문제
![](https://media.vlpt.us/images/cosmos/post/a7538bce-1997-4741-81b9-412b2b90a2b7/%E1%84%89%E1%85%B3%E1%84%8F%E1%85%B3%E1%84%85%E1%85%B5%E1%86%AB%E1%84%89%E1%85%A3%E1%86%BA%202022-04-04%20%E1%84%8B%E1%85%A9%E1%84%8C%E1%85%A5%E1%86%AB%2011.48.44.png)
풀이
- 2차원 list를 입력받은것을 전치행렬로 trans 한 후, count가 1인 요소만 더해서 반환하면 구현할수 있다.
코드
import sys
input = sys.stdin.readline
def solve(nums: list) -> str:
trans, filter_l, result = [], [], []
for x in range(3):
l = []
for y in nums:
l.append(y[x])
trans.append(l)
for x in trans:
l = []
for y in x:
if x.count(y) == 1:
l.append(y)
else:
l.append(0)
filter_l.append(l)
result = list(zip(filter_l[0], filter_l[1], filter_l[2]))
return '\n'.join(map(str, [sum(x) for x in result]))
if __name__ == '__main__':
n = int(input())
nums = [list(map(int, input().split())) for _ in range(n)]
print(solve(nums))
결과
![](https://media.vlpt.us/images/cosmos/post/8115e184-85ce-4135-adb8-e0b0cf7fc814/%E1%84%89%E1%85%B3%E1%84%8F%E1%85%B3%E1%84%85%E1%85%B5%E1%86%AB%E1%84%89%E1%85%A3%E1%86%BA%202022-04-04%20%E1%84%8B%E1%85%A9%E1%84%92%E1%85%AE%2012.48.36.png)
출처 & 깃허브
boj 5533
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