c = 3.14
k = 1.57
result = c * k
print(round(result,2))
k = 1.57
result = c * k
print(round(result,2))
2020-10-22
t = (0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9)
l = list(t)
print(l)
l = list(t)
print(l)
2020-10-20
# coding=utf-8
def sub_sum(l):
count = 0
s1 = 0
s2 = 0
for i in l:
if count%2==0:
s1+=i
count+=1
else:
s2+=i
count+=1
return s1,s2
l = [1,2,3,4,5]
s1,s2 = sub_sum(l)
print(s1,s2)
def sub_sum(l):
count = 0
s1 = 0
s2 = 0
for i in l:
if count%2==0:
s1+=i
count+=1
else:
s2+=i
count+=1
return s1,s2
l = [1,2,3,4,5]
s1,s2 = sub_sum(l)
print(s1,s2)
2020-10-19
def data_of_square(side):
C= 4*side
S=side * side
print('c is {},s is {}'.format(C,S))
return C,S
C,S= data_of_square(6)
print('周长={}'.format(C))
print('面积={}'.format(S))
C= 4*side
S=side * side
print('c is {},s is {}'.format(C,S))
return C,S
C,S= data_of_square(6)
print('周长={}'.format(C))
print('面积={}'.format(S))
2020-10-19
最赞回答 / 一地几毛
可以是可以，但是多此一举啊language = 'Phtyon'result = template.format (language)可以缩写成一句result = template.format ('Phtyon')你这个是多个变量要套进去才用得着这种形式
2020-10-19
最新回答 / qq_迷失在天堂里云_0
remove函数只能删除首个满足条件的数，不如换种方式
c=[2,5,2,3,3,4,8,'a','b','c','d',1.1] c2=[] for t in c: if isinstance(t,float) or isinstance(t,int): c2.append(t) print(c2)
2020-10-18
最新回答 / 慕无忌7227368
# Enter a coded = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}for key in d.keys(): for score in d[key]: print(key, score)
2020-10-18
# coding=utf-8
d = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}
template = "{}的第{}次成绩是{}分"
for key, value in d.items():
for index in range(len(value)):
print(template.format(key,index+1,value[index]))
d = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}
template = "{}的第{}次成绩是{}分"
for key, value in d.items():
for index in range(len(value)):
print(template.format(key,index+1,value[index]))
2020-10-18
