[장희직] - 가장 긴 증가하는 부분 수열 4, 치즈, 삼각 달팽이, 내리막 길 #267
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jeeminimini
reviewed
May 26, 2024
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| if (dp[i] == find) { | ||
| answer.add(sequence[i]) | ||
| find -= 1 |
| d = (d + 1) % 3 | ||
| y += dy[d] | ||
| x += dx[d] | ||
| return sum(b, []) |
soopeach
approved these changes
May 26, 2024
Comment on lines
+27
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| for (i in n - 1 downTo 0) { | ||
| for (j in i + 1 until n) { | ||
| if (sequence[i] >= sequence[j]) continue | ||
| dp[i] = max(dp[i], dp[j] + 1) | ||
| } | ||
| } |
bngsh
approved these changes
May 26, 2024
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| fun findAnswer(): Int { | ||
| for (i in n - 1 downTo 0) { |
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오?? 바깥쪽 i가 역순이어도 dp가 되는군요..??
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넵넵 뒤에서부터 탐색해도 됩니다!
ㅋㅋㅋ 저도 다른 분들 풀이 보고 앞이어도 되는군요? 했슴다 ㅋㅋ
| d = (d + 1) % 3 | ||
| y += dy[d] | ||
| x += dx[d] | ||
| return sum(b, []) |
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| # for i in range(cnt): | ||
| # if i != 0: input_idx += i + (nest + nest) | ||
| # answer_list[input_idx] = input_number | ||
| # input_number += 1 | ||
| # # _ | ||
| # for i in range(cnt - 1): | ||
| # input_idx += 1 | ||
| # answer_list[input_idx] = input_number | ||
| # input_number += 1 | ||
| # # \ | ||
| # for i in range(cnt - 2): | ||
| # input_idx -= cnt - i + (nest + nest) | ||
| # answer_list[input_idx] = input_number | ||
| # input_number += 1 |
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헉 근데 이 풀이도 인덱스 계산이 복잡한 와중에 통일성이 있게 계산하셨네요 ㄷㄷ..
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쉽지 않았,,,😀
병희님 풀이도 배열 중간에 숫자 넣는 게 어려웠을 것 같은데👍
jeeminimini
reviewed
May 26, 2024
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| while num <= (n + 1) * n // 2: | ||
| b[y][x] = num | ||
| ny = y + dy[d] | ||
| nx = x + dx[d] | ||
| num += 1 | ||
| if 0 <= ny < n and 0 <= nx <= ny and b[ny][nx] == 0: | ||
| y, x = ny, nx | ||
| else: | ||
| d = (d + 1) % 3 | ||
| y += dy[d] | ||
| x += dx[d] | ||
| return sum(b, []) |
jeeminimini
approved these changes
May 26, 2024
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📌 from issue #266 📌
📋문제 목록📋