已采納回答 / 因為是庫里呀
我給你舉個例子 最后一次?當a= 1000，滿足循環，仍進行一遍語句，最后一句是a=a+2,也就是這時候a= 1000+2=1002當a=1002，不滿足a?<=?1000，這時候你print(a)當然就是1002咯
2020-11-03
# Enter a code
num1 = 3.14
num2 = 1.57
result = num1 * num2 * 2
print(result) # ==>
num1 = 3.14
num2 = 1.57
result = num1 * num2 * 2
print(result) # ==>
2020-11-03
# Enter a code
n=0
s=0
while True:
n+=1
if n>1000:
break
if n%2==1:
continue
s+=n
print s
n=0
s=0
while True:
n+=1
if n>1000:
break
if n%2==1:
continue
s+=n
print s
2020-11-02
n=0
s=0
print n
while True:
if n>1000:
break
elif n%2==0:
s+=n
n+=1
print s
s=0
print n
while True:
if n>1000:
break
elif n%2==0:
s+=n
n+=1
print s
2020-11-02
i = 0
sum = 0
while True:
if i <= 1000:
sum += i
i += 2
else:
break
print(sum)
sum = 0
while True:
if i <= 1000:
sum += i
i += 2
else:
break
print(sum)
2020-10-29
L = [95.5, 85, 59, 66, 72]
L.sort(reverse = False)
print L[-3:]
L.sort(reverse = False)
print L[-3:]
2020-10-28
num = 1
sum = 1
while num <= 10:
sum = sum * num
num = num + 1
print(sum)
sum = 1
while num <= 10:
sum = sum * num
num = num + 1
print(sum)
2020-10-27
L = [1,2,3,4,5,6,7,8,9,10]
sum = 1
for x in L:
sum = sum * x
print(sum)
sum = 1
for x in L:
sum = sum * x
print(sum)
2020-10-27
s1 = set([1, 2, 3, 4, 5])
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
print(s1.isdisjoint(s2))
for item in s1:
if item in s2:
print(item)
s2 = set([1, 2, 3, 4, 5, 6, 7, 8, 9])
print(s1.isdisjoint(s2))
for item in s1:
if item in s2:
print(item)
2020-10-26
最新回答 / my藤上風鈴
# Enter a code# coding:utf-8names = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']scores = [89, 72, 88, 79, 99]templateNames = []templateScores = [89, 72, 88, 79, 99]scores.sort(reverse=True)print("降序排列后的成績：" + str(scores))index_Score = 0index_TemplateScore...
2020-10-25