Factors of an integer: Difference between revisions
Added an Algol W sample
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</pre>
=={{header|ALGOL W}}==
<lang algolw>begin
% return the factors of n ( n should be >= 1 ) in the array factor %
% the bounds of factor should be 0 :: len (len must be at least 1) %
% the number of factors will be returned in factor( 0 ) %
procedure getFactorsOf ( integer value n
; integer array factor( * )
; integer value len
) ;
begin
for i := 0 until len do factor( i ) := 0;
if n >= 1 and len >= 1 then begin
integer pos, lastFactor;
factor( 0 ) := factor( 1 ) := pos := 1;
% find the factors up to sqrt( n ) %
for f := 2 until truncate( sqrt( n ) ) + 1 do begin
if ( n rem f ) = 0 and pos <= len then begin
% found another factor and there's room to store it %
pos := pos + 1;
factor( 0 ) := pos;
factor( pos ) := f
end if_found_factor
end for_f;
% find the factors above sqrt( n ) %
lastFactor := factor( factor( 0 ) );
for f := factor( 0 ) step -1 until 1 do begin
integer newFactor;
newFactor := n div factor( f );
if newFactor > lastFactor and pos <= len then begin
% found another factor and there's room to store it %
pos := pos + 1;
factor( 0 ) := pos;
factor( pos ) := newFactor
end if_found_factor
end for_f;
end if_params_ok
end getFactorsOf ;
% prpocedure to test getFactorsOf %
procedure testFactorsOf( integer value n ) ;
begin
integer array factor( 0 :: 100 );
getFactorsOf( n, factor, 100 );
i_w := 1; s_w := 0; % set output format %
write( n, " has ", factor( 0 ), " factors:" );
for f := 1 until factor( 0 ) do writeon( " ", factor( f ) )
end testFactorsOf ;
% test the factorising %
for i := 1 until 100 do testFactorsOf( i )
end.</lang>
{{out}}
<pre>
1 has 1 factors: 1
2 has 2 factors: 1 2
3 has 2 factors: 1 3
4 has 3 factors: 1 2 4
...
96 has 12 factors: 1 2 3 4 6 8 12 16 24 32 48 96
97 has 2 factors: 1 97
98 has 6 factors: 1 2 7 14 49 98
99 has 6 factors: 1 3 9 11 33 99
100 has 9 factors: 1 2 4 5 10 20 25 50 100
</pre>
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