Largest proper divisor of n: Difference between revisions
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→{{header|Wren}}: Minor tidy
(→{{header|jq}}: slightly less naive) |
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;Task:a(1) = 1; for n > 1, a(n) = '''largest''' proper divisor of n, where '''n < 101 '''.
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">F lpd(n)
L(i) (n - 1 .< 0).step(-1)
I n % i == 0
R i
R 1
L(i) 1..100
print(‘#3’.format(lpd(i)), end' I i % 10 == 0 {"\n"} E ‘’)</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|Ada}}==
<syntaxhighlight lang="ada">with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
procedure Main is
subtype param_type is Integer range 1 .. 100;
function lpd (n : in param_type) return param_type is
result : param_type := 1;
begin
for divisor in reverse 1 .. n / 2 loop
if n rem divisor = 0 then
result := divisor;
exit;
end if;
end loop;
return result;
end lpd;
begin
for I in param_type loop
Put (Item => lpd (I), Width => 3);
if I rem 10 = 0 then
New_Line;
end if;
end loop;
end Main;</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|ALGOL 68}}==
<
INT largest proper divisor := 1;
FOR j FROM ( n OVER 2 ) BY -1 TO 2 WHILE largest proper divisor = 1 DO
Line 15 ⟶ 81:
IF n MOD 10 = 0 THEN print( ( newline ) ) FI
OD
</syntaxhighlight>
{{out}}
<pre>
Line 31 ⟶ 97:
=={{header|ALGOL W}}==
<
for j := n div 2 step -1 until 2 do begin
if n rem j = 0 then begin
Line 41 ⟶ 107:
foundLargestProperDivisor:
if n rem 10 = 0 then write()
end for_n.</
{{out}}
<pre>
Line 58 ⟶ 124:
=={{header|APL}}==
{{works with|Dyalog APL}}
<
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 74 ⟶ 140:
Most of this code is just to prepare the output for display. :D
<
if (n mod 2 = 0) then return n div 2
if (n mod 3 = 0) then return n div 3
Line 113 ⟶ 179:
end task
task(100)</
{{output}}
<
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 125 ⟶ 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 "</
Line 132 ⟶ 198:
Composing functionally, for rapid drafting and refactoring, with higher levels of code reuse:
<
Line 380 ⟶ 446:
item 1 of ((ca's NSArray's arrayWithObject:nsValue) as list)
end if
end unwrap</
{{Out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 392 ⟶ 458:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
=={{header|Arturo}}==
<syntaxhighlight 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
]</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|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f LARGEST_PROPER_DIVISOR_OF_N.AWK
# converted from C
Line 416 ⟶ 501:
}
}
</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
Line 430 ⟶ 514:
13 46 31 47 19 48 1 49 33 50
Largest proper divisor of n 1-100</pre>
=={{header|BASIC}}==
{{works with|QBasic|1.1}}
{{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 440 ⟶ 526:
60 NEXT J
70 IF I MOD 10=0 THEN PRINT
80 NEXT I</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 452 ⟶ 538:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
==={{header|ANSI BASIC}}===
{{trans|FreeBASIC}}
{{works with|Decimal BASIC}}
{{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
for j = i-1 to 1 step -1
If i % j = 0 then
print " "; j; " ";
exit for
end if
next j
if i % 10 = 0 then print
next i
end</syntaxhighlight>
==={{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}}==
<
let lpd(n) = valof
Line 464 ⟶ 934:
$( writed(lpd(i), 3)
if i rem 10=0 then wrch('*N')
$)</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 476 ⟶ 946:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
=={{header|BQN}}==
<syntaxhighlight lang="bqn">(1⌈´↕×0=↕|⊢)¨ ∘‿10⥊1+↕100</syntaxhighlight>
{{out}}
<pre>┌─
╵ 1 1 1 2 1 3 1 4 3 5
1 6 1 7 5 8 1 9 1 10
7 11 1 12 5 13 9 14 1 15
1 16 11 17 7 18 1 19 13 20
1 21 1 22 15 23 1 24 7 25
17 26 1 27 11 28 19 29 1 30
1 31 21 32 13 33 1 34 23 35
1 36 1 37 25 38 11 39 1 40
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
┘</pre>
=={{header|C}}==
<
unsigned int lpd(unsigned int n) {
Line 494 ⟶ 980:
}
return 0;
}</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 508 ⟶ 994:
=={{header|C++}}==
<
#include <iomanip>
#include <iostream>
Line 528 ⟶ 1,014:
<< (n % 10 == 0 ? '\n' : ' ');
}
}</
{{out}}
Line 545 ⟶ 1,031:
=={{header|Cowgol}}==
<
sub print3(n: uint8) is
Line 578 ⟶ 1,064:
end if;
i := i + 1;
end loop;</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 590 ⟶ 1,076:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
=={{header|D}}==
{{trans|C}}
<syntaxhighlight lang="d">
import std.stdio;
import std.range;
import std.algorithm;
uint lpd(uint n) {
if (n <= 1) {
return 1;
}
auto divisors = array(iota(1, n).filter!(i => n % i == 0));
return divisors.empty ? 1 : divisors[$ - 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}}==
<syntaxhighlight lang="draco">proc nonrec lpd(word n) word:
word d;
if n=1 then
1
else
d := n-1;
while n % d /= 0 do d := d-1 od;
d
fi
corp
proc nonrec main() void:
word n;
for n from 1 upto 100 do
write(lpd(n):3);
if n%10 = 0 then writeln() fi
od
corp</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|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#}}==
<
// 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>
{{out}}
<pre>
Line 602 ⟶ 1,251:
Real: 00:00:00.015
</pre>
=={{header|Factor}}==
{{works with|Factor|0.99 2021-02-05}}
<
math.ranges prettyprint sequences ;
Line 613 ⟶ 1,263:
: largest ( m -- n ) dup odd? [ odd ] [ 2/ ] if ;
100 [1,b] [ largest ] map 10 group simple-table.</
{{out}}
<pre>
Line 625 ⟶ 1,275:
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|Fermat}}==
<syntaxhighlight lang="fermat">Func Lpd(n) =
if n = 1 then Return(1) fi;
for i = n\2 to 1 by -1 do
if n|i = 0 then Return(i) fi;
od.;
for m = 1 to 100 do
!(Lpd(m):4);
if m|10=0 then ! fi;
od;</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|FOCAL}}==
<
01.20 Q
Line 643 ⟶ 1,317:
03.20 S A=N/10
03.30 I (FITR(A)-A)3.4;T !
03.40 R</
{{out}}
<pre>= 1= 1= 1= 2= 1= 3= 1= 4= 3= 5
Line 658 ⟶ 1,332:
=={{header|Forth}}==
{{works with|Gforth}}
<
n 1 and 0= if n 2/ exit then
3
Line 676 ⟶ 1,350:
main
bye</
{{out}}
Line 693 ⟶ 1,367:
=={{header|Fortran}}==
<
implicit none
integer i, lpd
Line 711 ⟶ 1,385:
20 lpd = i
end if
end function</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 723 ⟶ 1,397:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
=={{header|Frink}}==
Frink contains efficient routines for factoring numbers. It uses trial division, wheel factoring, and Pollard rho factoring.
<syntaxhighlight lang="frink">for n = 1 to 100
println[last[allFactors[n,true,false]]]</syntaxhighlight>
=={{header|Go}}==
<
import "fmt"
Line 751 ⟶ 1,431:
}
}
}</
{{out}}
Line 769 ⟶ 1,449:
=={{header|Haskell}}==
<
import Text.Printf (printf)
Line 779 ⟶ 1,459:
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . lpd <$> [1 .. 100]</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 798 ⟶ 1,478:
(Otherwise, the largest proper divisor will be 1 itself).
<
import Data.List.Split (chunksOf)
import Text.Printf (printf)
Line 813 ⟶ 1,493:
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . maxProperDivisors <$> [1 .. 100]</
<pre> 1 1 1 2 1 3 1 4 3 5
Line 827 ⟶ 1,507:
=={{header|J}}==
<syntaxhighlight lang
{{out}}
<pre>
lpd
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>
This works by prime factorization of n,
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.
=={{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 843 ⟶ 1,570:
Naive version:
<
def largestpd:
if . == 1 then 1
else . as $n
| first( range( ($n - ($n % 2)) /2; 0; -1) | (select($n % . == 0) ))
end;</
Slightly less naive:
<
if . == 1 then 1
else . as $n
Line 858 ⟶ 1,585:
else first( range( ($n - ($n % 11)) /11; 0; -1) | (select($n % . == 0) ))
end
end;</
<
def lpad($len):
tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
Line 869 ⟶ 1,596:
### The task
[range(1; 101) | largestpd]
| nwise(10) | map(lpad(2)) | join(" ")</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 883 ⟶ 1,610:
=={{header|Julia}}==
<
foreach(n -> print(rpad(largestpd(n), 3), n % 10 == 0 ? "\n" : ""), 1:100)
</
<pre>
1 1 1 2 1 3 1 4 3 5
Line 898 ⟶ 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}}==
<
INTERNAL FUNCTION(N)
Line 917 ⟶ 1,673:
VECTOR VALUES TABLE = $10(I3)*$
END OF PROGRAM </
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 931 ⟶ 1,687:
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<
{{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}}==
<syntaxhighlight lang="modula2">MODULE LargestProperDivisor;
FROM InOut IMPORT WriteCard, WriteLn;
VAR i: CARDINAL;
PROCEDURE lpd(n: CARDINAL): CARDINAL;
VAR i: CARDINAL;
BEGIN
IF n=1 THEN
RETURN 1;
END;
FOR i := n DIV 2 TO 1 BY -1 DO
IF n MOD i = 0 THEN
RETURN i;
END;
END;
END lpd;
BEGIN
FOR i := 1 TO 100 DO
WriteCard(lpd(i), 3);
IF i MOD 10 = 0 THEN
WriteLn();
END;
END;
END LargestProperDivisor.</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|Nim}}==
<
func largestProperDivisor(n: Positive): int =
Line 944 ⟶ 1,751:
for n in 1..100:
stdout.write ($n.largestProperDivisor).align(2), if n mod 10 == 0: '\n' else: ' '</
{{out}}
Line 959 ⟶ 1,766:
=={{header|Pascal}}==
<
program LarPropDiv;
Line 988 ⟶ 1,795:
Writeln;
end;
end.</
{{out}}
<pre>
Line 1,001 ⟶ 1,808:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50</pre>
===Alternative (sieve)===
Most solutions use a function that returns the largest proper divisor of an individual integer. This console program, written in Free Pascal, applies a sieve to the integers 1..100 (or other limit).
<syntaxhighlight lang="pascal">
program LPD;
(*
Displays largest proper divisor for each integer in range 1..limit.
Command line:
LPD limit items_per_line
or LPD limit // items_per_line defaults to 10
or LPD // limit defaults to 100
*)
{$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}}
<
use warnings;
use ntheory 'divisors';
Line 1,022 ⟶ 1,905:
while( my @batch = $iter->() ) {
printf '%3d', $_ == 1 ? 1 : max proper_divisors($_) for @batch; print "\n";
}</
{{out}}
<pre>GPD for 1 through 100:
Line 1,037 ⟶ 1,920:
=={{header|Phix}}==
<!--<
<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>
<!--</
{{out}}
<pre>
Line 1,054 ⟶ 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.
<!--<
<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 1,070 ⟶ 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>
<!--</
=={{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}}==
<syntaxhighlight lang="pli">largestProperDivisor: procedure options(main);
lpd: procedure(n) returns(fixed);
declare (n, i) fixed;
if n <= 1 then return(1);
do i=n-1 repeat(i-1) while(i>=1);
if mod(n,i)=0 then return(i);
end;
end lpd;
declare i fixed;
do i=1 to 100;
put edit(lpd(i)) (F(3));
if mod(i,10)=0 then put skip;
end;
end largestProperDivisor;</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/M}}==
<
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
Line 1,104 ⟶ 2,042:
END;
CALL EXIT;
EOF</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,118 ⟶ 2,056:
=={{header|Python}}==
<
for i in range(n-1,0,-1):
if n%i==0: return i
Line 1,124 ⟶ 2,062:
for i in range(1,101):
print("{:3}".format(lpd(i)), end=i%10==0 and '\n' or '')</
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,140 ⟶ 2,078:
Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:
<
from math import isqrt
Line 1,211 ⟶ 2,149:
# MAIN ---
if __name__ == '__main__':
main()</
{{Out}}
<pre> 1 1 1 2 1 3 1 4 3 5
Line 1,227 ⟶ 2,165:
<code>factors</code> is defined at [http://rosettacode.org/wiki/Factors_of_an_integer#Quackery Factors of an integer].
<
[ i^ 2 + factors
-2 peek join ]
echo</
{{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|R}}==
<syntaxhighlight lang="r">largest_proper_divisor <- function(n){
if(n == 1) return(1)
lpd = 1
for(i in seq(1, n-1, 1)){
if(n %% i == 0)
lpd = i
}
message(paste0("The largest proper divisor of ", n, " is ", lpd))
return(lpd)
}
#Verify
for (i in 1:100){ #about 10 seconds to calculate until 10000
largest_proper_divisor(i)
}
</syntaxhighlight>
{{out}}
<pre>
The largest proper divisor of 2 is 1
The largest proper divisor of 3 is 1
The largest proper divisor of 4 is 2
The largest proper divisor of 5 is 1
The largest proper divisor of 6 is 3
The largest proper divisor of 7 is 1
The largest proper divisor of 8 is 4
The largest proper divisor of 9 is 3
.
. blah blah blah
.
The largest proper divisor of 94 is 47
The largest proper divisor of 95 is 19
The largest proper divisor of 96 is 48
The largest proper divisor of 97 is 1
The largest proper divisor of 98 is 49
The largest proper divisor of 99 is 33
The largest proper divisor of 100 is 50
</pre>
=={{header|Raku}}==
Line 1,240 ⟶ 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"
for 0, 2**67 - 1 -> $add {
Line 1,253 ⟶ 2,234:
say (now - $start).fmt("%0.3f seconds\n");
}</
{{out}}
<pre>GPD for 1 through 100:
Line 1,285 ⟶ 2,266:
This addition made it about '''75%''' faster.
<
parse arg n cols . /*obtain optional argument from the CL.*/
if n=='' | n=="," then n= 101 /*Not specified? Then use the default.*/
Line 1,311 ⟶ 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.*/</
{{out|output|text= when using the default inputs:}}
<pre>
Line 1,330 ⟶ 2,311:
=={{header|Ring}}==
<syntaxhighlight lang="ring">? "working..."
limit = 100
? "Largest proper divisor up to " + limit + " are:"
see " 1 "
col = 1
for n = 2 to limit
for m = 1 to n
if n % m = 0 div = m ok
next
see "" + div + " "
if
next
? "done..."</syntaxhighlight>
{{out}}
<pre>working...
Largest proper divisor up to 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
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 {
}
}
}
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}}==
<syntaxhighlight lang="ruby">require 'prime'
def a(n)
return 1 if n == 1 || n.prime?
(n/2).downto(1).detect{|d| n.remainder(d) == 0}
end
(1..100).map{|n| a(n).to_s.rjust(3)}.each_slice(10){|slice| puts slice.join}
</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|Seed7}}==
<
const func integer: largestProperDivisor (in integer: number) is func
Line 1,401 ⟶ 2,486:
end if;
end for;
end func;</
{{out}}
<pre>
Line 1,414 ⟶ 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}}==
<
func largestProperDivisor(_ n : Int) -> Int? {
Line 1,439 ⟶ 2,542:
print(String(format: "%2d", largestProperDivisor(n)!),
terminator: n % 10 == 0 ? "\n" : " ")
}</
{{out}}
Line 1,453 ⟶ 2,556:
27 41 1 42 17 43 29 44 1 45
13 46 31 47 19 48 1 49 33 50
</pre>
=={{header|TI-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.
{| class="wikitable"
! Machine code
! Comment
|-
|
Lbl 0
1
STO 0
Lbl 1
RCL 0
SBR 9
Pause
1
SUM 0
100
x⮂t
RCL 0
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 {
return n / i
}
}
return 1
}
fn main() {
println("The largest proper divisors for numbers in the interval [1, 100] are:")
print(" 1 ")
for n := 2; n <= 100; n++ {
if n%2 == 0 {
print("${n/2:2} ")
} else {
print("${largest_proper_divisor(n):2} ")
}
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>
Line 1,458 ⟶ 2,693:
{{libheader|Wren-math}}
{{libheader|Wren-fmt}}
<
import "./fmt" for Fmt
System.print("The largest proper divisors for numbers in the interval [1, 100] are:")
Line 1,470 ⟶ 2,705:
}
if (n % 10 == 0) System.print()
}</
{{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
Line 1,484 ⟶ 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>
=={{header|X86 Assembly}}==
<
2 0000 .model tiny
3 0000 .code
Line 1,541 ⟶ 2,774:
52
53 end start
</syntaxhighlight>
{{out}}
Line 1,558 ⟶ 2,791:
=={{header|XPL0}}==
<
[for N:= 1 to 100 do
[D:= if N=1 then 1 else N/2;
Line 1,566 ⟶ 2,799:
if rem(N/10) = 0 then CrLf(0) else ChOut(0, ^ );
];
]</
{{out}}
|