Largest proper divisor of n: Difference between revisions
Add BCPL
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;Task:a(1) = 1; for n > 1, a(n) = '''largest''' proper divisor of n, where '''n < 101 '''.
<br><br>
=={{header|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>
{{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|Ring}}==
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