我覺得這個網特別好，我們那個年代根本沒有這么好的東西。大家免費學習，要有感恩的心。計算機這東西說復雜還是很復雜的，細節問題和版本有很大關系。
2020-05-25
for x in range(0,10):
for y in range(0,10):
for z in range(0,10):
if z == x:
print(str(x)+str(y)+str(z))
for y in range(0,10):
for z in range(0,10):
if z == x:
print(str(x)+str(y)+str(z))
2020-05-25
最贊回答 / fengyunzhu
我也遇到了這個問題，網上找的答案因為python中print函數需要返回值，如果你在print函數中所放的函數沒有返回值，那么print將會return None
2020-05-24
最新回答 / 慕UI1354896
sum?=?0? x?=?1? n?=?1? while?True: ????sum?=?sum?+?n ????n?=?n?*?2 ????x?=?x?+?1 ????if?x?>?20: ????????break print(sum)這個是隨意的，x當循環數也是可以的。
2020-05-24
s = '''Python was started in 1989 by "Guido".
Python is free and easy to learn.'''
print (s)
Python is free and easy to learn.'''
print (s)
2020-05-23
L = [75, 92, 59, 68]
sum = 0.0
for score in L:
sum+=score
print sum / 4
sum = 0.0
for score in L:
sum+=score
print sum / 4
2020-05-23
L = ['Adam', 'Lisa', 'Bart']
L.pop(2)
L.pop(0)
L.insert(0,'Bart')
L.append('Adam')
print L
復習一下前面幾節....
L.pop(2)
L.pop(0)
L.insert(0,'Bart')
L.append('Adam')
print L
復習一下前面幾節....
2020-05-23
s = set(['Adam', 'Lisa', 'Paul'])
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for l1 in L:
if l1 in s:
s.remove(l1)
else :
s.add(l1)
print s
L = ['Adam', 'Lisa', 'Bart', 'Paul']
for l1 in L:
if l1 in s:
s.remove(l1)
else :
s.add(l1)
print s
2020-05-22
# -*- coding: utf-8 -*-
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
x={}
for a, b in d.items():
a,b=b,a
x[a]=b
print(x)
for 循環寫了個自動的
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
x={}
for a, b in d.items():
a,b=b,a
x[a]=b
print(x)
for 循環寫了個自動的
2020-05-22