뺄 요소들은 append해서 전체 리스트에서 not in 해서 구하기보다 - 적용하면 됨
또한 set으로 하고 for문 돌리지 말고 sep='\n'하면 된다
백준 4673.self number
#길게 푼 방법
lst= [i for i in range(1,10000)]
new_lst=[]
for i in range(1,10000):
if i + sum(int(j) for j in str(i)) < 10000:
new_lst.append(i + sum(int(j) for j in str(i)))
answer=[]
for k in lst:
if k not in new_lst:
answer.append(k)
for i in answer:
print(i)
#시간 빠른 방법
def d(n):
output = n
for c in str(n):
output += int(c)
return output
dn2n = {}
for i in range(1, 10000):
dn2n[d(i)] = i
if i not in dn2n:
print(i)
#set이용,빼기 적용
def d(n):
return n + sum(map(int, str(n)))
def self_numbers(limit):
numbers = set(range(limit))
not_self_numbers = set()
for i in range(limit):
self_number = d(i)
if self_number < limit:
not_self_numbers.add(self_number)
return sorted(numbers - not_self_numbers)
result = self_numbers(10001)
print(*result, sep = '\n')
