Largest proper divisor of n: Difference between revisions
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=={{header|Raku}}== |
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A little odd to special case a(1) == 1 as technically, 1 doesn't have any proper divisors... but it matches '''[[oeis:A032742|OEIS A032742]]''' so whatever. |
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<lang perl6>use Prime::Factor; |
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say (flat 1, (2..100).map: *.&proper-divisors.sort.tail ).batch(10)».fmt("%2d").join: "\n";</lang> |
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{{out}} |
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<pre> 1 1 1 2 1 3 1 4 3 5 |
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1 6 1 7 5 8 1 9 1 10 |
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7 11 1 12 5 13 9 14 1 15 |
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1 16 11 17 7 18 1 19 13 20 |
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1 21 1 22 15 23 1 24 7 25 |
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13 46 31 47 19 48 1 49 33 50</pre> |
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=={{header|Ring}}== |
=={{header|Ring}}== |
Revision as of 16:01, 1 June 2021
- Task
- a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .
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
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
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
<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