
};
FindDisarium {
0 s.N = ;
s.I s.N, <Disarium s.N>: {
True = <Prout s.N> <FindDisarium <- s.I 1> <+ s.N 1>>;
False = <FindDisarium s.I <+ s.N 1>>;
};
};</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798</pre>
=={{header|RPL}}==
≪ DUP →STR → digits
≪ 0
1 digits SIZE '''FOR''' j
digits j DUP SUB STR→ j ^ +
'''NEXT'''
==
≫ ≫ '<span style="color:blue">DSRM?</span>' STO
≪ → max
≪ { } 0
'''WHILE''' OVER SIZE max < '''REPEAT'''
'''IF''' DUP <span style="color:blue">DSRM?</span> '''THEN''' SWAP OVER + SWAP '''END'''
1 +
'''END''' DROP
≫ ≫ '<span style="color:blue">DSRMN</span>' STO
18 <span style="color:blue">DSRMN</span>
{{out}}
<pre>
{ 0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 }
</pre>
=={{header|Ruby}}==
<syntaxhighlight lang="ruby">disariums = Enumerator.new do |y|
(0..).each do |n|
i = 0
y << n if n.digits.reverse.sum{|d| d ** (i+=1) } == n
end
end
puts disariums.take(19).to_a.join(" ")
</syntaxhighlight>
{{out}}
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|Rust}}==
<syntaxhighlight lang="rust">fn power(n: i32, exp: i32) -> i32 {
let mut result = 1;
for _i in 0..exp {
result *= n;
}
return result;
}
fn is_disarium(num: i32) -> bool {
let mut n = num;
let mut sum = 0;
let mut i = 1;
let len = num.to_string().len();
while n > 0 {
sum += power(n % 10, len as i32 - i + 1);
n /= 10;
i += 1
}
return sum == num;
}
fn main() {
let mut i = 0;
let mut count = 0;
while count <= 18 {
if is_disarium(i) {
print!("{} ", i);
count += 1;
}
i += 1;
}
println!("{}", " ")
}
</syntaxhighlight>
<pre>0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798</pre>
=={{header|Ring}}==
<syntaxhighlight lang="ring">
i = 0
count = 0
while count < 19
if is_disarium(i)
see "" + i + " "
count++
ok
i++
end
see nl
func pow (base, exp)
result = 1
for i = 0 to exp - 1
result *= base
next
return result
func is_disarium (num)
n = "" + num
sum = 0
for i = 1 to len(n)
sum += pow (n[i] % 10, i)
next
return sum = num
</syntaxhighlight>
{{out}}
<pre>
$ ring ./disarium.ring
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|Scala}}==
<syntaxhighlight lang="scala">object Disarium extends App {
def power(base: Int, exp: Int): Int = {
var result = 1
for (i <- 1 to exp) {
result *= base
}
return result
}
def is_disarium(num: Int): Boolean = {
val digits = num.toString.split("")
var sum = 0
for (i <- 0 to (digits.size - 1)) {
sum += power(digits(i).toInt, i + 1)
}
return num == sum
}
var i = 0
var count = 0
while (count < 19) {
if (is_disarium(i)) {
count += 1
printf("%d ", i)
}
i += 1
}
println("")
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
</pre>
=={{header|SETL}}==
<syntaxhighlight lang="setl">program disarium_numbers;
loop for n in [0..2700000] | disarium n do
print(n);
end loop;
op disarium(n);
k := n;
digits := [[k mod 10, k div:= 10](1) : until k=0];
p := #digits+1;
powsum := +/[d ** (p -:= 1) : d in digits];
return powsum = n;
end op;
end program;</syntaxhighlight>
{{out}}
<pre>0
1
2
3
4
5
6
7
8
9
89
135
175
518
598
1306
1676
2427
2646798</pre>
=={{header|Sidef}}==
<
n.digits.flip.sum_kv{|k,d| d**(k+1) } == n
}
say 18.by(is_disarium)</
{{out}}
<pre>
Line 1,255 ⟶ 3,305:
</pre>
=={{header|
<syntaxhighlight lang="tcl">proc is_disarium {num} {
set n num
set sum 0
set i 1
set ch 1
foreach char [split $num {}] {
scan $char %d ch
set sum [ expr ($sum + $ch ** $i)]
incr i
}
return [ expr $num == $sum ? 1 : 0]
}
set i 0
set count 0
while { $count < 19 } {
if [ is_disarium $i ] {
puts -nonewline "${i} "
incr count
}
incr i
}
puts ""
</syntaxhighlight>
{{out}}
<pre>
$ time tclsh ./disarium.tcl
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798
real 1m59,638s
user 1m56,328s
sys 0m0,234s
}
</pre>
=={{header|V (Vlang)}}==
Recommend to build first `v -prod disarium.v` and then run `./disarium`
{{trans|Go}}
<syntaxhighlight lang="v (vlang)">import strconv
const dmax = 20 // maximum digits
Line 1,444 ⟶ 3,487:
println('')
}
}</
{{out}}
Line 1,509 ⟶ 3,552:
Found the first 20 Disarium numbers.
</pre>
=={{header|VTL-2}}==
{{Trans|ALGOL W}}
Finds the first 18 Disarium numbers - computes a table of digit powers up to the fourth power.
<syntaxhighlight lang="vtl2">1000 N=1
1010 D=0
1020 :N*10+D)=D
1030 D=D+1
1040 #=D<10*1020
1050 N=2
1060 :N*10)=0
1070 D=1
1080 :N*10+D)=:N-1*10+D)*D
1090 D=D+1
1100 #=D<10*1080
1120 N=N+1
1130 #=N<5*1060
2000 C=0
2010 T=10
2020 L=1
2030 N=0
2040 #=N=T=0*2070
2050 T=T*10
2060 L=L+1
2070 V=N
2080 P=L
2090 S=0
2100 V=V/10
2110 S=S+:P*10+%
2120 P=P-1
2130 #=V>1*(S-1<N)*2100
2140 #=S=N=0*2180
2150 C=C+1
2160 $=32
2170 ?=N
2180 N=N+1
2190 #=C<18*2040
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
</pre>
Line 1,518 ⟶ 3,603:
As a possible optimization, I tried caching all possible digit powers but there was no perceptible difference in running time for numbers up to 7 digits long.
<
var limit = 19
Line 1,535 ⟶ 3,620:
}
System.print("The first 19 Disarium numbers are:")
System.print(disarium)</
{{out}}
Line 1,551 ⟶ 3,636:
So, if the Phix example requires 48 minutes to find the 20th number, it would probably take Wren the best part of a day to do the same which is far longer than I have patience for.
<
var LIMIT = 19 // maximum number of Disariums to find
Line 1,686 ⟶ 3,771:
}
System.print()
}</
{{out}}
Line 1,739 ⟶ 3,824:
I haven't bothered to search all 20 digits numbers up to the unsigned 64 limit as this would take far longer and, of course, be fruitless in any case.
<
var DMAX = 20 // maxmimum digits
Line 1,900 ⟶ 3,985:
}
System.print()
}</
{{out}}
Line 1,973 ⟶ 4,058:
=={{header|XPL0}}==
1.35 seconds on Pi4.
<
int N, N0, D(10), A(10), I, J, Sum;
[N0:= N;
Line 1,999 ⟶ 4,084:
N:= N+1;
];
]</
{{out}}
| |||