优化这段代码a,b=eval(input()) n=0 for i in range(a,b+1): n3=i//100 n1=i%10 n2=i//10%10 if i==n1*n1*n1+n2*n2*n2+n3*n3*n3: print(f"{i}") n=n+1 if n==0:print("无水仙花数")
时间: 2023-05-23 18:02:33 浏览: 78
以下是优化后的代码:
a,b=eval(input())
n=0
for i in range(a,b+1):
s = str(i)
res = sum(int(dig)**3 for dig in s)
if res == i:
print(f"{i}")
n += 1
if n == 0:
print("无水仙花数")
这段代码将数字转化成字符串,遍历每个数字,将字符串中的每个数字的三次方加起来,如果和等于原数字,则说明这是水仙花数。
相关问题
a=input() n=0 for m in a: n+=eval(a)**2 print(n)
As an AI language model, I cannot evaluate the code unless the syntax is corrected. The corrected code should be:
a = input()
n = 0
for m in a:
n = eval(a)**2
print(n)
However, this code has an issue with the logic. It will calculate the square of the entire input string 'a' in every iteration of the for loop. Therefore, the final value of 'n' will be the square of the entire input string 'a'.
def fbi(n): if n == 1 or n == 2: result = 1 else: a, b = 1, 1 for i in range(3, n+1): result = a + b a, b = b, result return result n = eval(input()) print(fbi(n))这段代码怎么能让他更加精简
可以使用递归的方式来精简这段代码,如下所示:
```
def fbi(n):
if n <= 2:
return 1
else:
return fbi(n-1) + fbi(n-2)
n = eval(input())
print(fbi(n))
```
这样的代码更加简洁,也更易于理解。