s1 = 'ABC'
s2 = '123'
s3 = 'xyz'
n = 0 #设置循环次数初始值
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
n =n + 1
print(n) #打印出最终循环的次数
s2 = '123'
s3 = 'xyz'
n = 0 #设置循环次数初始值
for x in s1:
for y in s2:
for z in s3:
print(x+y+z)
n =n + 1
print(n) #打印出最终循环的次数
3天前
# Enter a code
num = 0
sum = 0
while True:
if num > 1000:
break
if num <= 1000:
sum = sum + num
num = num + 2
print(sum)
num = 0
sum = 0
while True:
if num > 1000:
break
if num <= 1000:
sum = sum + num
num = num + 2
print(sum)
num = 0
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum)
sum = 0
while True:
if num > 1000:
break
sum = sum + num
num = num + 2
print(sum)
num = 1
result = 1
while num <= 10:
result = result * num
num = num + 1
print(result)
result = 1
while num <= 10:
result = result * num
num = num + 1
print(result)
4天前
最新回答 / Pusheng_tDHINZ
作为一个计算器，用于控制循环体中代码执行的条件。
L = [75, 92, 59, 68, 99]
sum = 0.0
ab = len(L)
for i in L:
sum = sum + i
print(sum/ab)
sum = 0.0
ab = len(L)
for i in L:
sum = sum + i
print(sum/ab)
5天前
age = 4
if age >= 18:
print("adult")
elif age >= 6:
print("teenager")
elif age >=3:
print("kid")
else:
print("baby")
if age >= 18:
print("adult")
elif age >= 6:
print("teenager")
elif age >=3:
print("kid")
else:
print("baby")
result = 0
num = 0
while True:
if num <= 1000:
result += num
num += 2
else:
break
print(result)
num = 0
while True:
if num <= 1000:
result += num
num += 2
else:
break
print(result)
2025-03-03
d = {'Alice': [50, 61, 66], 'Bob': [80, 61, 66], 'Candy': [88, 75, 90]}
for k,v in d.items():
for v in d[k]:
print(k,v)
for k,v in d.items():
for v in d[k]:
print(k,v)
2025-03-03
print('方式1')
tp = 'Life is {},you need {}'
print (tp.format('short','python'))
print('方式2')
tp1= 'Life is {0},you need {1}'
result1=tp1.format('short','python')
print(result1)
tp = 'Life is {},you need {}'
print (tp.format('short','python'))
print('方式2')
tp1= 'Life is {0},you need {1}'
result1=tp1.format('short','python')
print(result1)
2025-02-28
a = r'''
"to be,or not to be":that is the question
whether it's nobler in hte mind to suffer!'''
print (a)
"to be,or not to be":that is the question
whether it's nobler in hte mind to suffer!'''
print (a)
2025-02-28
L1 = [1, 2, 3]
L2 = [5, 3, 2]
L3 = [7, 3, 2]
def s(L):
A = L[0] * L[1] + L[1] * L[2]+ L[0] * L[2]
return A
print(s(L1), s(L2), s(L3))
L2 = [5, 3, 2]
L3 = [7, 3, 2]
def s(L):
A = L[0] * L[1] + L[1] * L[2]+ L[0] * L[2]
return A
print(s(L1), s(L2), s(L3))
2025-02-28
a = 'python'
print('hello,', a or 'world')
#a为非空字符串，所以为True，或运算，a为True则返回a
b = ''
print('hello,', b or 'world')
#b为空字符串，所以为False，b为False，或运算，有一个为True则结果为True，所以返回后面的word
print('hello,', a or 'world')
#a为非空字符串，所以为True，或运算，a为True则返回a
b = ''
print('hello,', b or 'world')
#b为空字符串，所以为False，b为False，或运算，有一个为True则结果为True，所以返回后面的word
2025-02-27
3.1415926 浮点数
'Learn Python in imooc.' 字符串
100 整数
0b1101 二进制整数
'Learn Python in imooc.' 字符串
100 整数
0b1101 二进制整数
2025-02-27