Deceptive numbers: Difference between revisions
Added the missing code!!!!
(Created Nim solution.) |
(Added the missing code!!!!) |
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Line 472:
{{trans|Python}}
This algorithm doesn’t need a multiprecision library, but we have to define the modular exponentiation as is is not provided by the standard library.
<syntaxhighlight lang="Nim">import std/[math, strutils]
func pow(a, n: Natural; m: Positive): Natural =
var a = a mod m
var n = n
if a > 0:
result = 1
while n > 0:
if (n and 1) != 0:
result = (result * a) mod m
n = n shr 1
a = (a * a) mod m
func sqrt(n: Natural): Natural = Natural(sqrt(float(n)))
func isDeceptive(n: Natural): bool =
if (n and 1) != 0 and n mod 3 != 0 and n mod 5 != 0 and pow(10, n - 1, n) == 1:
for d in countup(7, sqrt(n), 6):
if n mod d == 0 or n mod (d + 4) == 0:
return true
result = false
var count = 0
var n = 7
while true:
if n.isDeceptive:
inc count
stdout.write align($n, 6)
stdout.write if count mod 10 == 0: '\n' else: ' '
if count == 100: break
inc n
</syntaxhighlight>
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