质数个数_python3
KiKi求质数个数
http://www.nowcoder.com/questionTerminal/d3a404ee0f8d41f98bf4707035d91086
N = 1001
count = [0]*N
l = list(range(2, N))
def countP(n):
for i in range(2, n):
count[i] = count[i-1] + 1 if i in l else count[i-1]
for j in range(count[i]):
if i * l[j] >= n:
break
l.remove(i*l[j])
if not i % l[j]:
break
countP(N)
print(count[1000] - count[99])
#print(143)
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