最新回答 / 送東野
根據短路運算原理，True or True進行運算時，在看到第一個True時就可以確定結果是True，不需要關注后面參與運算的是True還是False，所以只輸出第一個True。
2023-07-22
template='Life is {},you need {}'
result=template.format('short','python')
print(result)
template='Life is {s},you need {p}'
short='short'
python='python'
result=template.format(s=short,p=python)
print(result)
result=template.format('short','python')
print(result)
template='Life is {s},you need {p}'
short='short'
python='python'
result=template.format(s=short,p=python)
print(result)
2023-07-18
# Enter a code
def classmate(**kwargs):
for i in range(0,len(kwargs.get('names'))):
print('{} ,{} ,{}'.format(kwargs.get('names')[i],kwargs.get('gender')[i],kwargs.get('age')[i]))
classmate(names = ['Alice', 'Bob', 'Candy'], gender = ['girl', 'boy', 'girl'], age = [16, 17, 15])
def classmate(**kwargs):
for i in range(0,len(kwargs.get('names'))):
print('{} ,{} ,{}'.format(kwargs.get('names')[i],kwargs.get('gender')[i],kwargs.get('age')[i]))
classmate(names = ['Alice', 'Bob', 'Candy'], gender = ['girl', 'boy', 'girl'], age = [16, 17, 15])
2023-07-16
def greet(x=None):
if x != None:
print("Hello,World")
else:
print('hello')
s1=greet(22)
print(s1)
if x != None:
print("Hello,World")
else:
print('hello')
s1=greet(22)
print(s1)
2023-07-15
可以這樣就行了
def fact(n):
if n==1:
return 1
return n + fact(n - 1)
s2 =fact(100)
print(s2)
def fact(n):
if n==1:
return 1
return n + fact(n - 1)
s2 =fact(100)
print(s2)
2023-07-15
>>> print(r'''"To be,or not to be":that is the question.
... Whether it's nobler in the mind to suffer.''')
"To be,or not to be":that is the question.
Whether it's nobler in the mind to suffer.
>>>
... Whether it's nobler in the mind to suffer.''')
"To be,or not to be":that is the question.
Whether it's nobler in the mind to suffer.
>>>
2023-07-14
s='special string:(該重復符號了，用\隔開）\',",(又該重復\符號了）\\（表示\本身），（又和\\重復了）\\\\（兩個本身）,（又重復\）\\n,\\t'
2023-07-14
s1 = set([1, 2, 3, 4, 5])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
for intem in s2:
if intem in s1:
s1.isdisjoint("")
else:
print(intem)
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
for intem in s2:
if intem in s1:
s1.isdisjoint("")
else:
print(intem)
2023-07-14
正解：
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
n = 0
for a in L:
if a in S:
S.remove(a)
L.pop(n)
n += 1
S.update(L)
print(S)
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
n = 0
for a in L:
if a in S:
S.remove(a)
L.pop(n)
n += 1
S.update(L)
print(S)
2023-07-13
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for i in range(len(L)):
if L[i] in S:
S.remove(L[i])
else:
S.add(L[i])
print(S)
S = set([1, 3, 5, 7, 9, 11])
for i in range(len(L)):
if L[i] in S:
S.remove(L[i])
else:
S.add(L[i])
print(S)
2023-07-12
L = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
S = set([1, 3, 5, 7, 9, 11])
for i in range(len(L)):
if L[i] in S:
S.remove(L[i])
print(S)
S = set([1, 3, 5, 7, 9, 11])
for i in range(len(L)):
if L[i] in S:
S.remove(L[i])
print(S)
2023-07-12
已采納回答 / Danny_L
def my_sumA(a):? ? cc = 100? ? s=1? ? ? ? ? ? ?? ? while a < cc:? ? ? ? a += 1? ? ? ? s=s+a? ? return sprint(my_sumA(1))
2023-07-02