38 lines
No EOL
1.5 KiB
Text
38 lines
No EOL
1.5 KiB
Text
Algorithm to calculate the first 1000 prime numbers using the Sieve of Eratosthenes:
|
|
|
|
1. Initialize an empty list to store prime numbers.
|
|
2. Set a limit for the sieve. Since the 1000th prime is 7919, a safe upper bound is 10000.
|
|
3. Create a boolean array `is_prime` of size `limit + 1` and initialize all entries as `True`. Set `is_prime[0]` and `is_prime[1]` to `False` as 0 and 1 are not prime numbers.
|
|
4. For each number `p` starting from 2, do the following:
|
|
a. If `is_prime[p]` is `True`, it means `p` is a prime number.
|
|
b. Append `p` to the list of prime numbers.
|
|
c. Mark all multiples of `p` (starting from `p*p`) as `False` in the `is_prime` array.
|
|
5. Continue the process until the list of prime numbers contains 1000 elements.
|
|
6. Return the list of the first 1000 prime numbers.
|
|
|
|
Pseudocode:
|
|
|
|
function calculate_first_1000_primes():
|
|
limit = 10000
|
|
is_prime = [True] * (limit + 1)
|
|
is_prime[0] = is_prime[1] = False
|
|
primes = []
|
|
|
|
for p in range(2, limit + 1):
|
|
if is_prime[p]:
|
|
primes.append(p)
|
|
if len(primes) == 1000:
|
|
break
|
|
for multiple in range(p * p, limit + 1, p):
|
|
is_prime[multiple] = False
|
|
|
|
return primes
|
|
|
|
# Validation Step:
|
|
# Ensure the length of the primes list is 1000 and the last element is 7919.
|
|
|
|
primes = calculate_first_1000_primes()
|
|
assert len(primes) == 1000, "The list does not contain 1000 primes."
|
|
assert primes[-1] == 7919, "The 1000th prime is incorrect."
|
|
|
|
# If assertions pass, the list of primes is correct. |