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Largest proper divisor of n: Difference between revisions

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Largest proper divisor of n (view source)

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Line 7:
{{trans|Python}}
 
<langsyntaxhighlight lang="11l">F lpd(n)
L(i) (n - 1 .< 0).step(-1)
I n % i == 0
Line 14:
 
L(i) 1..100
print(‘#3’.format(lpd(i)), end' I i % 10 == 0 {"\n"} E ‘’)</langsyntaxhighlight>
 
{{out}}
Line 31:
 
=={{header|Ada}}==
<langsyntaxhighlight Adalang="ada">with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
 
Line 55:
end if;
end loop;
end Main;</langsyntaxhighlight>
{{out}}
<pre>
Line 71:
 
=={{header|ALGOL 68}}==
<langsyntaxhighlight lang="algol68">FOR n TO 100 DO # show the largest proper divisors for n = 1..100 #
INT largest proper divisor := 1;
FOR j FROM ( n OVER 2 ) BY -1 TO 2 WHILE largest proper divisor = 1 DO
Line 81:
IF n MOD 10 = 0 THEN print( ( newline ) ) FI
OD
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 97:
 
=={{header|ALGOL W}}==
<langsyntaxhighlight lang="algolw">for n := 1 until 100 do begin % show the largest proper divisors for n = 1..100 %
for j := n div 2 step -1 until 2 do begin
if n rem j = 0 then begin
Line 107:
foundLargestProperDivisor:
if n rem 10 = 0 then write()
end for_n.</langsyntaxhighlight>
{{out}}
<pre>
Line 124:
=={{header|APL}}==
{{works with|Dyalog APL}}
<langsyntaxhighlight lang="apl">(⌈/1,(⍸0=¯1↓⍳|⊢))¨10 10⍴⍳100</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 140:
Most of this code is just to prepare the output for display. :D
 
<langsyntaxhighlight lang="applescript">on largestProperDivisor(n)
if (n mod 2 = 0) then return n div 2
if (n mod 3 = 0) then return n div 3
Line 179:
end task
 
task(100)</langsyntaxhighlight>
 
{{output}}
<langsyntaxhighlight lang="applescript">"1:1 2:1 3:1 4:2 5:1 6:3 7:1 8:4 9:3 10:5
11:1 12:6 13:1 14:7 15:5 16:8 17:1 18:9 19:1 20:10
21:7 22:11 23:1 24:12 25:5 26:13 27:9 28:14 29:1 30:15
Line 191:
71:1 72:36 73:1 74:37 75:25 76:38 77:11 78:39 79:1 80:40
81:27 82:41 83:1 84:42 85:17 86:43 87:29 88:44 89:1 90:45
91:13 92:46 93:31 94:47 95:19 96:48 97:1 98:49 99:33 100:50 "</langsyntaxhighlight>
 
 
Line 198:
Composing functionally, for rapid drafting and refactoring, with higher levels of code reuse:
 
<langsyntaxhighlight lang="applescript">use framework "Foundation"
 
 
Line 446:
item 1 of ((ca's NSArray's arrayWithObject:nsValue) as list)
end if
end unwrap</langsyntaxhighlight>
{{Out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 461:
=={{header|Arturo}}==
 
<langsyntaxhighlight lang="rebol">loop split.every:10 [1] ++ map 2..100 'x -> last chop factors x 'row [
print map to [:string] row 'r -> pad r 5
]</langsyntaxhighlight>
 
{{out}}
Line 479:
 
=={{header|AWK}}==
<syntaxhighlight lang="awk">
<lang AWK>
# syntax: GAWK -f LARGEST_PROPER_DIVISOR_OF_N.AWK
# converted from C
Line 501:
}
}
</syntaxhighlight>
</lang>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
Line 515 ⟶ 514:
13 46 31 47 19 48 1 49 33 50
 
Largest proper divisor of n 1-100</pre>
</pre>
 
=={{header|BASIC}}==
{{works with|QBasic|1.1}}
<lang basic>10 DEFINT A-Z
{{works with|QuickBasic|4.5}}
<syntaxhighlight lang="basic">10 DEFINT A-Z
20 FOR I=1 TO 100
30 IF I=1 THEN PRINT " 1";: GOTO 70
Line 526:
60 NEXT J
70 IF I MOD 10=0 THEN PRINT
80 NEXT I</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 539:
13 46 31 47 19 48 1 49 33 50</pre>
 
==={{header|BASIC256ANSI BASIC}}===
{{trans|FreeBASIC}}
{{works with|Decimal BASIC}}
<lang BASIC256>print "El mayor divisor propio de n es:\n"
{{works with|IS-BASIC}}
<syntaxhighlight lang="basic">
100 REM Largest proper divisor of n
110 PRINT "The largest proper divisor of n is:"
120 PRINT
130 PRINT USING " ## ##": 1, 1;
140 FOR I = 3 TO 100
150 FOR J = I - 1 TO 1 STEP -1
160 IF MOD(I, J) = 0 THEN
170 PRINT USING "###": J;
180 EXIT FOR
190 END IF
200 NEXT J
210 IF MOD(I, 10) = 0 THEN PRINT
220 NEXT I
230 END
</syntaxhighlight>
{{out}}
<pre>
The largest proper divisor of n is:
 
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
==={{header|Applesoft BASIC}}===
{{works with|Quite BASIC}}
Solution [[#Quite_BASIC|Quite BASIC]] work without changes.
 
==={{header|BASIC256}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="basic256">print "El mayor divisor propio de n es:\n"
print " 1 1 ";
for i = 3 to 100
Line 552 ⟶ 592:
if i % 10 = 0 then print
next i
end</langsyntaxhighlight>
 
==={{header|Chipmunk Basic}}===
{{works with|Chipmunk Basic|3.6.4}}
{{trans|FreeBASIC}}
<syntaxhighlight lang="qbasic">10 print "El mayor divisor propio de n es:"
20 print chr$(10)+" 1 1";
30 for i = 3 to 100
40 for j = i-1 to 1 step -1
50 if i mod j = 0 then print using "###";j; : exit for
60 next j
70 if i mod 10 = 0 then print
80 next i
90 end</syntaxhighlight>
 
==={{header|Craft Basic}}===
<syntaxhighlight lang="basic">print "Largest proper divisor of n is:"
print tab, "1", tab, "1",
 
for i = 3 to 100
 
for j = i - 1 to 1 step -1
 
if i mod j = 0 then
 
print tab, j,
break j
 
endif
 
wait
 
next j
 
if i mod 10 = 0 then
 
print
 
endif
 
next i</syntaxhighlight>
{{out| Output}}<pre>
Largest proper divisor of n is:
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
==={{header|FreeBASIC}}===
<syntaxhighlight lang="freebasic">Print !"El mayor divisor propio de n es:\n"
Print " 1 1";
For i As Byte = 3 To 100
For j As Byte = i-1 To 1 Step -1
If i Mod j = 0 Then Print Using "###"; j; : Exit For
Next j
If i Mod 10 = 0 Then Print
Next i
Sleep</syntaxhighlight>
{{out}}
<pre>El mayor divisor propio de n es:
 
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
==={{header|FutureBasic}}===
<syntaxhighlight lang="futurebasic">NSUInteger i, j
 
print " 1 1";
for i = 3 to 100
for j = i - 1 to 1 step - 1
if i mod j = 0 then print using "###"; j; : exit for
next
if i mod 10 = 0 then print
next
 
HandleEvents</syntaxhighlight>
{{output}}
<pre> 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
==={{header|GW-BASIC}}===
<syntaxhighlight lang="gwbasic">10 PRINT 1;
20 FOR I = 1 TO 101
30 FOR D = I\2 TO 1 STEP -1
40 IF I MOD D = 0 THEN PRINT D; : GOTO 60
50 NEXT D
60 NEXT I</syntaxhighlight>
 
==={{header|Gambas}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="vbnet">Public Sub Main()
Print "El mayor divisor propio de n es:\n"
Print " 1 1";
For i As Byte = 3 To 100
For j As Byte = i - 1 To 1 Step -1
If i Mod j = 0 Then
Print Format$(j, "###");
Break
End If
Next
If i Mod 10 = 0 Then Print
Next
End</syntaxhighlight>
 
==={{header|Minimal BASIC}}===
{{works with|BASICA}}
{{trans|FreeBASIC}}
{{works with|IS-BASIC}}
<syntaxhighlight lang="qbasic">100 PRINT "El mayor divisor propio de n es:"
110 PRINT
120 PRINT " 1 1 ";
130 FOR i = 3 TO 100
140 FOR j = i-1 TO 1 STEP -1
150 IF i-INT(i/j)*j = 0 THEN 200
160 NEXT j
170 IF i-INT(i/10)*10 = 0 THEN 220
180 NEXT i
190 GOTO 240
200 PRINT j;
210 GOTO 170
220 PRINT
230 GOTO 180
240 END</syntaxhighlight>
 
==={{header|MSX Basic}}===
{{works with|MSX BASIC|any}}
{{works with|GW-BASIC}}
{{trans|FreeBASIC}}
<syntaxhighlight lang="qbasic">10 PRINT "El mayor divisor propio de n es:"
20 PRINT CHR$(10) + " 1 1";
30 FOR I = 3 TO 100
40 FOR J = I-1 TO 1 STEP -1
50 IF I MOD J = 0 THEN PRINT USING "###"; J; : GOTO 70
60 NEXT J
70 IF I MOD 10 = 0 THEN PRINT
80 NEXT I
90 END</syntaxhighlight>
 
==={{header|Palo Alto Tiny BASIC}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="basic">
10 REM LARGEST PROPER DIVISOR OF N
20 PRINT "THE LARGEST PROPER DIVISOR OF N IS:"
30 PRINT;PRINT #3,1,1,
40 FOR I=3 TO 100
50 FOR J=I-1 TO 1 STEP -1
60 IF I=(I/J)*J PRINT #3,J,;GOTO 80
70 NEXT J
80 IF I=(I/10)*10 PRINT
90 NEXT I
100 STOP
</syntaxhighlight>
{{out}}
<pre>
THE LARGEST PROPER DIVISOR OF N IS:
 
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
==={{header|PureBasic}}===
<syntaxhighlight lang="purebasic">Procedure.i lpd(v.i)
For i=v/2 To 1 Step -1
If v%i=0 : ProcedureReturn i : EndIf
Next
ProcedureReturn 1
EndProcedure
 
If OpenConsole("")
For i=1 To 100
Print(RSet(Str(lpd(i)),3))
If i%10=0 : PrintN("") : EndIf
Next
Input()
EndIf</syntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
==={{header|Quite BASIC}}===
{{works with|Applesoft BASIC}}
{{trans|FreeBASIC}}
<syntaxhighlight lang="qbasic">100 PRINT "El mayor divisor propio de n es:"
110 PRINT : PRINT " 1 1";
120 FOR i = 3 TO 100
130 FOR j = i-1 TO 1 STEP -1
140 LET a = i-INT(i/j)*j
150 IF a = 0 THEN GOTO 210
160 IF a = 0 THEN GOTO 180
170 NEXT j
180 IF i-INT(i/10)*10 = 0 THEN PRINT
190 NEXT i
200 END
210 IF j < 10 THEN PRINT " "; j;
220 IF j >= 10 THEN PRINT " "; j;
230 GOTO 160</syntaxhighlight>
 
==={{header|Run BASIC}}===
<syntaxhighlight lang="vbnet">print "Largest proper divisor of n is:"
print chr$(10)+" 1 1";
for i = 3 to 100
for j = i-1 to 1 step -1
if i mod j = 0 then print using("###",j); : goto [exit]
next j
[exit]
if i mod 10 = 0 then print
next i
end</syntaxhighlight>
 
==={{Header|Tiny BASIC}}===
<syntaxhighlight lang="qbasic">REM Rosetta Code problem: https://rosettacode.org/wiki/Largest_proper_divisor_of_n
REM by Jjuanhdez, 05/2023
 
REM Largest proper divisor of n
 
PRINT "El mayor divisor propio de n es:"
PRINT ""
PRINT "1"
PRINT "1"
LET I = 3
10 IF I = 101 THEN GOTO 40
LET J = I-1
20 IF J = 0 THEN GOTO 30
LET A = I-(I/J)*J
IF A = 0 THEN PRINT J
IF A = 0 THEN GOTO 30
LET J = J-1
GOTO 20
30 IF I-(I/10)*10 = 0 THEN PRINT ""
LET I = I+1
GOTO 10
40 END
</syntaxhighlight>
 
==={{header|True BASIC}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="qbasic">PRINT "El mayor divisor propio de n es:"
PRINT
PRINT " 1 1";
FOR i = 3 To 100
FOR j = i-1 To 1 Step -1
IF remainder(i, j) = 0 Then
PRINT Using$("###", j);
EXIT FOR
END IF
NEXT j
IF remainder(i, 10) = 0 Then PRINT
NEXT i
END</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
 
==={{header|XBasic}}===
{{works with|Windows XBasic}}
{{trans|FreeBASIC}}
<syntaxhighlight lang="qbasic">PROGRAM "progname"
VERSION "0.0000"
 
DECLARE FUNCTION Entry ()
 
FUNCTION Entry ()
PRINT "El mayor divisor propio de n es:\n"
PRINT " 1 1 ";
FOR i = 3 TO 100
FOR j = i-1 TO 1 STEP -1
IF i MOD j = 0 THEN
PRINT FORMAT$("###", j);
EXIT FOR
END IF
NEXT j
IF i MOD 10 = 0 THEN PRINT
NEXT i
END FUNCTION
END PROGRAM</syntaxhighlight>
 
==={{header|Yabasic}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="yabasic">print "El mayor divisor propio de n es:\n"
print " 1 1 ";
for i = 3 to 100
for j = i-1 to 1 step -1
If mod(i, j) = 0 then print j using "##"; : break : fi
next j
if mod(i, 10) = 0 then print : fi
next i
print
end</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
 
=={{header|BCPL}}==
<langsyntaxhighlight lang="bcpl">get "libhdr"
 
let lpd(n) = valof
Line 566 ⟶ 934:
$( writed(lpd(i), 3)
if i rem 10=0 then wrch('*N')
$)</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 580 ⟶ 948:
 
=={{header|BQN}}==
<langsyntaxhighlight lang="bqn">(1⌈´↕×0=↕|⊢)¨ ∘‿10⥊1+↕100</langsyntaxhighlight>
{{out}}
<pre>┌─
Line 596 ⟶ 964:
 
=={{header|C}}==
<langsyntaxhighlight lang="c">#include <stdio.h>
 
unsigned int lpd(unsigned int n) {
Line 612 ⟶ 980:
}
return 0;
}</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 626 ⟶ 994:
 
=={{header|C++}}==
<langsyntaxhighlight lang="cpp">#include <cassert>
#include <iomanip>
#include <iostream>
Line 646 ⟶ 1,014:
<< (n % 10 == 0 ? '\n' : ' ');
}
}</langsyntaxhighlight>
 
{{out}}
Line 663 ⟶ 1,031:
 
=={{header|Cowgol}}==
<langsyntaxhighlight lang="cowgol">include "cowgol.coh";
 
sub print3(n: uint8) is
Line 696 ⟶ 1,064:
end if;
i := i + 1;
end loop;</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 708 ⟶ 1,076:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
=={{header|D}}==
 
{{trans|C}}
 
<syntaxhighlight lang="d">
import std.stdio;
import std.range;
import std.algorithm;
 
uint lpd(uint n) {
if (n <= 1) {
return 1;
}
 
auto divisors = array(iota(1, n).filter!(i => n % i == 0));
 
return divisors.empty ? 1 : divisors[$ - 1];
}
 
void main() {
foreach (i; 1 .. 101) {
writef("%3d", lpd(i));
 
if (i % 10 == 0) {
writeln();
}
}
}
</syntaxhighlight>
 
{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Dart}}==
<syntaxhighlight lang="dart">
import "dart:io";
 
num largest_proper_divisor(int n) {
assert(n > 0);
if ((n & 1) == 0) return n >> 1;
for (int p = 3; p * p <= n; p += 2) {
if (n % p == 0) return n / p;
}
return 1;
}
 
void main() {
print("El mayor divisor propio de n es:");
for (int n = 1; n < 101; ++n) {
stdout.write(largest_proper_divisor(n));
print(largest_proper_divisor(n) + n % 10 == 0 ? "\n" : " ");
}
}
</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
 
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
 
 
<syntaxhighlight lang="Delphi">
function FindProperDivisor(N: Integer): integer;
{Find the highest proper divisor}
{i.e. The highest number that evenly divides in N}
begin
if N=1 then Result:=1
else for Result:=N-1 downto 1 do
if (N mod Result)=0 then break;
end;
 
 
 
procedure AllProperDivisors(Memo: TMemo);
{Show all proper divisors for number 1..100}
var I: integer;
var S: string;
begin
S:='';
for I:=1 to 100 do
begin
S:=S+Format('%3d',[FindProperDivisor(I)]);
if (I mod 10)=0 then S:=S+#$0D#$0A;
end;
Memo.Text:=S;
end;
 
 
</syntaxhighlight>
{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
 
=={{header|Draco}}==
<langsyntaxhighlight lang="draco">proc nonrec lpd(word n) word:
word d;
if n=1 then
Line 727 ⟶ 1,211:
if n%10 = 0 then writeln() fi
od
corp</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 739 ⟶ 1,223:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
=={{header|EasyLang}}==
<syntaxhighlight>
func lpdiv v .
r = 1
for i = 2 to v div 2
if v mod i = 0
r = i
.
.
return r
.
for i = 1 to 100
write lpdiv i & " "
.
</syntaxhighlight>
 
=={{header|F_Sharp|F#}}==
<langsyntaxhighlight lang="fsharp">
// Largest proper divisor of n: Nigel Galloway. June 2nd., 2021
let fN g=let rec fN n=let i=Seq.head n in match(g/i,g%i) with (1,_)->1 |(n,0)->n |_->fN(Seq.tail n) in fN(Seq.initInfinite((+)2))
seq{yield 1; yield! seq{2..100}|>Seq.map fN}|>Seq.iter(printf "%d "); printfn ""
</syntaxhighlight>
</lang>
{{out}}
<pre>
Line 754 ⟶ 1,254:
=={{header|Factor}}==
{{works with|Factor|0.99 2021-02-05}}
<langsyntaxhighlight lang="factor">USING: grouping kernel math math.bitwise math.functions
math.ranges prettyprint sequences ;
 
Line 763 ⟶ 1,263:
: largest ( m -- n ) dup odd? [ odd ] [ 2/ ] if ;
 
100 [1,b] [ largest ] map 10 group simple-table.</langsyntaxhighlight>
{{out}}
<pre>
Line 779 ⟶ 1,279:
 
=={{header|Fermat}}==
<langsyntaxhighlight lang="fermat">Func Lpd(n) =
if n = 1 then Return(1) fi;
for i = n\2 to 1 by -1 do
Line 788 ⟶ 1,288:
!(Lpd(m):4);
if m|10=0 then ! fi;
od;</langsyntaxhighlight>
{{out}}<pre>
1 1 1 2 1 3 1 4 3 5
Line 803 ⟶ 1,303:
 
=={{header|FOCAL}}==
<langsyntaxhighlight lang="focal">01.10 F N=1,100;D 2;D 3
01.20 Q
 
Line 817 ⟶ 1,317:
03.20 S A=N/10
03.30 I (FITR(A)-A)3.4;T !
03.40 R</langsyntaxhighlight>
{{out}}
<pre>= 1= 1= 1= 2= 1= 3= 1= 4= 3= 5
Line 832 ⟶ 1,332:
=={{header|Forth}}==
{{works with|Gforth}}
<langsyntaxhighlight lang="forth">: largest-proper-divisor { n -- n }
n 1 and 0= if n 2/ exit then
3
Line 850 ⟶ 1,350:
 
main
bye</langsyntaxhighlight>
 
{{out}}
Line 867 ⟶ 1,367:
 
=={{header|Fortran}}==
<langsyntaxhighlight lang="fortran"> program LargestProperDivisors
implicit none
integer i, lpd
Line 885 ⟶ 1,385:
20 lpd = i
end if
end function</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
 
=={{header|FreeBASIC}}==
<lang freebasic>Print !"El mayor divisor propio de n es:\n"
Print " 1 1";
For i As Byte = 3 To 100
For j As Byte = i-1 To 1 Step -1
If i Mod j = 0 Then Print Using "###"; j; : Exit For
Next j
If i Mod 10 = 0 Then Print
Next i
Sleep</lang>
{{out}}
<pre>El mayor divisor propio de n es:
 
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
Line 926 ⟶ 1,401:
Frink contains efficient routines for factoring numbers. It uses trial division, wheel factoring, and Pollard rho factoring.
 
<langsyntaxhighlight lang="frink">for n = 1 to 100
println[last[allFactors[n,true,false]]]</langsyntaxhighlight>
 
=={{header|Go}}==
<langsyntaxhighlight lang="go">package main
 
import "fmt"
Line 956 ⟶ 1,431:
}
}
}</langsyntaxhighlight>
 
{{out}}
Line 972 ⟶ 1,447:
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|GW-BASIC}}==
<lang gwbasic>10 PRINT 1;
20 FOR I = 1 TO 101
30 FOR D = I\2 TO 1 STEP -1
40 IF I MOD D = 0 THEN PRINT D; : GOTO 60
50 NEXT D
60 NEXT I</lang>
 
=={{header|Haskell}}==
<langsyntaxhighlight lang="haskell">import Data.List.Split (chunksOf)
import Text.Printf (printf)
 
Line 992 ⟶ 1,459:
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . lpd <$> [1 .. 100]</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,011 ⟶ 1,478:
(Otherwise, the largest proper divisor will be 1 itself).
 
<langsyntaxhighlight lang="haskell">import Data.List (find)
import Data.List.Split (chunksOf)
import Text.Printf (printf)
Line 1,026 ⟶ 1,493:
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . maxProperDivisors <$> [1 .. 100]</langsyntaxhighlight>
 
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,040 ⟶ 1,507:
 
=={{header|J}}==
<syntaxhighlight lang ="j"> lpd =: (1 |}.!.1 ])&.q:</langsyntaxhighlight>
{{out}}
<pre>
lpd 1+>: i.100 5 20
1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10
1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 ...</pre>
7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 50
</pre>
 
This works by prime factorization of n, replacingremoving the largestsmallest prime factor with 1, and then taking the product of this new list of prime factors. In J terms, we work "under" factorization, in analogy to a medical operation where the patient is "under" anaethesiaanesthesia: (put patient to sleep) rearrange guts (wake patient up).
 
The core logic only concerns itself with a list of prime factors; it is never even aware of the original input integer, nor the final integer result. In fact, note that the final multiplication is implicit and never spelled out by the programmer; product is the inverse of factorization, and we requested to work "under" factorization, thus J's algebra knows to apply the inverse of factorization (i.e. taking the product) as the final step.
 
=={{header|Java}}==
<syntaxhighlight lang="java">
public final class LargestProperDivisor {
 
public static void main(String[] aArgs) {
for ( int n = 1; n < 101; n++ ) {
System.out.print(String.format("%2d%s", largestProperDivisor(n), ( n % 10 == 0 ? "\n" : " " )));
}
}
private static int largestProperDivisor(int aNumber) {
if ( aNumber < 1 ) {
throw new IllegalArgumentException("Argument must be >= 1: " + aNumber);
}
if ( ( aNumber & 1 ) == 0 ) {
return aNumber >> 1;
}
for ( int p = 3; p * p <= aNumber; p += 2 ) {
if ( aNumber % p == 0 ) {
return aNumber / p;
}
}
return 1;
}
 
}
</syntaxhighlight>
{{ out }}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|jq}}==
Line 1,056 ⟶ 1,570:
 
Naive version:
<langsyntaxhighlight lang="jq"># (1|largestpd) is 1 as per the task definition
def largestpd:
if . == 1 then 1
else . as $n
| first( range( ($n - ($n % 2)) /2; 0; -1) | (select($n % . == 0) ))
end;</langsyntaxhighlight>
 
Slightly less naive:
<langsyntaxhighlight lang="jq">def largestpd:
if . == 1 then 1
else . as $n
Line 1,071 ⟶ 1,585:
else first( range( ($n - ($n % 11)) /11; 0; -1) | (select($n % . == 0) ))
end
end;</langsyntaxhighlight>
<langsyntaxhighlight lang="jq"># For neatness
def lpad($len):
tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
Line 1,082 ⟶ 1,596:
### The task
[range(1; 101) | largestpd]
| nwise(10) | map(lpad(2)) | join(" ")</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,096 ⟶ 1,610:
 
=={{header|Julia}}==
<langsyntaxhighlight lang="julia">largestpd(n) = (for i in n÷2:-1:1 (n % i == 0) && return i; end; 1)
 
foreach(n -> print(rpad(largestpd(n), 3), n % 10 == 0 ? "\n" : ""), 1:100)
</langsyntaxhighlight>{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
Line 1,111 ⟶ 1,625:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Lua}}==
{{Trans|ALGOL 68}}
<syntaxhighlight lang="lua">
for n = 1, 100 do -- show the largest proper divisors for n = 1..100
local largestProperDivisor, j = 1, math.floor( n / 2 )
while j >= 2 and largestProperDivisor == 1 do
if n % j == 0 then
largestProperDivisor = j
end
j = j - 1
end
io.write( string.format( "%3d", largestProperDivisor ) )
if n % 10 == 0 then io.write( "\n" ) end
end
</syntaxhighlight>
{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|MAD}}==
<langsyntaxhighlight MADlang="mad"> NORMAL MODE IS INTEGER
INTERNAL FUNCTION(N)
Line 1,130 ⟶ 1,673:
VECTOR VALUES TABLE = $10(I3)*$
END OF PROGRAM </langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,144 ⟶ 1,687:
 
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<langsyntaxhighlight Mathematicalang="mathematica">Last[Prepend[DeleteCases[Divisors[#], #], 1]] & /@ Range[100]</langsyntaxhighlight>
{{out}}
<pre>{1,1,1,2,1,3,1,4,3,5,1,6,1,7,5,8,1,9,1,10,7,11,1,12,5,13,9,14,1,15,1,16,11,17,7,18,1,19,13,20,1,21,1,22,15,23,1,24,7,25,17,26,1,27,11,28,19,29,1,30,1,31,21,32,13,33,1,34,23,35,1,36,1,37,25,38,11,39,1,40,27,41,1,42,17,43,29,44,1,45,13,46,31,47,19,48,1,49,33,50}</pre>
 
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">
lpd(n):=if n=1 then 1 else listify(divisors(n))[length(divisors(n))-1]$
 
makelist(lpd(i),i,100);
</syntaxhighlight>
{{out}}
<pre>
[1,1,1,2,1,3,1,4,3,5,1,6,1,7,5,8,1,9,1,10,7,11,1,12,5,13,9,14,1,15,1,16,11,17,7,18,1,19,13,20,1,21,1,22,15,23,1,24,7,25,17,26,1,27,11,28,19,29,1,30,1,31,21,32,13,33,1,34,23,35,1,36,1,37,25,38,11,39,1,40,27,41,1,42,17,43,29,44,1,45,13,46,31,47,19,48,1,49,33,50]
</pre>
 
=={{header|Modula-2}}==
<langsyntaxhighlight lang="modula2">MODULE LargestProperDivisor;
FROM InOut IMPORT WriteCard, WriteLn;
 
Line 1,175 ⟶ 1,729:
END;
END;
END LargestProperDivisor.</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,189 ⟶ 1,743:
 
=={{header|Nim}}==
<langsyntaxhighlight Nimlang="nim">import math, strutils
 
func largestProperDivisor(n: Positive): int =
Line 1,197 ⟶ 1,751:
 
for n in 1..100:
stdout.write ($n.largestProperDivisor).align(2), if n mod 10 == 0: '\n' else: ' '</langsyntaxhighlight>
 
{{out}}
Line 1,212 ⟶ 1,766:
 
=={{header|Pascal}}==
<langsyntaxhighlight lang="pascal">
program LarPropDiv;
 
Line 1,241 ⟶ 1,795:
Writeln;
end;
end.</langsyntaxhighlight>
{{out}}
<pre>
Line 1,254 ⟶ 1,808:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
===Alternative (sieve)===
Most solutions use a function that returns the largest proper divisor of an individual integer. This console program, written in Free Pascal, applies a sieve to the integers 1..100 (or other limit).
<syntaxhighlight lang="pascal">
program LPD;
(*
Displays largest proper divisor for each integer in range 1..limit.
Command line:
LPD limit items_per_line
or LPD limit // items_per_line defaults to 10
or LPD // limit defaults to 100
*)
{$mode objfpc}{$H+}
 
uses SysUtils;
var
limit, items_per_line, nr_items, j, p : integer;
a : array of integer;
begin
// Set up defaults
limit := 100;
items_per_line := 10;
// Overwrite defaults with command-line parameters, if present
if ParamCount > 0 then
limit := SysUtils.StrToInt( ParamStr(1));
if ParamCount > 1 then
items_per_line := SysUtils.StrToInt( ParamStr(2));
WriteLn( 'Largest proper divisors 1..', limit);
// Dynamic arrays are 0-based. To keep it simple, we ignore a[0]
// and use a[j] for the integer j, 1 <= j <= limit
SetLength( a, limit + 1);
for j := 1 to limit do a[j] := 1; // stays at 1 if j is 1 or prime
 
// Sieve; if j is composite then a[j] := smallest prime factor of j
p := 2; // p = next prime
while p*p < limit do begin
j := 2*p;
while j <= limit do begin
if a[j] = 1 then a[j] := p;
inc( j, p);
end;
repeat
inc(p);
until (p > limit) or (a[p] = 1);
end;
 
// If j is composite, divide j by its smallest prime factor
for j := 1 to limit do
if a[j] > 1 then a[j] := j div a[j];
 
// Write the array to the console
nr_items := 0;
for j := 1 to limit do begin
Write( a[j]:5);
inc( nr_items);
if nr_items = items_per_line then begin
WriteLn;
nr_items := 0;
end;
end;
if nr_items > 0 then WriteLn;
end.
</syntaxhighlight>
{{out}}
<pre>
Largest proper divisors 1..100
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Perl}}==
{{libheader|ntheory}}
<langsyntaxhighlight lang="perl">use strict;
use warnings;
use ntheory 'divisors';
Line 1,275 ⟶ 1,905:
while( my @batch = $iter->() ) {
printf '%3d', $_ == 1 ? 1 : max proper_divisors($_) for @batch; print "\n";
}</langsyntaxhighlight>
{{out}}
<pre>GPD for 1 through 100:
Line 1,290 ⟶ 1,920:
 
=={{header|Phix}}==
<!--<langsyntaxhighlight Phixlang="phix">(phixonline)-->
<span style="color: #7060A8;">puts</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join_by</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">apply</span><span style="color: #0000FF;">(</span><span style="color: #004600;">true</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">sprintf</span><span style="color: #0000FF;">,{{</span><span style="color: #008000;">"%3d"</span><span style="color: #0000FF;">},</span><span style="color: #7060A8;">vslice</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">apply</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">apply</span><span style="color: #0000FF;">(</span><span style="color: #004600;">true</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">factors</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">tagset</span><span style="color: #0000FF;">(</span><span style="color: #000000;">100</span><span style="color: #0000FF;">),-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">}),</span><span style="color: #7060A8;">reverse</span><span style="color: #0000FF;">),</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)}),</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">10</span><span style="color: #0000FF;">,</span><span style="color: #008000;">""</span><span style="color: #0000FF;">))</span>
<!--</langsyntaxhighlight>-->
{{out}}
<pre>
Line 1,307 ⟶ 1,937:
</pre>
Alternative, same output, optimised: obviously checking for a factor from 2 up is going to be significantly faster than n-1 down... or of course as above collecting all of them and extracting/discarding all but the last, at least on much larger numbers, that is, and I suppose you could (perhaps) improve even further on this by only checking primes.
<!--<langsyntaxhighlight Phixlang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">100</span> <span style="color: #008080;">do</span>
Line 1,323 ⟶ 1,953:
<span style="color: #008080;">if</span> <span style="color: #7060A8;">remainder</span><span style="color: #0000FF;">(</span><span style="color: #000000;">n</span><span style="color: #0000FF;">,</span><span style="color: #000000;">10</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span> <span style="color: #7060A8;">puts</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"\n"</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<!--</langsyntaxhighlight>-->
 
=={{header|Picat}}==
<syntaxhighlight lang="picat">main =>
foreach(I in 1..100)
printf("%2d%s",a(I), cond(I mod 10 == 0,"\n", " "))
end.
 
a(1) = 1.
a(N) = Div =>
a(N,N//2,Div).
a(N,I,I) :-
N mod I == 0.
a(N,I,Div) :-
a(N,I-1,Div).</syntaxhighlight>
 
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
 
=={{header|PL/I}}==
<langsyntaxhighlight lang="pli">largestProperDivisor: procedure options(main);
lpd: procedure(n) returns(fixed);
declare (n, i) fixed;
Line 1,340 ⟶ 1,997:
if mod(i,10)=0 then put skip;
end;
end largestProperDivisor;</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,354 ⟶ 2,011:
 
=={{header|PL/M}}==
<langsyntaxhighlight lang="plm">100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
Line 1,385 ⟶ 2,042:
END;
CALL EXIT;
EOF</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
 
=={{header|PureBasic}}==
<lang PureBasic>Procedure.i lpd(v.i)
For i=v/2 To 1 Step -1
If v%i=0 : ProcedureReturn i : EndIf
Next
ProcedureReturn 1
EndProcedure
 
If OpenConsole("")
For i=1 To 100
Print(RSet(Str(lpd(i)),3))
If i%10=0 : PrintN("") : EndIf
Next
Input()
EndIf</lang>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,426 ⟶ 2,056:
 
=={{header|Python}}==
<langsyntaxhighlight lang="python">def lpd(n):
for i in range(n-1,0,-1):
if n%i==0: return i
Line 1,432 ⟶ 2,062:
 
for i in range(1,101):
print("{:3}".format(lpd(i)), end=i%10==0 and '\n' or '')</langsyntaxhighlight>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,448 ⟶ 2,078:
Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:
 
<langsyntaxhighlight lang="python">'''Largest proper divisor of'''
 
from math import isqrt
Line 1,519 ⟶ 2,149:
# MAIN ---
if __name__ == '__main__':
main()</langsyntaxhighlight>
{{Out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,535 ⟶ 2,165:
<code>factors</code> is defined at [http://rosettacode.org/wiki/Factors_of_an_integer#Quackery Factors of an integer].
 
<langsyntaxhighlight Quackerylang="quackery">' [ 1 ] 99 times
[ i^ 2 + factors
-2 peek join ]
echo</langsyntaxhighlight>
 
{{out}}
Line 1,545 ⟶ 2,175:
 
=={{header|R}}==
<langsyntaxhighlight Rlang="r">largest_proper_divisor <- function(n){
if(n == 1) return(1)
Line 1,563 ⟶ 2,193:
largest_proper_divisor(i)
}
</syntaxhighlight>
</lang>
 
{{out}}
Line 1,591 ⟶ 2,221:
 
Note: this example is ridiculously overpowered (and so, somewhat inefficient) for a range of 1 to 100, but will easily handle large numbers without modification.
<syntaxhighlight lang="raku" perl6line>use Prime::Factor;
 
for 0, 2**67 - 1 -> $add {
Line 1,604 ⟶ 2,234:
 
say (now - $start).fmt("%0.3f seconds\n");
}</langsyntaxhighlight>
{{out}}
<pre>GPD for 1 through 100:
Line 1,636 ⟶ 2,266:
 
This addition made it about &nbsp; '''75%''' &nbsp; faster.
<langsyntaxhighlight lang="rexx">/*REXX program finds the largest proper divisors of all numbers (up to a given limit). */
parse arg n cols . /*obtain optional argument from the CL.*/
if n=='' | n=="," then n= 101 /*Not specified? Then use the default.*/
Line 1,662 ⟶ 2,292:
if x//k==0 then return k /*Remainder=0? Got largest proper div.*/
end /*k*/
return 1 /*If we get here, then X is a prime.*/</langsyntaxhighlight>
{{out|output|text=&nbsp; when using the default inputs:}}
<pre>
Line 1,681 ⟶ 2,311:
 
=={{header|Ring}}==
<langsyntaxhighlight lang="ring">? "working..."
limit = 100
? "Largest proper divisor up to " + limit + " are:"
Line 1,695 ⟶ 2,325:
if col++ % 10 = 0 ? "" ok
next
? "done..."</langsyntaxhighlight>
{{out}}
<pre>working...
Line 1,710 ⟶ 2,340:
13 46 31 47 19 48 1 49 33 50
done...</pre>
 
=={{header|RPL}}==
≪ '''IF''' DUP 1 ≠ '''THEN''' DUP '''DO''' 1 - '''UNTIL''' DUP2 MOD NOT '''END''' SWAP DROP '''END''' ≫ '<span style="color:blue">LPROPDIV</span>' STO
≪ { } 1 100 '''FOR''' j j <span style="color:blue">LPROPDIV</span> + '''NEXT''' ≫ EVAL
{{out}}
<pre>
1: { 1 1 1 2 1 3 1 4 3 5 1 6 1 7 5 8 1 9 1 10 7 11 1 12 5 13 9 14 1 15 1 16 11 17 7 18 1 19 13 20 1 21 1 22 15 23 1 24 7 25 17 26 1 27 11 28 19 29 1 30 1 31 21 32 13 33 1 34 23 35 1 36 1 37 25 38 11 39 1 40 27 41 1 42 17 43 29 44 1 45 13 46 31 47 19 48 1 49 33 50 }
</pre>
 
=={{header|Rust}}==
 
{{trans|Go}}
 
<syntaxhighlight lang="rust">
fn largest_proper_divisor(n: i32) -> i32 {
for i in 2..=(n as f64).sqrt() as i32 {
if n % i == 0 {
return n / i;
}
}
}
 
fn main() {
println!("The largest proper divisors for numbers in the interval [1, 100] are:");
print!(" 1 ");
for n in 2..=100 {
if n % 2 == 0 {
print!("{:2} ", n / 2);
} else {
print!("{:2} ", largest_proper_divisor(n));
}
if n % 10 == 0 {
println!();
}
}
}
</syntaxhighlight>
 
{{out}}
<pre>
The largest proper divisors for numbers in the interval [1, 100] are:
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
 
{{trans|C}}
 
<syntaxhighlight lang="rust">
fn lpd(n: u32) -> u32 {
if n <= 1 {
1
} else {
(1..n).rev().find(|&i| n % i == 0).unwrap_or(1)
}
}
 
fn main() {
for i in 1..=100 {
print!("{:3}", lpd(i));
 
if i % 10 == 0 {
println!();
}
}
}
</syntaxhighlight>
 
{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Ruby}}==
<langsyntaxhighlight lang="ruby">require 'prime'
 
def a(n)
Line 1,720 ⟶ 2,440:
 
(1..100).map{|n| a(n).to_s.rjust(3)}.each_slice(10){|slice| puts slice.join}
</syntaxhighlight>
</lang>
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,735 ⟶ 2,455:
 
=={{header|Seed7}}==
<langsyntaxhighlight lang="seed7">$ include "seed7_05.s7i";
 
const func integer: largestProperDivisor (in integer: number) is func
Line 1,766 ⟶ 2,486:
end if;
end for;
end func;</langsyntaxhighlight>
{{out}}
<pre>
Line 1,779 ⟶ 2,499:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Sidef}}==
<syntaxhighlight lang="ruby">1..100 -> map {|n| proper_divisors(n).tail \\ 1 }.slices(10).each {|a|
a.map{ '%3s' % _ }.join(' ').say
}</syntaxhighlight>
{{out}}
<pre>
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
 
=={{header|Swift}}==
<langsyntaxhighlight lang="swift">import Foundation
 
func largestProperDivisor(_ n : Int) -> Int? {
Line 1,804 ⟶ 2,542:
print(String(format: "%2d", largestProperDivisor(n)!),
terminator: n % 10 == 0 ? "\n" : " ")
}</langsyntaxhighlight>
 
{{out}}
Line 1,821 ⟶ 2,559:
 
 
=={{header|True BASICTI-57}}==
TI-57 calculators could display only one number at a time. The program below pauses almost a second to show the largest proper divisor of each integer between 1 and 100 - which would take 6 or 7 minutes on a genuine machine - then displays <code>0</code> and stops.
{{trans|FreeBASIC}}
{| class="wikitable"
<lang qbasic>PRINT "El mayor divisor propio de n es:"
! Machine code
PRINT
! Comment
PRINT " 1 1";
|-
FOR i = 3 To 100
|
FOR j = i-1 To 1 Step -1
Lbl 0
IF remainder(i, j) = 0 Then
1
PRINT Using$("###", j);
STO 0
EXIT FOR
Lbl 1
END IF
RCL 0
NEXT j
SBR 9
IF remainder(i, 10) = 0 Then PRINT
Pause
NEXT i
1
END</lang>
SUM 0
{{out}}
100
<pre>
x⮂t
Igual que la entrada de FreeBASIC.
RCL 0
</pre>
INV x≥t
GTO 1
CLR
R/S
RST
Lbl 9
x⮂t
1
x⮂t
x=t
INV SBR
STO 1
STO 2
C.t
Lbl 8
1
INV SUM 2
RCL 1
/
RCL 2
-
CE
Int
=
INV x=t
GTO 8
RCL 2
INV SBR
|
program task()
r0 = 1
loop
get a(r0)
display a(r0)
r0 += 1
t = 100
while r0 < t
end loop
clear display
end program
subroutine a(x)
t = 1
if x = t
return x
r1 = x
r2 = x
t = 0
loop
r2 -= 1
get r1
get r1/r2
get int(r1/r2)
get mod(r1,r2)
while mod(r1,r2) <> 0
end loop
return r2
end subroutine
|}
 
=={{header|V (Vlang)}}==
{{trans|go}}
<syntaxhighlight lang="v (vlang)">fn largest_proper_divisor(n int) int {
for i := 2; i*i <= n; i++ {
if n%i == 0 {
Line 1,865 ⟶ 2,673:
}
}
}</langsyntaxhighlight>
 
{{out}}
Line 1,881 ⟶ 2,689:
13 46 31 47 19 48 1 49 33 50
</pre>
 
 
=={{header|Wren}}==
{{libheader|Wren-math}}
{{libheader|Wren-fmt}}
<langsyntaxhighlight ecmascriptlang="wren">import "./math" for Int
import "./fmt" for Fmt
 
System.print("The largest proper divisors for numbers in the interval [1, 100] are:")
Line 1,898 ⟶ 2,705:
}
if (n % 10 == 0) System.print()
}</langsyntaxhighlight>
 
{{out}}
<pre>The largest proper divisors for numbers in the interval [1, 100] are:
<pre>
The largest proper divisors for numbers in the interval [1, 100] are:
1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
Line 1,912 ⟶ 2,718:
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50 </pre>
</pre>
 
 
=={{header|Yabasic}}==
{{trans|FreeBASIC}}
<lang yabasic>print "El mayor divisor propio de n es:\n"
print " 1 1 ";
for i = 3 to 100
for j = i-1 to 1 step -1
If mod(i, j) = 0 then print j using "##"; : break : fi
next j
if mod(i, 10) = 0 then print : fi
next i
print
end</lang>
{{out}}
<pre>
Igual que la entrada de FreeBASIC.
</pre>
 
 
=={{header|X86 Assembly}}==
<langsyntaxhighlight lang="asm"> 1 ;Assemble with: tasm, tlink /t
2 0000 .model tiny
3 0000 .code
Line 1,988 ⟶ 2,774:
52
53 end start
</syntaxhighlight>
</lang>
 
{{out}}
Line 2,005 ⟶ 2,791:
 
=={{header|XPL0}}==
<langsyntaxhighlight XPL0lang="xpl0">int N, D;
[for N:= 1 to 100 do
[D:= if N=1 then 1 else N/2;
Line 2,013 ⟶ 2,799:
if rem(N/10) = 0 then CrLf(0) else ChOut(0, ^ );
];
]</langsyntaxhighlight>
 
{{out}}
PureFox
9,476

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