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Largest proper divisor of n

From Rosetta Code
Task
Largest proper divisor of n
You are encouraged to solve this task according to the task description, using any language you may know.
Task
a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .



ALGOL 68[edit]

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
IF n MOD j = 0 THEN
largest proper divisor := j
FI
OD;
print( ( whole( largest proper divisor, -3 ) ) );
IF n MOD 10 = 0 THEN print( ( newline ) ) FI
OD
 
Output:
  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

ALGOL W[edit]

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
writeon( i_w := 3, s_w := 0, j );
goto foundLargestProperDivisor
end if_n_rem_j_eq_0
end for_j;
writeon( i_w := 3, s_w := 0, 1 );
foundLargestProperDivisor:
if n rem 10 = 0 then write()
end for_n.
Output:
  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

APL[edit]

Works with: Dyalog APL
(⌈/1,(⍸0=¯1↓⍳|⊢))¨10 10⍴⍳100
Output:
 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

AppleScript[edit]

Most of this code is just to prepare the output for display.  :D

on largestProperDivisor(n)
if (n mod 2 = 0) then return n div 2
if (n mod 3 = 0) then return n div 3
if (n < 5) then return missing value
repeat with i from 5 to (n ^ 0.5 div 1) by 6
if (n mod i = 0) then return n div i
if (n mod (i + 2) = 0) then return n div (i + 2)
end repeat
return 1
end largestProperDivisor
 
on task(max)
script o
property LPDs : {}
property output : {}
end script
 
set w to (count (max as text)) * 2 + 3
set padding to text 1 thru (w - 2) of " "
set end of o's LPDs to "1:1" & padding
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
set c to 1
repeat with n from 2 to max
set end of o's LPDs to text 1 thru w of ((n as text) & ":" & largestProperDivisor(n) & padding)
set c to c + 1
if (c mod 10 is 0) then
set end of o's output to o's LPDs as text
set o's LPDs to {}
end if
end repeat
if (c mod 10 > 0) then set end of o's output to o's LPDs as text
set AppleScript's text item delimiters to linefeed
set o's output to o's output as text
set AppleScript's text item delimiters to astid
 
return o's output
end task
 
task(100)
Output:
"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
31:1 32:16 33:11 34:17 35:7 36:18 37:1 38:19 39:13 40:20
41:1 42:21 43:1 44:22 45:15 46:23 47:1 48:24 49:7 50:25
51:17 52:26 53:1 54:27 55:11 56:28 57:19 58:29 59:1 60:30
61:1 62:31 63:21 64:32 65:13 66:33 67:1 68:34 69:23 70:35
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 "


Functional[edit]

Composing functionally, for rapid drafting and refactoring, with higher levels of code reuse:

use framework "Foundation"
 
 
--------------- LARGEST PROPER DIVISOR OF N --------------
 
-- maxProperDivisors :: Int -> Int
on maxProperDivisors(n)
script p
on |λ|(x)
0 = n mod x
end |λ|
end script
 
set mb to find(p, enumFromTo(2, sqrt(n)))
 
if missing value is mb then
1
else
n div mb
end if
end maxProperDivisors
 
 
--------------------------- TEST -------------------------
on run
set xs to map(maxProperDivisors, enumFromTo(1, 100))
set w to length of (maximum(xs) as string)
 
unlines(map(unwords, ¬
chunksOf(10, ¬
map(compose(justifyRight(w, space), str), xs))))
end run
 
 
------------------------- GENERIC ------------------------
 
-- chunksOf :: Int -> [a] -> [[a]]
on chunksOf(k, xs)
script
on go(ys)
set ab to splitAt(k, ys)
set a to item 1 of ab
if {} ≠ a then
{a} & go(item 2 of ab)
else
a
end if
end go
end script
result's go(xs)
end chunksOf
 
 
-- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
on compose(f, g)
script
property mf : mReturn(f)
property mg : mReturn(g)
on |λ|(x)
mf's |λ|(mg's |λ|(x))
end |λ|
end script
end compose
 
 
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
lst
else
{}
end if
end enumFromTo
 
 
-- find :: (a -> Bool) -> [a] -> (a | missing value)
on find(p, xs)
tell mReturn(p)
set lng to length of xs
repeat with i from 1 to lng
if |λ|(item i of xs) then return item i of xs
end repeat
missing value
end tell
end find
 
 
-- justifyRight :: Int -> Char -> String -> String
on justifyRight(n, cFiller)
script
on |λ|(strText)
if n > length of strText then
text -n thru -1 of ((replicate(n, cFiller) as text) & strText)
else
strText
end if
end |λ|
end script
end justifyRight
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
 
-- maximum :: Ord a => [a] -> a
on maximum(xs)
set ca to current application
unwrap((ca's NSArray's arrayWithArray:(xs))'s ¬
valueForKeyPath:"@max.self")
end maximum
 
 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
 
-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> String -> String
on replicate(n, s)
-- Egyptian multiplication - progressively doubling a list,
-- appending stages of doubling to an accumulator where needed
-- for binary assembly of a target length
script p
on |λ|({n})
n ≤ 1
end |λ|
end script
 
script f
on |λ|({n, dbl, out})
if (n mod 2) > 0 then
set d to out & dbl
else
set d to out
end if
{n div 2, dbl & dbl, d}
end |λ|
end script
 
set xs to |until|(p, f, {n, s, ""})
item 2 of xs & item 3 of xs
end replicate
 
 
-- splitAt :: Int -> [a] -> ([a], [a])
on splitAt(n, xs)
if n > 0 and n < length of xs then
if class of xs is text then
{items 1 thru n of xs as text, ¬
items (n + 1) thru -1 of xs as text}
else
{items 1 thru n of xs, items (n + 1) thru -1 of xs}
end if
else
if n < 1 then
{{}, xs}
else
{xs, {}}
end if
end if
end splitAt
 
 
-- sqrt :: Num -> (Num | missing value)
on sqrt(n)
if 0 ≤ n then
n ^ (1 / 2)
else
missing value
end if
end sqrt
 
 
-- str :: a -> String
on str(x)
x as string
end str
 
 
-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
 
 
-- until :: (a -> Bool) -> (a -> a) -> a -> a
on |until|(p, f, x)
set v to x
set mp to mReturn(p)
set mf to mReturn(f)
repeat until mp's |λ|(v)
set v to mf's |λ|(v)
end repeat
v
end |until|
 
 
-- unwords :: [String] -> String
on unwords(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, space}
set s to xs as text
set my text item delimiters to dlm
return s
end unwords
 
 
-- unwrap :: NSValue -> a
on unwrap(nsValue)
if nsValue is missing value then
missing value
else
set ca to current application
item 1 of ((ca's NSArray's arrayWithObject:nsValue) as list)
end if
end unwrap
Output:
 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

AWK[edit]

 
# syntax: GAWK -f LARGEST_PROPER_DIVISOR_OF_N.AWK
# converted from C
BEGIN {
start = 1
stop = 100
for (i=start; i<=stop; i++) {
printf("%3d%1s",largest_proper_divisor(i),++count%10?"":"\n")
}
printf("\nLargest proper divisor of n %d-%d\n",start,stop)
exit(0)
}
function largest_proper_divisor(n, i) {
if (n <= 1) {
return(1)
}
for (i=n-1; i>0; i--) {
if (n % i == 0) {
return(i)
}
}
}
 
Output:
  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

Largest proper divisor of n 1-100

BASIC[edit]

10 DEFINT A-Z
20 FOR I=1 TO 100
30 IF I=1 THEN PRINT " 1";: GOTO 70
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
Output:
  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

BCPL[edit]

get "libhdr"
 
let lpd(n) = valof
for i = n<=1 -> 1, n-1 to 1 by -1
if n rem i=0 resultis i
 
let start() be
for i=1 to 100
$( writed(lpd(i), 3)
if i rem 10=0 then wrch('*N')
$)
Output:
  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

C[edit]

#include <stdio.h>
 
unsigned int lpd(unsigned int n) {
if (n<=1) return 1;
int i;
for (i=n-1; i>0; i--)
if (n%i == 0) return i;
}
 
int main() {
int i;
for (i=1; i<=100; i++) {
printf("%3d", lpd(i));
if (i % 10 == 0) printf("\n");
}
return 0;
}
Output:
  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

C++[edit]

#include <cassert>
#include <iomanip>
#include <iostream>
 
int 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;
}
 
int main() {
for (int n = 1; n < 101; ++n) {
std::cout << std::setw(2) << largest_proper_divisor(n)
<< (n % 10 == 0 ? '\n' : ' ');
}
}
Output:
 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

Cowgol[edit]

include "cowgol.coh";
 
sub print3(n: uint8) is
print_char(' ');
if n>9 then
print_char('0' + n/10);
else
print_char(' ');
end if;
print_char('0' + n%10);
end sub;
 
sub lpd(n: uint8): (r: uint8) is
if n <= 1 then
r := 1;
else
r := n - 1;
while r > 0 loop
if n % r == 0 then
break;
end if;
r := r - 1;
end loop;
end if;
end sub;
 
var i: uint8 := 1;
while i <= 100 loop
print3(lpd(i));
if i%10 == 0 then
print_nl();
end if;
i := i + 1;
end loop;
Output:
  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

F#[edit]

 
// 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 ""
 
Output:
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
Real: 00:00:00.015

Factor[edit]

Works with: Factor version 0.99 2021-02-05
USING: grouping kernel math math.bitwise math.functions
math.ranges prettyprint sequences ;
 
: odd ( m -- n )
dup 3 /i 1 - next-odd 1 -2 <range>
[ divisor? ] with find nip ;
 
: largest ( m -- n ) dup odd? [ odd ] [ 2/ ] if ;
 
100 [1,b] [ largest ] map 10 group simple-table.
Output:
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

FOCAL[edit]

01.10 F N=1,100;D 2;D 3
01.20 Q
 
02.10 S V=1
02.20 I (1-N)2.3;R
02.30 S V=N-1
02.40 S A=N/V
02.50 I (FITR(A)-A)2.6;R
02.60 S V=V-1
02.70 G 2.4
 
03.10 T %2,V
03.20 S A=N/10
03.30 I (FITR(A)-A)3.4;T !
03.40 R
Output:
=  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

Forth[edit]

Works with: Gforth
: largest-proper-divisor { n -- n }
n 1 and 0= if n 2/ exit then
3
begin
dup dup * n <=
while
dup n swap /mod swap
0= if nip exit else drop then
2 +
repeat drop 1 ;
 
: main
101 1 do
i largest-proper-divisor 2 .r
i 10 mod 0= if cr else space then
loop ;
 
main
bye
Output:
 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

Fortran[edit]

       program LargestProperDivisors
implicit none
integer i, lpd
do 10 i=1, 100
write (*,'(I3)',advance='no') lpd(i)
10 if (i/10*10 .eq. i) write (*,*)
end program
 
integer function lpd(n)
implicit none
integer n, i
if (n .le. 1) then
lpd = 1
else
do 10 i=n-1, 1, -1
10 if (n/i*i .eq. n) goto 20
20 lpd = i
end if
end function
Output:
  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

Go[edit]

package main
 
import "fmt"
 
func largestProperDivisor(n int) int {
for i := 2; i*i <= n; i++ {
if n%i == 0 {
return n / i
}
}
return 1
}
 
func main() {
fmt.Println("The largest proper divisors for numbers in the interval [1, 100] are:")
fmt.Print(" 1 ")
for n := 2; n <= 100; n++ {
if n%2 == 0 {
fmt.Printf("%2d ", n/2)
} else {
fmt.Printf("%2d ", largestProperDivisor(n))
}
if n%10 == 0 {
fmt.Println()
}
}
}
Output:
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  

Haskell[edit]

import Data.List.Split (chunksOf)
import Text.Printf (printf)
 
lpd :: Int -> Int
lpd 1 = 1
lpd n = head [x | x <- [n -1, n -2 .. 1], n `mod` x == 0]
 
main :: IO ()
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . lpd <$> [1 .. 100]
Output:
  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

Or, considering the two smallest proper divisors:

(If there are two, then the largest proper divisor will be N divided by the first proper divisor that is not 1)

(Otherwise, the largest proper divisor will be 1 itself).

import Data.List (find)
import Data.List.Split (chunksOf)
import Text.Printf (printf)
 
maxProperDivisors :: Int -> Int
maxProperDivisors n =
case find ((0 ==) . rem n) [2 .. root] of
Nothing -> 1
Just x -> quot n x
where
root = (floor . sqrt) (fromIntegral n :: Double)
 
main :: IO ()
main =
(putStr . unlines . map concat . chunksOf 10) $
printf "%3d" . maxProperDivisors <$> [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

J[edit]

   lpd =:  (1 |.!.1 ])&.q:
Output:
   lpd 1+i.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 ...

This works by prime factorization of n, replacing the largest 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" anaethesia: (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.

jq[edit]

Works with jq

Works with gojq, the Go implementation of jq

Naive version:

# (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;

Slightly less naive:

def largestpd:
if . == 1 then 1
else . as $n
| (first(2,3,5,7 | select($n % . == 0)) // null) as $div
| if $div then $n/$div
else first( range( ($n - ($n % 11)) /11; 0; -1) | (select($n % . == 0) ))
end
end;
# For neatness
def lpad($len):
tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
 
def nwise($n):
def w: if length <= $n then . else .[:$n], (.[$n:]|w) end;
w;
 
### The task
[range(1; 101) | largestpd]
| nwise(10) | map(lpad(2)) | join(" ")
Output:
 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

Julia[edit]

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)
 
Output:
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

MAD[edit]

            NORMAL MODE IS INTEGER
 
INTERNAL FUNCTION(N)
ENTRY TO LPD.
WHENEVER N.LE.1, FUNCTION RETURN 1
THROUGH TEST, FOR D=N-1, -1, D.L.1
TEST WHENEVER N/D*D.E.N, FUNCTION RETURN D
END OF FUNCTION
 
THROUGH SHOW, FOR I=1, 10, I.GE.100
SHOW PRINT FORMAT TABLE,
0 LPD.(I), LPD.(I+1), LPD.(I+2), LPD.(I+3),
1 LPD.(I+4), LPD.(I+5), LPD.(I+6), LPD.(I+7),
2 LPD.(I+8), LPD.(I+9)
 
VECTOR VALUES TABLE = $10(I3)*$
END OF PROGRAM
Output:
  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

Mathematica / Wolfram Language[edit]

Last[Prepend[DeleteCases[Divisors[#], #], 1]] & /@ Range[100]
Output:
{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}

Nim[edit]

import math, strutils
 
func largestProperDivisor(n: Positive): int =
for d in 2..sqrt(float(n)).int:
if n mod d == 0: return n div d
result = 1
 
for n in 1..100:
stdout.write ($n.largestProperDivisor).align(2), if n mod 10 == 0: '\n' else: ' '
Output:
 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

Pascal[edit]

 
program LarPropDiv;
 
function LargestProperDivisor(n:NativeInt):NativeInt;
//searching upwards to save time for example 100
//2..sqrt(n) aka 1..10 instead downwards n..sqrt(n) 100..10
var
i,j: NativeInt;
Begin
i := 2;
repeat
If n Mod i = 0 then
Begin
LargestProperDivisor := n DIV i;
EXIT;
end;
inc(i);
until i*i > n;
LargestProperDivisor := 1;
end;
var
n : Uint32;
begin
for n := 1 to 100 do
Begin
write(LargestProperDivisor(n):4);
if n mod 10 = 0 then
Writeln;
end;
end.
Output:
   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

Perl[edit]

Library: ntheory
use strict;
use warnings;
use ntheory 'divisors';
use List::AllUtils <max natatime>;
 
sub proper_divisors {
my $n = shift;
return 1 if $n == 0;
my @d = divisors($n);
pop @d;
@d;
}
 
my @range = 1 .. 100;
print "GPD for $range[0] through $range[-1]:\n";
my $iter = natatime 10, @range;
while( my @batch = $iter->() ) {
printf '%3d', $_ == 1 ? 1 : max proper_divisors($_) for @batch; print "\n";
}
Output:
GPD for 1 through 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

Phix[edit]

puts(1,join_by(apply(true,sprintf,{{"%3d"},vslice(apply(apply(true,factors,{tagset(100),-1}),reverse),1)}),1,10,""))
Output:
  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

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.

with javascript_semantics
for n=1 to 100 do
    integer p = 2,
            lim = floor(sqrt(n)),
            hf = 1
    while p<=lim do
        if remainder(n,p)=0 then
            hf = n/p
            exit
        end if
        p += 1
    end while 
    printf(1,"%3d",hf)
    if remainder(n,10)=0 then puts(1,"\n") end if
end for

PL/M[edit]

100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PUT$CHAR: PROCEDURE (CH); DECLARE CH BYTE; CALL BDOS(2,CH); END PUT$CHAR;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
 
PRINT$3: PROCEDURE (N);
DECLARE N BYTE;
CALL PUT$CHAR(' ');
IF N>9
THEN CALL PUT$CHAR('0' + N/10);
ELSE CALL PUT$CHAR(' ');
CALL PUT$CHAR('0' + N MOD 10);
END PRINT$3;
 
LPD: PROCEDURE (N) BYTE;
DECLARE (N, I) BYTE;
IF N <= 1 THEN RETURN 1;
I = N-1;
DO WHILE I >= 1;
IF N MOD I = 0 THEN RETURN I;
I = I - 1;
END;
END LPD;
 
DECLARE I BYTE;
DO I=1 TO 100;
CALL PRINT$3(LPD(I));
IF I MOD 10 = 0 THEN CALL PRINT(.(13,10,'$'));
END;
CALL EXIT;
EOF
Output:
  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

Python[edit]

def lpd(n):
for i in range(n-1,0,-1):
if n%i==0: return i
return 1
 
for i in range(1,101):
print("{:3}".format(lpd(i)), end=i%10==0 and '\n' or '')
Output:
  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


Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:

'''Largest proper divisor of'''
 
from math import isqrt
 
 
# maxProperDivisors :: Int -> Int
def maxProperDivisors(n):
'''The largest proper divisor of n.
'''

secondDivisor = find(
lambda x: 0 == (n % x)
)(
range(2, 1 + isqrt(n))
)
return 1 if None is secondDivisor else (
n // secondDivisor
)
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Test for values of n in [1..100]'''
 
xs = [
maxProperDivisors(n) for n
in range(1, 1 + 100)
]
 
colWidth = 1 + len(str(max(xs)))
 
print(
'\n'.join([
''.join(row) for row in
chunksOf(10)([
str(x).rjust(colWidth, " ") for x in xs
])
])
)
 
 
# ----------------------- GENERIC ------------------------
 
# chunksOf :: Int -> [a] -> [[a]]
def chunksOf(n):
'''A series of lists of length n, subdividing the
contents of xs. Where the length of xs is not evenly
divisible, the final list will be shorter than n.
'''

def go(xs):
return (
xs[i:n + i] for i in range(0, len(xs), n)
) if 0 < n else None
return go
 
 
# find :: (a -> Bool) -> [a] -> (a | None)
def find(p):
'''Just the first element in the list that matches p,
or None if no elements match.
'''

def go(xs):
try:
return next(x for x in xs if p(x))
except StopIteration:
return None
return go
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
  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

Quackery[edit]

factors is defined at Factors of an integer.

' [ 1 ] 99 times
[ i^ 2 + factors
-2 peek join ]
echo
Output:
[ 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 ]

Raku[edit]

A little odd to special case a(1) == 1 as technically, 1 doesn't have any proper divisors... but it matches OEIS A032742 so whatever.

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.

use Prime::Factor;
 
for 0, 2**67 - 1 -> $add {
my $start = now;
 
my $range = $add + 1 .. $add + 100;
 
say "GPD for {$range.min} through {$range.max}:";
 
say ( $range.hyper(:14batch).map: {$_ == 1 ?? 1 !! $_ %% 2 ?? $_ / 2 !! .&proper-divisors.max} )\
.batch(10)».fmt("%{$add.chars + 1}d").join: "\n";
 
say (now - $start).fmt("%0.3f seconds\n");
}
Output:
GPD for 1 through 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
0.071 seconds

GPD for 147573952589676412928 through 147573952589676413027:
  73786976294838206464   49191317529892137643   73786976294838206465                      1   73786976294838206466   21081993227096630419   73786976294838206467   49191317529892137645   73786976294838206468    8680820740569200761
  73786976294838206469    5088756985850910791   73786976294838206470   49191317529892137647   73786976294838206471   13415813871788764813   73786976294838206472   29514790517935282589   73786976294838206473   49191317529892137649
  73786976294838206474    6416258808246800563   73786976294838206475     977310944302492801   73786976294838206476   49191317529892137651   73786976294838206477   29514790517935282591   73786976294838206478      87167130885810049
  73786976294838206479   49191317529892137653   73786976294838206480   21081993227096630423   73786976294838206481    7767050136298758577   73786976294838206482   49191317529892137655   73786976294838206483    2784414199805215339
  73786976294838206484   11351842506898185613   73786976294838206485   49191317529892137657   73786976294838206486     939961481462907089   73786976294838206487   29514790517935282595   73786976294838206488   49191317529892137659
  73786976294838206489      82581954443019817   73786976294838206490        147258377885867   73786976294838206491   49191317529892137661   73786976294838206492   29514790517935282597   73786976294838206493   13415813871788764817
  73786976294838206494   49191317529892137663   73786976294838206495                      1   73786976294838206496    2202596307308603179   73786976294838206497   49191317529892137665   73786976294838206498    5088756985850910793
  73786976294838206499                      1   73786976294838206500   49191317529892137667   73786976294838206501   21081993227096630429   73786976294838206502   29514790517935282601   73786976294838206503   49191317529892137669
  73786976294838206504   13415813871788764819   73786976294838206505     103270785577100359   73786976294838206506   49191317529892137671   73786976294838206507   29514790517935282603   73786976294838206508   21081993227096630431
  73786976294838206509   49191317529892137673   73786976294838206510   11351842506898185617   73786976294838206511                      1   73786976294838206512   49191317529892137675   73786976294838206513    1305964182209525779
0.440 seconds

REXX[edit]

Logic was added to the   LPDIV   function so that for   odd   numbers,   only odd divisors were used.

This addition made it about   75%   faster.

/*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.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= length(n) + 1 /*W: the length of any output column. */
@LPD = "largest proper divisor of N, where N < " n
idx = 1 /*initialize the index (output counter)*/
say ' index │'center(@LPD, 1 + cols*(w+1) ) /*display the title for the output. */
say '───────┼'center("" , 1 + cols*(w+1), '─') /* " " sep " " " */
$= /*a list of largest proper divs so far.*/
do j=1 for n-1; $= $ right(LPDIV(j), w) /*add a largest proper divisor ──► list*/
if j//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
 
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
say '───────┴'center("" , 1 + cols*(w+1), '─') /*display the foot separator. */
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LPDIV: procedure; parse arg x; if x<4 then return 1 /*obtain X; test if X < 4. */
odd= x // 2; beg= x % 2 /*use BEG as the first divisor.*/
if odd then if beg//2==0 then beg= beg - 1 /*Is X odd? Then make BEG odd.*/
do k=beg by -1-odd /*begin at halfway point and decrease. */
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.*/
output   when using the default inputs:
 index │  largest proper divisor of  N,  where  N  <  101
───────┼───────────────────────────────────────────────────
   1   │    1    1    1    2    1    3    1    4    3    5
  11   │    1    6    1    7    5    8    1    9    1   10
  21   │    7   11    1   12    5   13    9   14    1   15
  31   │    1   16   11   17    7   18    1   19   13   20
  41   │    1   21    1   22   15   23    1   24    7   25
  51   │   17   26    1   27   11   28   19   29    1   30
  61   │    1   31   21   32   13   33    1   34   23   35
  71   │    1   36    1   37   25   38   11   39    1   40
  81   │   27   41    1   42   17   43   29   44    1   45
  91   │   13   46   31   47   19   48    1   49   33   50
───────┴───────────────────────────────────────────────────

Ring[edit]

 
see "working..." + nl
see "Largest proper divisor of n are:" + nl
see "1 "
row = 1
limit = 100
 
for n = 2 to limit
for m = 1 to n-1
if n%m = 0
div = m
ok
next
row = row + 1
see "" + div + " "
if row%10 = 0
see nl
ok
next
 
see "done..." + nl
 
Output:
working...
Largest proper divisor of n 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...

Seed7[edit]

$ include "seed7_05.s7i";
 
const func integer: largestProperDivisor (in integer: number) is func
result
var integer: divisor is 0;
begin
if not odd(number) then
divisor := number >> 1;
else
divisor := number div 3 - 1;
if odd(divisor) then
divisor +:= 2;
else
incr(divisor);
end if;
while number rem divisor <> 0 do
divisor -:= 2;
end while;
end if;
end func;
 
const proc: main is func
local
var integer: n is 0;
begin
for n range 1 to 100 do
write(largestProperDivisor(n) lpad 3);
if n rem 10 = 0 then
writeln;
end if;
end for;
end func;
Output:
  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

Swift[edit]

import Foundation
 
func largestProperDivisor(_ n : Int) -> Int? {
guard n > 0 else {
return nil
}
if (n & 1) == 0 {
return n >> 1
}
var p = 3
while p * p <= n {
if n % p == 0 {
return n / p
}
p += 2
}
return 1
}
 
for n in (1..<101) {
print(String(format: "%2d", largestProperDivisor(n)!),
terminator: n % 10 == 0 ? "\n" : " ")
}
Output:
 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

Wren[edit]

Library: Wren-math
Library: Wren-fmt
import "/math" for Int
import "/fmt" for Fmt
 
System.print("The largest proper divisors for numbers in the interval [1, 100] are:")
System.write(" 1 ")
for (n in 2..100) {
if (n % 2 == 0) {
Fmt.write("$2d ", n / 2)
} else {
Fmt.write("$2d ", Int.properDivisors(n)[-1])
}
if (n % 10 == 0) System.print()
}
Output:
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  

X86 Assembly[edit]

      1                              ;Assemble with: tasm, tlink /t
2 0000 .model tiny
3 0000 .code
4 org 100h ;.com program starts here
5
6 ;bl = D = divisor
7 ;bh = N
8
9 0100 B7 01 start: mov bh, 1 ;for N:= 1 to 100 do
10 0102 8A DF d10: mov bl, bh ;D:= if N=1 then 1 else N/2
11 0104 D0 EB shr bl, 1
12 0106 75 02 jne d15
13 0108 FE C3 inc bl
14 010A d15:
15 010A 8A C7 d20: mov al, bh ;while rem(N/D) # 0 do
16 010C 98 cbw ; ah:= 0 (extend sign of al into ah)
17 010D F6 FB idiv bl ; al:= ax/bl; ah:= remainder
18 010F 84 E4 test ah, ah
19 0111 74 04 je d30
20 0113 FE CB dec bl ; D--
21 0115 EB F3 jmp d20
22 0117 d30:
23 0117 8A C3 mov al, bl ;output number in D
24 0119 D4 0A aam 10 ;ah:= al/10; al:= remainder
25 011B 50 push ax ;save low digit in remainder
26 011C 8A C4 mov al, ah ;get high digit
27 011E 04 20 add al, 20h ;if zero make it a space char
28 0120 84 E4 test ah, ah
29 0122 74 02 je d50
30 0124 04 10 add al, 10h ;else make it into ASCII digit
31 0126 CD 29 d50: int 29h ;output high digit or space
32 0128 58 pop ax ;get low digit
33 0129 04 30 add al, 30h ;make it ASCII
34 012B CD 29 int 29h ;output low digit
35
36 012D B0 20 mov al, 20h ;output space char
37 012F CD 29 int 29h
38
39 0131 8A C7 mov al, bh ;if remainder(N/10) = 0 then CR LF
40 0133 D4 0A aam 10 ;ah:= al/10; al:= remainder
41 0135 3C 00 cmp al, 0
42 0137 75 08 jne next
43 0139 B0 0D mov al, 0Dh ;CR
44 013B CD 29 int 29h
45 013D B0 0A mov al, 0Ah ;LF
46 013F CD 29 int 29h
47
48 0141 FE C7 next: inc bh ;next N
49 0143 80 FF 64 cmp bh, 100
50 0146 7E BA jle d10
51 0148 C3 ret
52
53 end start
 
Output:
 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

XPL0[edit]

int N, D;
[for N:= 1 to 100 do
[D:= if N=1 then 1 else N/2;
while rem(N/D) do D:= D-1;
if D<10 then ChOut(0, ^ );
IntOut(0, D);
if rem(N/10) = 0 then CrLf(0) else ChOut(0, ^ );
];
]
Output:
 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