print r'''"To be, or not to be":that is the question.
Whether it's nobler in the mind to suffer.'''
中间不可以加空格,不然报错的
Whether it's nobler in the mind to suffer.'''
中间不可以加空格,不然报错的
2016-04-11
sum = 0
x = 1
while True:
x = x + 2
if x > 100:
break
sum=sum+x
sum=sum+1
print sum
x = 1
while True:
x = x + 2
if x > 100:
break
sum=sum+x
sum=sum+1
print sum
2016-04-11
x1 = 1
d = 3
n = 100
x100 = x1+(n-1)*d
s =n*(x1+x100)/2
print s
d = 3
n = 100
x100 = x1+(n-1)*d
s =n*(x1+x100)/2
print s
2016-04-11
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for key in d:
print key,':',d[key]
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
for key in d:
print key,':',d[key]
2016-04-11
d = {
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam',':',d['Adam']
print 'Lisa',':', d['Lisa']
print 'Bart',':', d['Bart']
'Adam': 95,
'Lisa': 85,
'Bart': 59
}
print 'Adam',':',d['Adam']
print 'Lisa',':', d['Lisa']
print 'Bart',':', d['Bart']
2016-04-11
print [x*100+y*10+z for x in range(1,10) for y in range(0,10) for z in range(1,10) if x==z ]
# 这个才是标准答案,两头的range是1,10 中间的range是0,10
# 这个才是标准答案,两头的range是1,10 中间的range是0,10
2016-04-11
最新回答 / 清波
不管list 中是什么元素, 想要删除的话, 都可以使用 L.pop(index) 来删除, 就比如说题主的例子:L=['我','爱','吃饭']L.pop(1) ## 即可删除
2016-04-11