最新回答 / 幸福的棉花糖
在交互式环境中，执行上述代码后，会直接输出 3.14，而不需要显式调用 print()。这是因为 Python 的交互式环境会将表达式的计算结果作为返回值自动显示。然而，在脚本文件（如 .py 文件）中运行相同的代码时，如果没有使用 print()，则不会输出任何内容，因为脚本模式不会自动打印表达式的返回值。因此，在脚本中需要显式使用 print() 来显示结果
2025-04-29
h='{0} {1} {2}, {3} {4} {5}'
g=h.format('life','is','short','you','need','Python')
print(g)
g=h.format('life','is','short','you','need','Python')
print(g)
2025-04-21
h='{a} {b} {c}, {d} {e} {f}'
a1 ='list'
b1 = 'is'
c1 ='short'
d1 = 'you'
e1 = 'need'
f1 = 'Python'
g=h.format(a=a1,b=b1,c=c1,d=d1,e=e1,f=f1)
print(g)
a1 ='list'
b1 = 'is'
c1 ='short'
d1 = 'you'
e1 = 'need'
f1 = 'Python'
g=h.format(a=a1,b=b1,c=c1,d=d1,e=e1,f=f1)
print(g)
2025-04-21
最赞回答 / qq_慕慕1036804
T = ((1 + 2), ((1 + 2),), ('a' + 'b'), (1,), (1, 2, 3, 4, 5))count = 0for element in T: if isinstance(element, tuple):#使用 isinstance() 函数来检查当前的 element 是否为元组类型。isinstance() 函数的第一个参数是要检查的对象，第二个参数是要检查的类型。如果 element 是元组类型，条件判断结果为 True，则执行下一行代码。 cou...
2025-04-11
最新回答 / 慕粉2021077471
注释是中文（#指定顺序，#指定名字）报错。修改成拼音或者在代码前加入：#coding=utf-8。# Enter a code#coding=utf-8#指定顺序
2025-03-24
Alice_score = ['Alice','chinese',92,'math',75,'english',99]
print(Alice_score)
print(Alice_score)
2025-03-18
评论区这个程序可以参考：
T = ((1+2), ((1+2),), ('a'+'b'), (1, ), (1,2,3,4,5))
for a in T:
print(type(a))
结果：
<class 'int'>
<class 'tuple'>
<class 'str'>
<class 'tuple'>
<class 'tuple'>
T = ((1+2), ((1+2),), ('a'+'b'), (1, ), (1,2,3,4,5))
for a in T:
print(type(a))
结果：
<class 'int'>
<class 'tuple'>
<class 'str'>
<class 'tuple'>
<class 'tuple'>
2025-03-18
s1 = 'ABC'
s2 = '123'
s3 = 'xyz'
n = 0 #设置循环次数初始值
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
n =n + 1
print(n) #打印出最终循环的次数
s2 = '123'
s3 = 'xyz'
n = 0 #设置循环次数初始值
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
n =n + 1
print(n) #打印出最终循环的次数
2025-03-12
# Enter a code
num = 0
sum = 0
while True:
if num > 1000:
break
if num <= 1000:
sum = sum + num
num = num + 2
print(sum)
num = 0
sum = 0
while True:
if num > 1000:
break
if num <= 1000:
sum = sum + num
num = num + 2
print(sum)
2025-03-12
num = 0
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum)
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum)
2025-03-11
num = 1
result = 1
while num <= 10:
result = result * num
num = num + 1
print(result)
result = 1
while num <= 10:
result = result * num
num = num + 1
print(result)
2025-03-11
L = [75, 92, 59, 68, 99]
sum = 0.0
ab = len(L)
for i in L:
sum = sum + i
print(sum/ab)
sum = 0.0
ab = len(L)
for i in L:
sum = sum + i
print(sum/ab)
2025-03-10