L = [95.5, 85, 59, 66, 72]
L.sort()
i=L[-3:]
print (i)
L.sort()
i=L[-3:]
print (i)
2021-10-28
s1='ABC'
s2='123'
s3='xyz'
for x in s1:
for y in s2:
for z in s3:
print (x+y+z)
s2='123'
s3='xyz'
for x in s1:
for y in s2:
for z in s3:
print (x+y+z)
2021-10-28
s1='\"这是一句中英文混合的Python字符串\":'
s2='Hello World!'
print(s1+s2)
s2='Hello World!'
print(s1+s2)
2021-10-28
#play 1
assign ='Life {0},you {1}.'
result= assign.format('is short','need Python')
print (result)
#play 2
assign ='Life {i},you {n}.'
is_short='is short'
need_Python='need Python'
result= assign.format(i=is_short,n=need_Python)
print (result)
assign ='Life {0},you {1}.'
result= assign.format('is short','need Python')
print (result)
#play 2
assign ='Life {i},you {n}.'
is_short='is short'
need_Python='need Python'
result= assign.format(i=is_short,n=need_Python)
print (result)
2021-10-28
# Enter a code
L1=[1,2,3]
L2=[5,3,2]
L3=[7,3,2]
L=[L1,L2,L3]
s1=(L[0][0]*L[0][1]+L[0][0]*L[0][2]+L[0][1]*L[0][2])*2
s2=(L[1][0]*L[1][1]+L[1][0]*L[1][2]+L[1][1]*L[1][2])*2
s3=(L[2][0]*L[2][1]+L[2][0]*L[2][2]+L[2][1]*L[2][2])*2
print(s1,s2,s3)
L1=[1,2,3]
L2=[5,3,2]
L3=[7,3,2]
L=[L1,L2,L3]
s1=(L[0][0]*L[0][1]+L[0][0]*L[0][2]+L[0][1]*L[0][2])*2
s2=(L[1][0]*L[1][1]+L[1][0]*L[1][2]+L[1][1]*L[1][2])*2
s3=(L[2][0]*L[2][1]+L[2][0]*L[2][2]+L[2][1]*L[2][2])*2
print(s1,s2,s3)
2021-10-27
s ='Life {w},you {e}'
d=s.format(w='is short',e='need python.')
print(d)
d=s.format(w='is short',e='need python.')
print(d)
2021-10-25
最新回答 / 慕运维3303162
在Python的交互式命令行写程序，好处是一下就能得到结果，坏处是没法保存，下次还想运行的时候，还得再敲一遍。所以，实际开发的时候，我们总是使用一个文本编辑器来写代码，写完了，保存为一个文件，这样，程序就可以反复运行了。
2021-10-24
最赞回答 / 戚薇000
L1 = [1,2,3]#changL2 = [5,3,2]#kuanL3 = [7,3,2]#gaoi = 0all_L = [L1,L2,L3]for L in all_L: c = 2*(all_L[0][i]*all_L[1][i]+all_L[1][i]*all_L[2][i]+all_L[0][i]*all_L[2][i]) i = i + 1 print (c)
2021-10-19
最新回答 / 明天就暴富
你这个num的累加，只有在if条件满足的时候才会进行，如果if条件不满足，就没有累加的操作，比如一开始num=0，if语句不满足条件，就直接执行sum=sum+num，输出sum=0，回到while判断，此时num没有增加...
2021-10-19
