[Weekly / BOJ_2579_계단오르기] 코드작성 (36ms/31120kb) #33
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minjeeki
reviewed
Sep 25, 2024
| dp[1] = lst[0] # dp[1]값 지정 | ||
| dp[2] = dp[1] + lst[1] # dp[2]값 지정 | ||
| for i in range(3, N + 1): | ||
| dp[i] = max(dp[i-3] + lst[i-1] + lst[i-2], dp[i-2] + lst[i-1]) |
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인덱스 조절의 신 경민이는 lst와 dp의 인덱스를 다르더라도 야무지게 짤 수 있는 능력자야 역시
| for i in range(3, N + 1): | ||
| dp[i] = max(dp[i-3] + lst[i-1] + lst[i-2], dp[i-2] + lst[i-1]) | ||
| # lst의 0번인덱스가 dp의 1번 인덱스와 매칭되므로 | ||
| print(dp[-1]) |
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dp[-1] 인덱스가 어떻게 동작하는거지 한참 생각했네. 역시 파이썬 최고
| if N == 1: # 계단 1개일때 | ||
| print(lst[0]) | ||
| elif N == 2: # 계단 2개 일때 | ||
| print(lst[0]+lst[1]) |
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[화요일] '[계단오르기]' 풀이 요약
문제 푼 날짜 : [2024년09월24일]
시간 : [실행 시간]36ms
메모리 : [메모리]31120KB
코드 길이 : [코드 길이]309B
시도 횟수 : [시도 횟수]2회
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