Anonymous user
Long primes: Difference between revisions
→{{header|Nim}}
Alextretyak (talk | contribs) m (→{{header|11l}}) |
|||
Line 1,328:
64000 --> 2430
</pre>
=={{header|Nim}}==
{{trans|Kotlin}}
<lang Nim>import strformat
func sieve(limit: int): seq[int] =
var composite = newSeq[bool](limit + 1)
var p = 3
var p2 = p * p
while p2 < limit:
if not composite[p]:
for n in countup(p2, limit, 2 * p):
composite[n] = true
inc p, 2
p2 = p * p
for n in countup(3, limit, 2):
if not composite[n]:
result.add n
func period(n: int): int =
## Find the period of the reciprocal of "n".
var r = 1
for i in 1..(n + 1):
r = 10 * r mod n
let r1 = r
while true:
r = 10 * r mod n
inc result
if r == r1: break
let primes = sieve(64000)
var longPrimes: seq[int]
for prime in primes:
if prime.period() == prime - 1:
longPrimes.add prime
const Numbers = [500, 1000, 2000, 4000, 8000, 16000, 32000, 64000]
var index, count = 0
var totals = newSeq[int](Numbers.len)
for longPrime in longPrimes:
if longPrime > Numbers[index]:
totals[index] = count
inc index
inc count
totals[^1] = count
echo &"The long primes up to {Numbers[0]} are:"
for i in 0..<totals[0]:
stdout.write ' ', longPrimes[i]
stdout.write '\n'
echo "\nThe number of long primes up to:"
for i, total in totals:
echo &" {Numbers[i]:>5} is {total}"</lang>
{{out}}
<pre>The long primes up to 500 are:
7 17 19 23 29 47 59 61 97 109 113 131 149 167 179 181 193 223 229 233 257 263 269 313 337 367 379 383 389 419 433 461 487 491 499
The number of long primes up to:
500 is 35
1000 is 60
2000 is 116
4000 is 218
8000 is 390
16000 is 716
32000 is 1300
64000 is 2430</pre>
=={{header|Pascal}}==
| |||