or为或运算,有true记3为true,a='python',a不为空及true,根据短路计算,直接输出hello,world;
2023-03-06
L = [75, 92, 59, 68, 99]
sum=0
count=0
for i in L:
sum=sum + i
count=count+1
print(sum/count)
sum=0
count=0
for i in L:
sum=sum + i
count=count+1
print(sum/count)
2023-03-04
num = 0
sum = 0
while num <= 1000:
num = num + 1
if num % 2 <> 0:
continue
sum = sum + num
print(sum)
sum = 0
while num <= 1000:
num = num + 1
if num % 2 <> 0:
continue
sum = sum + num
print(sum)
2023-03-02
num = 0
sum = 0
while True:
if num > 1000:
break
if num % 2 == 0:
sum = sum + num
num = num + 1
print(sum)
sum = 0
while True:
if num > 1000:
break
if num % 2 == 0:
sum = sum + num
num = num + 1
print(sum)
2023-02-28
L = ['Alice', 66, 'Bob', True, 'False', 100]
for i in range(0, len(L), 2):
print(L[i])
这样也行
for i in range(0, len(L), 2):
print(L[i])
这样也行
2023-02-23
name=['Alice', 'Bob', 'Candy', 'David', 'Ellena']
names = [];
scores=[89,72,88,79,99]
score=[89,72,88,79,99]
scores.sort()
scores.reverse()
x = 0
y = 0
for ls in scores:
y=0;
for l in score:
if ls == l:
names.append(name[y])
y = y+1
x = x+1
print(names)
names = [];
scores=[89,72,88,79,99]
score=[89,72,88,79,99]
scores.sort()
scores.reverse()
x = 0
y = 0
for ls in scores:
y=0;
for l in score:
if ls == l:
names.append(name[y])
y = y+1
x = x+1
print(names)
2023-02-08
最赞回答 / qq_慕函数1463992
s1 = set([1,10]) 2个值也就是 for a in s1: 循环2次了。一次if a in s2: print(a)一次else: print('none')
2023-02-06