查找第n个素数算法[关闭]

问题描述 投票:-4回答:1
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?
python-3.x algorithm primes
1个回答
-2
投票
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
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