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[2023-09-15] wooyeol #208 #229
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ksumini
approved these changes
Sep 15, 2023
| # 작은 소수 값이 먼저 나오며 target 값에서 작은 소수 값을 제외한 값이 소수일 경우 -> 골드바흐의 파티션 | ||
| if (p_num <= target - p_num) and (target - p_num) in prime_numbers: | ||
| # 만약 두 소수의 차이가 최소일 경우 정답 업데이트 | ||
| if target - (2 * p_num) < min_gap: |
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아..!
(target - p_num) - p_num = target - (2 * p_num)이라
두 소수의 차이를 이렇게 표현하신 거군요!!
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아 제가 노트에 쓴 걸 옮겨적다보니 축약해서 적고 설명도 안 적었네요 ㅎㅎ 말씀해주신 내용이 맞습니다!
| # 소수 중에서 | ||
| for p_num in prime_numbers.keys(): | ||
| # 작은 소수 값이 먼저 나오며 target 값에서 작은 소수 값을 제외한 값이 소수일 경우 -> 골드바흐의 파티션 | ||
| if (p_num <= target - p_num) and (target - p_num) in prime_numbers: |
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오 전체적으로 주석이 깔끔하고 이 부분을 딕셔너리로 처리할 생각을 정말 좋은 아이디어라고 생각해요!!
시간도 O(1) 밖에 들지 않고..! 좋은 풀이 감사합니다 우열님!👍👍
zsmalla
approved these changes
Sep 15, 2023
| for idx, target in enumerate(targets): | ||
| min_gap = 100001 | ||
| # 소수 중에서 | ||
| for p_num in prime_numbers.keys(): |
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정렬되어있다면, 소수 크기(p_num)가 target을 넘어가면 break하는 로직을 추가해도 좋을 것 같습니다.
limstonestone
approved these changes
Sep 15, 2023
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PR Summary
골드바흐의 추측
풀이시간
20:44 ~ 21:28 (44분)
문제 조건
4 <= N <= 10,000
2 <= target <= 10,000인 짝수
시간 복잡도 :
O(값 입력 + 에라토스테네스의 체 만들기 + 골드바흐의 파티션 찾기)
접근법
무슨 알고리즘으로 풀이 할 수 있을까? -> 구현