有一种方法可以通过key来获取对应的value，这种方法不会引起错误，dict本身提供get方法，把key当作参数传递给get方法，就可以获取对应的value，当key不存在时，也不会报错，而是返回None。
2023-06-11
已采纳回答 / 慕工程9338430
这是个死循环当num为奇数时，跳过了，下面的都不会执行，直接到下次而下一次还是执行奇数，就会这样一直循环，所以运行不出来试试这个num = 1result = 0while num <= 1000: if num % 2 == 1: num += 1 continue result += num num += 1print(result)
2023-06-06
# Enter a code
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
n1=0
for name in L:
if L[n1]=='Candy':
L.pop(n1)
else:
n1+=1
n2=0
for name in L:
if L[n2]=='David':
L.pop(n2)
else:
n2+=1
print(L)
L = ['Alice', 'Bob', 'Candy', 'David', 'Ellena']
n1=0
for name in L:
if L[n1]=='Candy':
L.pop(n1)
else:
n1+=1
n2=0
for name in L:
if L[n2]=='David':
L.pop(n2)
else:
n2+=1
print(L)
2023-06-05
num = 0
sum = 0
while True:
num += 1
if num % 2 != 0:
continue
if num > 1000:
break
sum = sum + num
print(sum)
sum = 0
while True:
num += 1
if num % 2 != 0:
continue
if num > 1000:
break
sum = sum + num
print(sum)
2023-05-24
L = ['Alice', 66, 'Bob', True, 'False', 100]
for n in range(len(L)):
if (n+1) % 2 == 0:
print (L[n])
for n in range(len(L)):
if (n+1) % 2 == 0:
print (L[n])
2023-05-24
num=0
sum=0
while num<=1000:
if num%2==0:
sum=num+sum
num=num+1
print(sum)
sum=0
while num<=1000:
if num%2==0:
sum=num+sum
num=num+1
print(sum)
2023-05-24
# Enter a code
num=1
result=0
while num<1001:
if num%2==1:
num+=1
continue
result+=num
num+=1
print(result)
num=1
result=0
while num<1001:
if num%2==1:
num+=1
continue
result+=num
num+=1
print(result)
2023-05-15
