[소병희] - 작업, 키 순서, 구간 합 구하기 5, 도넛과 막대 그래프 #218
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jhg3410
approved these changes
Mar 10, 2024
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| fun solve() = with(System.`in`.bufferedReader()) { | ||
| val (n, m) = readLine().split(" ").map { it.toInt() } | ||
| val table = Array(n + 1) { IntArray(n + 1) } | ||
| val sb = StringBuilder() |
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| repeat(n) { i -> | ||
| readLine().split(" ").forEachIndexed { j, v -> | ||
| table[i+1][j+1] = table[i+1][j] + v.toInt() | ||
| } | ||
| } | ||
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| repeat(n) { i -> | ||
| repeat(n) { j -> | ||
| table[i+1][j+1] += table[i][j+1] | ||
| } | ||
| } |
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| for(b in 0 until n) { | ||
| if (adj[a][b] != 0) cnt++ | ||
| } |
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앞뒤를 모두 체크하셔서 이 부분에서 저랑 차이가 있네요!
저는 [a][b] || [b][a] 이렇게 둘 다 확인이 필요했어요!
soopeach
approved these changes
Mar 10, 2024
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| repeat(n) { i -> | ||
| readLine().split(" ").forEachIndexed { j, v -> | ||
| table[i+1][j+1] = table[i+1][j] + v.toInt() |
There was a problem hiding this comment.
내 위치에 이전 위치를 더하는 것이 아니고 다음 위치를 기준으로 연산하면 요렇게 처리가 가능하군요!
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| 2 -> { | ||
| adj[a][b] = 1 | ||
| adj[b][a] = -1 | ||
| } | ||
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| -2 -> { | ||
| adj[a][b] = -1 | ||
| adj[b][a] = 1 | ||
| } |
jeeminimini
approved these changes
Mar 10, 2024
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| repeat(n) { i -> | ||
| readLine().split(" ").forEachIndexed { j, v -> | ||
| table[i+1][j+1] = table[i+1][j] + v.toInt() |
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| 2 -> { | ||
| adj[a][b] = 1 | ||
| adj[b][a] = -1 | ||
| } | ||
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| -2 -> { | ||
| adj[a][b] = -1 | ||
| adj[b][a] = 1 | ||
| } |
bngsh
changed the title
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📌 from issue #214 📌
📋문제 목록📋