# coding: utf-8
a = '这是一句中英文混合的Python字符串'
b = 'Hello World!'
print(a +': '+ b)
chinese = '这是一句中英文混合的Python字符串'
english = 'Hello World!'
print (chinese + ':' + english)
s = r'''这是一句中英文混合的Python字符串:Hello World!'''
print(s)
a = '这是一句中英文混合的Python字符串'
b = 'Hello World!'
print(a +': '+ b)
chinese = '这是一句中英文混合的Python字符串'
english = 'Hello World!'
print (chinese + ':' + english)
s = r'''这是一句中英文混合的Python字符串:Hello World!'''
print(s)
2023-03-14
a = r'''"To be, or not to be":
that is the question.
Whether it's nobler in the mind to suffer.'''
print(a)
that is the question.
Whether it's nobler in the mind to suffer.'''
print(a)
2023-03-14
def square_of_sum(L):
a = [i**2 for i in L]
result = sum(a)
return result
L = [1,2,3,4]
x = square_of_sum(L)
print(x)
a = [i**2 for i in L]
result = sum(a)
return result
L = [1,2,3,4]
x = square_of_sum(L)
print(x)
2023-03-13
if not isinstance(x, int) or not isinstance(x, float):
这段代码应该使用逻辑运算符 and 而不是 or。这样,只有当 x 既不是整数也不是浮点数时,才会打印错误信息并返回 None
这段代码应该使用逻辑运算符 and 而不是 or。这样,只有当 x 既不是整数也不是浮点数时,才会打印错误信息并返回 None
2023-03-13
# Enter a code
N=['Alice', 'Bob', 'Candy', 'David', 'Ellena']
N[1]=N[3]
N.pop()
N.insert(0,'ELLENA')
print(N)
N=['Alice', 'Bob', 'Candy', 'David', 'Ellena']
N[1]=N[3]
N.pop()
N.insert(0,'ELLENA')
print(N)
2023-03-07
or为或运算,有true记3为true,a='python',a不为空及true,根据短路计算,直接输出hello,world;
2023-03-06
L = [75, 92, 59, 68, 99]
sum=0
count=0
for i in L:
sum=sum + i
count=count+1
print(sum/count)
sum=0
count=0
for i in L:
sum=sum + i
count=count+1
print(sum/count)
2023-03-04
num = 0
sum = 0
while num <= 1000:
num = num + 1
if num % 2 <> 0:
continue
sum = sum + num
print(sum)
sum = 0
while num <= 1000:
num = num + 1
if num % 2 <> 0:
continue
sum = sum + num
print(sum)
2023-03-02
num = 0
sum = 0
while True:
if num > 1000:
break
if num % 2 == 0:
sum = sum + num
num = num + 1
print(sum)
sum = 0
while True:
if num > 1000:
break
if num % 2 == 0:
sum = sum + num
num = num + 1
print(sum)
2023-02-28
L = ['Alice', 66, 'Bob', True, 'False', 100]
for i in range(0, len(L), 2):
print(L[i])
这样也行
for i in range(0, len(L), 2):
print(L[i])
这样也行
2023-02-23