2024/124 백준 32929 UOS 문자열 https://www.acmicpc.net/problem/32929 1234567891011121314def sol(n): n -= 1 if n % 3 == 0: return "U" elif n % 3 == 1: return "O" elif n % 3 == 2: return "S" if __name__ == '__main__': n = int(input()) print(sol(n)) cs넘나 쉬운것.. 2024. 12. 18. 백준 27058 Message Decowding https://www.acmicpc.net/problem/27058 12345678910111213141516171819202122232425def sol(encryption, message ): answer = "" for char in message: is_upper = False if char.isupper() : is_upper = True if char == " ": answer += " " continue decoded_char = encryption[ord(char.lower()) - 97] if is_upper: answer += deco.. 2024. 12. 16. 백준 4775 Spelling Be https://www.acmicpc.net/problem/4775 1234567891011121314151617181920212223242526if __name__ == '__main__': num_word = int(input()) word_dict = {} cnt = 1 #word_dict = [input() for i in range(n)] for i in range(num_word): word_dict[input()] = 1 n = int(input()) for i in range(1, n+1): not_exist = [] while True: s = input() if s == .. 2024. 12. 3. 백준 18698 The Walking Adam https://www.acmicpc.net/problem/18698 1234567891011def sol(steps): first_fall_down = steps.split("D") return len(first_fall_down[0]) if __name__ == '__main__': n = int(input()) for i in range(n): steps = input() print(sol(steps))cs사실 이 문제는 U를 선형탐색하면서 조건에 따라 카운팅 해도 되지만그럼 길어지므로 차라리 D를 기준으로 쪼개는게 더 빠를것 같아요 2024. 12. 3. 이전 1 다음