Largest proper divisor of n: Difference between revisions

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map (printf "%3d") $
map (printf "%3d") $
map lpd [1..100]</lang>
map lpd [1..100]</lang>
{{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}}==
<lang MAD> 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 </lang>
{{out}}
{{out}}
<pre> 1 1 1 2 1 3 1 4 3 5
<pre> 1 1 1 2 1 3 1 4 3 5

Revision as of 16:30, 1 June 2021

Largest proper divisor of n is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task
a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .



APL

Works with: Dyalog APL

<lang apl>(⌈/1,(⍸0=¯1↓⍳|⊢))¨10 10⍴⍳100</lang>

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

BASIC

<lang basic>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</lang>

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

<lang bcpl>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')
   $)</lang>
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

<lang c>#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;

}</lang>

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

<lang focal>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</lang>

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

<lang fortran> 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</lang>
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

Haskell

<lang haskell>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 $ 
   map (printf "%3d") $ 
   map lpd [1..100]</lang>
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

<lang MAD> 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 </lang>
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

<lang python>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 )</lang>
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

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. <lang perl6>use Prime::Factor;

say (flat 1, (2..100).map: *.&proper-divisors.sort.tail ).batch(10)».fmt("%2d").join: "\n";</lang>

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

Ring

<lang ring> 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 </lang>

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...

Wren

Library: Wren-math
Library: Wren-fmt

<lang ecmascript>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()

}</lang>

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
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