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Commit
e81e3671
authored
Sep 11, 2022
by
BellCodeEditor
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diy1.py
diy1.py
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e81e3671
#定义一个函数sum_numbers()
#要求这个函数能接收一个名为num的整型参数
#要求这个函数能计算1+2+3+……+num的结果
def
sum_numbers
(
num
):
#递归的出口
'''
递归可以理解为一个循环,只不过是这个循环的效率低
num是可以随便传入参数的,可以是100也可以是1000
当循环到最后num==1证明这个循环走到头了,当这个值返回以后这个循环也就不执行了
'''
if
num
==
1
:
return
1
#数字的累加
temp
=
sum_numbers
(
num
-
1
)
return
num
+
temp
result
=
sum_numbers
(
100
)
print
()
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