def sushu(n): for i in range(2,n): if n%i==0: return False elif n<=1: return False else: return True ls=[] for i in range(1,21): if sushu(i)==True: ls.append(i) print(ls)
代码存在逻辑错误和效率问题。在函数sushu中,n<=1的情况应该放在最开始判断,否则当n为1时会进入for循环,而for循环里的除法取余操作会导致错误结果。for循环用于检查n是否为素数,如果在循环过程中发现n能被i整除,应立即返回False表示n不是素数。但如果n不能被2到n-1之间的任何数整除,则n是素数,返回True。对于列表生成部分,可以使用列表推导式简化代码。 修正后的代码如下: ```python def sushu(n): if n <= 1: return False for i in range(2, int(n 0.5) + 1): # 只需遍历到sqrt(n) if n % i == 0: return False return True ls = [i for i in range(1, 21) if sushu(i)] print(ls) ``` 三条需要注意的逻辑点: 1. 判断一个数是否为素数时,只需检查到该数平方根即可。 2. 素数定义是从大于1的自然数中去除能被自身和1整除外的其他数整除的数。 3. 使用列表推导式可以使代码更加简洁高效。 [2025-01-09 14:51:20 | AI写代码神器 | 287点数解答]
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