# 在if语句中Python等价于&&（逻辑和）

# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
#  a-front + b-front + a-back + b-back
def front_back(a, b):
if len(a) % 2 == 0 && len(b) % 2 == 0:
return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):]
else:
#todo! Not yet done. :P
return

IF条件中出现错误。 我究竟做错了什么？

Python使用andor条件。

if foo == 'abc' and bar == 'bac' or zoo == '123':
# do something

• 在Python中使用andor进行逻辑运算。
• 使用4个空格来缩进，而不是2个。稍后你会感谢你自己，因为你的代码看起来和其他人的代码差不多。 有关更多详细信息，请参阅PEP 8

def front_back(a, b):
return a[:(len(a)+1)//2]+b[:(len(b)+1)//2]+a[(len(a)+1)//2:]+b[(len(b)+1)//2:]

def front_back(a, b):

if len(a) % 2 == 0 and len(b) % 2 == 0:
print a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):]

elif len(a) % 2 == 1 and len(b) % 2 == 0:
print a[:(len(a)/2)+1] + b[:(len(b)/2)] + a[(len(a)/2)+1:] + b[(len(b)/2):]

elif len(a) % 2 == 0 and len(b) % 2 == 1:
print a[:(len(a)/2)] + b[:(len(b)/2)+1] + a[(len(a)/2):] + b[(len(b)/2)+1:]

else :
print a[:(len(a)/2)+1] + b[:(len(b)/2)+1] + a[(len(a)/2)+1:] + b[(len(b)/2)+1:]

### 看看这个有趣的例子，

def AND(a,b):
return (a and b) #using and operator

def OR(a,b):
return (a or b)  #using or operator

print AND(False, False)
print OR(True, False)

### 这将输出：

False
True

x是偶数？ x％2 == 0

x是奇数？ 不是x％2 == 0

x是奇数？ x＆1

x是偶数？ 不是x＆1（不是奇数）

def front_back(a, b):
if not len(a) & 1 and not len(b) & 1:
return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):]
else:
#todo! Not yet done. :P
return

def find_local_py_scripts():
import os # does not cost if already imported
for entry in os.scandir('.'):
# find files ending with .py
if entry.is_file() and entry.name.endswith(".py") :
print("- ", entry.name)
find_local_py_scripts()