Largest proper divisor of n: Difference between revisions
(Realize in F#) |
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=={{header|REXX}}== |
=={{header|REXX}}== |
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Logic was added to the '''LPDIV''' function so that for odd numbers, only odd divisors were used. |
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This addition made it about '''75%''' faster. |
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<lang rexx>/*REXX program finds the largest proper divisors of all numbers (up to a given limit). */ |
<lang rexx>/*REXX program finds the largest proper divisors of all numbers (up to a given limit). */ |
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parse arg n cols . /*obtain optional argument from the CL.*/ |
parse arg n cols . /*obtain optional argument from the CL.*/ |
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/*──────────────────────────────────────────────────────────────────────────────────────*/ |
/*──────────────────────────────────────────────────────────────────────────────────────*/ |
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LPDIV: procedure; parse arg x; if x<4 then return 1 /*obtain X; test if X < 4. */ |
LPDIV: procedure; parse arg x; if x<4 then return 1 /*obtain X; test if X < 4. */ |
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odd= x // 2; beg= x % 2 /*use BEG as the first divisor.*/ |
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if odd then if beg//2==0 then beg= beg - 1 /*Is X odd? Then make BEG odd.*/ |
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do k=beg by -1-odd /*begine at halfway point and decrease.*/ |
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if x//k==0 then return k /*Remainder=0? Got largest proper div.*/ |
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end /*k*/ |
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return 1 /*If we get here, then X is a prime.*/</lang> |
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{{out|output|text= when using the default inputs:}} |
{{out|output|text= when using the default inputs:}} |
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<pre> |
<pre> |
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Revision as of 16:29, 2 June 2021
- Task
- a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .
ALGOL 68
<lang algol68>FOR n TO 100 DO # show the largest proper divisors for n = 1..100 #
INT largest proper divisor := 1;
FOR j FROM ( n OVER 2 ) BY -1 TO 2 WHILE largest proper divisor = 1 DO
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 </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
ALGOL W
<lang algolw>for n := 1 until 100 do begin % show the largest proper divisors for n = 1..100 %
for j := n div 2 step -1 until 2 do begin
if n rem j = 0 then begin
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.</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
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
C++
<lang cpp>#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' : ' ');
}
}</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
Cowgol
<lang cowgol>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;</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
F#
<lang fsharp> // Largest proper divisor of n: Nigel Galloway. June 2nd., 2021 let fN g=let rec fN n=let i=Seq.head n in match(g/i,g%i) with (1,_)->1 |(n,0)->n |_->fN(Seq.tail n) in fN(Seq.initInfinite((+)2)) seq{yield 1; yield! seq{2..100}|>Seq.map fN}|>Seq.iter(printf "%d "); printfn "" </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 Real: 00:00:00.015
Factor
<lang factor>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.</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
Forth
<lang forth>: 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</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) $
printf "%3d" . 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
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).
<lang haskell>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]</lang>
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
<lang julia>largestpd(n) = (for i in n÷2:-1:1 (n % i == 0) && return i; end; 1)
foreach(n -> print(rpad(largestpd(n), 3), n % 10 == 0 ? "\n" : ""), 1:100)
</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
Pascal
<lang pascal> 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.</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
Phix
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... at least on much larger numbers, that is.
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
<lang plm>100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; 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</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
Or, reducing the search space, formatting more flexibly (to allow for experiments with larger ranges) and composing functionally:
<lang python>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()</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.
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. <lang perl6>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");
}</lang>
- 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
Logic was added to the LPDIV function so that for odd numbers, only odd divisors were used.
This addition made it about 75% faster. <lang rexx>/*REXX program finds the largest proper divisors of all numbers (up to a given limit). */ parse arg n cols . /*obtain optional argument from the CL.*/ if n== | n=="," then n= 101 /*Not specified? Then use the default.*/ 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 /*begine 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.*/</lang>
- 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
<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...
Swift
<lang swift>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" : " ")
}</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
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
X86 Assembly
<lang asm> 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
</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
XPL0
<lang XPL0>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, ^ ); ];
]</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