使用python求10万内的所有素数的个数

#求10万内的所有素数(9592)

print(2)

count = 1

for i in range(3,100000):

   for j in range(2,i):

          if i%j ==0:

                 break

          if j==i-1:

                 print(i)

优化:

count = 1

for i in range(3,100000,2):#跳过所有偶数

for j in range(2,i):

    if i%j ==0:

        break

    if j==i-1:

        count +=1

print(count)

再优化:

count = 1

for i in range(3,100000,2):

for j in range(2,int(i**0.5)+1):#便利到i的开平方

    if i%j ==0:

        break

else:

    count +=1

print(count)

再再优化:

#由于数学成绩有限,方法来自网上

count = 2 #大于等于5的素数一定和6的倍数相邻,所以2、3不在循环内统计,

n = 100000

for num in range(4,n):

if num%6 != 1 and num%6 !=5:

    continue

else:

    snum = int(num**0.5+1)

    for i in range(5,snum):

        if not num%i:

            break

    else:

        count +=1

print(count)

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转载自blog.51cto.com/6300167/2345467