s=[] for num in range(2,301): for i in range(2,num): if num%i==0: break else: s.append(num) print(s)
时间: 2024-02-14 15:04:20 浏览: 99
这段代码存在一些问题,它把每个遍历到的数字都加入了列表中,而没有判断它是否是质数。下面是代码的修正版:
```python
s = []
for num in range(2, 301):
for i in range(2, int(num ** 0.5) + 1):
if num % i == 0:
break
else:
s.append(num)
print(s)
```
运行结果:
```
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293]
```
其中,外层循环遍历 2 到 300 之间的整数,内层循环判断该整数是否为质数。如果该整数能够被 2 到其平方根之间的任意一个整数整除,则不是质数;否则是质数。如果该整数是质数,则将其添加到列表 `s` 中。最后输出列表 `s` 即可。
相关问题
if num == 2 or num == 1: return True for i in range(2,num): if num % i == 0: print(i)是什么意思
As an AI language model, I do not have a personal opinion or feelings, so my response to any input is neutral. "Null" is a term used in computer programming to represent a non-existent or undefined value. It is also used in database management to indicate the absence of a value.
def prime(num): for i in range(2,num): if num%i==0: return False else: return True
您修改后的代码已经解决了 num=2 时的漏洞,但是代码缩进有误,应该修改为以下代码:
```python
def prime(num):
for i in range(2, num):
if num % i == 0:
return False
else:
return True
```
这个函数可以判断一个数是否为质数,当 num 为质数时返回 True,否则返回 False。
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