Using the Sieve of Eratosthenes find the first n primes numbers.

1 2 3 4 5 6 7 8 9 10 11 12 | ```
def prime(n):
def z(x):
if x :return True
return False
num1=range(0,n+1); num2=int(n**0.5) +1
for k in range(2,num2):
num0=range(k,n+1,k);del num0[0]
for i in num0:
num1[i]=0
return filter(z, num1)
print prime(102)
``` |

Hello, Your code is very concise. But I don't think it's efficient. The time cost of your code is n

(n*0.5)My code is not good either, but I think its time cost is better than yours. For example, when I set n = 1000000, your code cost 15 seconds while mine cost 12 seconds on my computer.

The time cost of your code is n*(n**0.5)

Liu you are right now is n(n**0.5)/2 thanks