a = [2]
ferro = [163151, 177259, 145862, 146366, 107166, 137073, 192293, 124514, 155307, 107583, 121753]
i = 3
def eso(x):
for j in a:
if x % j == 0:
return
a.append(x)
while len(a)<193000:
eso(i)
i += 2
n = 7
for h in ferro:
print(a[h-1],' ')
Ferro是n的值的列表。另外,我需要一个快速的代码来返回前200万个质数的列表。您能解释一下这段代码的逻辑吗?
x=int(input("please enter the number: "))
def eso(x):
i = 0
a=[]
for e in range(2,x):
a.append(e)
for j in a:
if x % j == 0:
i=1
if i == 1:
print(x,"is not a prime no.")
else:
print(x,"is a prime no.")
eso(x)
尝试使用此程序检查数字是否为质数