Data commit

Task/Long-primes/Python/long-primes.py

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+def sieve(limit):
+    primes = []
+    c = [False] * (limit + 1) # composite = true
+    # no need to process even numbers
+    p = 3
+    while True:
+        p2 = p * p
+        if p2 > limit: break
+        for i in range(p2, limit, 2 * p): c[i] = True
+        while True:
+            p += 2
+            if not c[p]: break
+
+    for i in range(3, limit, 2):
+        if not c[i]: primes.append(i)
+    return primes
+
+# finds the period of the reciprocal of n
+def findPeriod(n):
+    r = 1
+    for i in range(1, n): r = (10 * r) % n
+    rr = r
+    period = 0
+    while True:
+        r = (10 * r) % n
+        period += 1
+        if r == rr: break
+    return period
+
+primes = sieve(64000)
+longPrimes = []
+for prime in primes:
+    if findPeriod(prime) == prime - 1:
+        longPrimes.append(prime)
+numbers = [500, 1000, 2000, 4000, 8000, 16000, 32000, 64000]
+count = 0
+index = 0
+totals = [0] * len(numbers)
+for longPrime in longPrimes:
+    if longPrime > numbers[index]:
+        totals[index] = count
+        index += 1
+    count += 1
+totals[-1] = count
+print('The long primes up to 500 are:')
+print(str(longPrimes[:totals[0]]).replace(',', ''))
+print('\nThe number of long primes up to:')
+for (i, total) in enumerate(totals):
+    print('  %5d is %d' % (numbers[i], total))