전체 글1543 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. 백준 32498 Call for Problems https://www.acmicpc.net/problem/32498 1234if __name__ == '__main__': n = int(input()) nums = [1 for num in range(n) if int(input()) % 2 == 1] print(len(nums))Colored by Color Scriptercs 2024. 11. 27. 백준 32384 사랑은 고려대입니다 https://www.acmicpc.net/problem/32384 12345678910def sol(n): answer = ["LoveisKoreaUniversity" for i in range(n)] print(*answer) if __name__ == '__main__': n = int(input()) sol(n) Colored by Color Scriptercs조건문이나 다른방식으로 출력해도 되는데가장 pythonic 하게 풀려면 list comprehension 과 unpacking 해서 출력하면 깔끔하게 해결가능 2024. 11. 27. leetcode 3330. Find the Original Typed String I https://leetcode.com/problems/find-the-original-typed-string-i/description/ 1234567891011class Solution: def possibleStringCount(self, word: str) -> int: answer = 1 current_alphabet = word[0] for i in range(1, len(word)): if word[i] == word[i - 1]: answer += 1 else: current_alphabet = word.. 2024. 10. 27. 백준 32260 A + B https://www.acmicpc.net/problem/32260 123456#include "aplusb.h" int sum(int A, int B) { return A + B ;} cs 2024. 10. 27. 백준 32154 SUAPC 2024 Winter https://www.acmicpc.net/problem/32154 1234567891011121314151617181920def sol(rank_num): answer = {1: [11, ["A", "B", "C", "D", "E", "F", "G", "H", "J", "L", "M"]] , 2: [9, ["A", "C", "E", "F", "G", "H", "I", "L", "M"]] , 3: [9, ["A", "C", "E", "F", "G", "H", "I", "L", "M"]] , 4: [9, ["A", "B", "C", "E", "F", "G", "H", "L", "M"]] , 5: [8, ["A", "C", "E", "F", "G", ".. 2024. 10. 27. 3300. Minimum Element After Replacement With Digit Sum https://leetcode.com/problems/minimum-element-after-replacement-with-digit-sum/123456class Solution: def minElement(self, nums: List[int]) -> int: answer = [sum(map(int, str(nums[i]))) for i in range(len(nums))] return min(answer)Colored by Color Scriptercslist comprehension과 각 자리수 합을 구하기 위해서원본 데이터 타입을 문자열로 타입 캐스팅해서 사용합니다. 2024. 10. 5. leetcode 3210. Find the Encrypted String https://leetcode.com/problems/find-the-encrypted-string/description/12345678class Solution: def getEncryptedString(self, s: str, k: int) -> str: answer = "" for i in range(len(s)): answer += s[(i+k)%len(s)] return answerColored by Color Scriptercs문자열 s 를 rotate 하는 느낌으로 구현 2024. 9. 23. 3194. Minimum Average of Smallest and Largest Elements https://leetcode.com/problems/minimum-average-of-smallest-and-largest-elements/description/12345678910class Solution: def minimumAverage(self, nums: List[int]) -> float: nums.sort() averages = [] for i in range(len(nums)//2): averages.append((nums[i] + nums[-(i+1)])/2) return min(averages) Colored by Color Scriptercs 2024. 9. 21. 이전 1 2 3 4 ··· 129 다음