getNum(n)。他们试图用一些论据来称呼它。最后他们得到了这本日记：
getNum(300) #-> returns 2
getNum(90783) #-> returns 4
getNum(123321) #-> returns 0
getNum(89282350306) #-> returns 8
getNum(3479283469) #-> returns 5
考古学家完全被这一挑战所困。他们都绝望了，但后来……你是超级棒的程序员。你能理解这个函数的奥秘并重写它吗？
题目难度：简单https://www.codewars.com/kata/55217af7ecb43366f8000f76
def get_num(n):
# your code
return
assert get_num(300) == 2
assert get_num(90783) == 4
assert get_num(123321) == 0
assert get_num(89282350306) == 8
assert get_num(3479283469) == 5
def get_num(n):
sum=0
for i in str(n):
#0,6,9中都包含1个0，sum+1
if i in ('0','6','9'):
sum+=1
#8中包含2个0，sum+2
elif i=='8':
sum+=2
#其他数字中不包含0，sum+0
else:
sum+=0
return sum
assert get_num(300) == 2
assert get_num(90783) == 4
assert get_num(123321) == 0
assert get_num(89282350306) == 8
assert get_num(3479283469) == 5
xiaopeng
2022 年8 月 9 日 05:53
3
def get_num(n):
nums = [1, 0, 0, 0, 0, 0, 1, 0, 2, 1]
total = 0
for i in str(n):
total += nums[int(i)]
return total
2 个赞
joyoko
2022 年8 月 10 日 01:52
4
def get_num(n):
# your code
result = 0
for i in str(n):
if i in ['0', '6', '9']:
result += 1
elif i == '8':
result += 2
return result
assert get_num(300) == 2
assert get_num(90783) == 4
assert get_num(123321) == 0
assert get_num(89282350306) == 8
assert get_num(3479283469) == 5
joyoko
2022 年8 月 10 日 01:56
5
def get_num(n):
return sum({'0': 1, '6': 1, '8': 2, '9': 1}.get(d, 0) for d in str(n))
def get_num(n):
num_list=[1,0,0,0,0,0,1,0,2,1]
s=0
for i in str(n):
s+=num_list[int(i)]
return s
assert get_num(300) == 2
assert get_num(90783) == 4
assert get_num(123321) == 0
assert get_num(89282350306) == 8
assert get_num(3479283469) == 5
在一座古城的废墟中，一群考古学家发现了一个神秘的功能，其中有很多洞，称为