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lesson10-2
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Commit
4d3c3dc2
authored
Nov 03, 2024
by
BellCodeEditor
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30 additions
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17 deletions
diy1.py
quiz.py
diy1.py
View file @
4d3c3dc2
# 灭霸打了一个响指,宇宙一半生物都灰飞烟灭。
# 剩下的复仇者联盟成员们依旧没有放弃反击灭霸的机会,他们决定利用最后一次行动机会,去把灭霸手里的宝石偷回来。
# 如果偷回的宝石数是4颗及以上,便获得了打败灭霸的力量;如果偷回的宝石数是1-3颗,他们可以全员出动,殊死一搏;如果偷回的宝石数是0颗,只能尝试呼叫惊奇队长。
# 最终,他们因为实力相差太大,1颗宝石都没有偷回来。
# 悟空读了上面的故事,写出一段代码,在一颗宝石都没偷回来的赋值下,进行条件判断,并产生对应的结果:
# 如果偷回的宝石数是4颗及以上,输出结果:获得了打败灭霸的力量,反杀稳了
# 如果偷回的宝石数是1-3颗,输出结果:可以全员出动,殊死一搏
# 如果偷回的宝石数是0颗,输出结果:没办法了,只能尝试呼叫惊奇队长
# 但是运行错误了,请你帮助悟空消灭bug!
def
new_input
():
total
=
[]
while
True
:
money
=
input
(
'请输入(q退出)'
)
if
money
==
"q"
:
break
else
:
try
:
money
=
int
(
money
)
except
:
pint
(
"请输入一个数字"
)
else
:
total
.
append
(
money
)
finally
:
print
(
"-"
*
30
)
return
total
def
cout
(
lianbion
):
num
=
0
for
i
in
lianbion
:
umn
=
umn
+
i
:
return
unm
result
=
new_input
()
print
(
result
)
if
stonenumber
>=
4
:
print
(
'获得了打败灭霸的力量,反杀稳了'
)
elif
1
<=
stonenum
<=
3
:
print
(
'可以全员出动,殊死一搏'
)
else
:
print
(
'没办法了,只能尝试呼叫惊奇队长'
)
quiz.py
View file @
4d3c3dc2
def
new_input
():
#自定义函数
Total
=
[]
#建立一个空的列表用来保存数据
while
True
:
#无限循环
unif
=
input
(
"请输入菜品价格(输入
Q
为退出)"
):
#获取用户输入,并打印出提示语
if
unif
==
"
Q
"
:
#判断输入的是否为Q,如果是:
unif
=
input
(
"请输入菜品价格(输入
q
为退出)"
):
#获取用户输入,并打印出提示语
if
unif
==
"
q
"
:
#判断输入的是否为Q,如果是:
break
#退出程序
else
:
#否则
Total
.
append
(
unif
)
#将输入的内容添加到列表中
...
...
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