[BOJ] 마법사 상어와 토네이도 / 골드 3 / 80분 실패

https://www.acmicpc.net/problem/20057

Author: YoonYn9915

YoonYn9915/implementation/2025-02-16-[백준]-#20057-마법사 상어와 토네이도.py

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+# 모래 계산하는 함수
+def recount(time, dx, dy, direction):
+    global ans, s_x, s_y
+
+    # y좌표 계산 & x좌표 갱신
+    for _ in range(time):
+        s_x += dx
+        s_y += dy
+        if s_y < 0:  # 범위 밖이면 stop
+            break
+
+        # 3. a, out_sand
+        total = 0  # a 구하기 위한 변수
+        for dx, dy, z in direction:
+            nx = s_x + dx
+            ny = s_y + dy
+            if z == 0:  # a(나머지)
+                new_sand = sand[s_x][s_y] - total
+            else:  # 비율
+                new_sand = int(sand[s_x][s_y] * z)
+                total += new_sand
+
+            if 0 <= nx < N and 0 <= ny < N:   # 인덱스 범위이면 값 갱신
+                sand[nx][ny] += new_sand
+            else:  # 범위 밖이면 ans 카운트
+                ans += new_sand
+
+
+N = int(input())
+sand = [list(map(int, input().split())) for _ in range(N)]
+
+# 2. 방향별 모래 비율 위치
+left = [(1, 1, 0.01), (-1, 1, 0.01), (1, 0, 0.07), (-1, 0, 0.07), (1, -1, 0.1),
+         (-1, -1, 0.1), (2, 0, 0.02), (-2, 0, 0.02), (0, -2, 0.05), (0, -1, 0)]
+right = [(x, -y, z) for x, y, z in left]
+down = [(-y, x, z) for x, y, z in left]
+up = [(y, x, z) for x, y, z in left]
+
+s_x, s_y = N//2, N//2  # 시작좌표(x좌표)
+ans = 0  # out_sand
+
+# 1.토네이도 회전 방향(y위치)
+for i in range(1, N + 1):
+    if i % 2:
+        recount(i, 0, -1, left)
+        recount(i, 1, 0, down)
+    else:
+        recount(i, 0, 1, right)
+        recount(i, -1, 0, up)
+
+print(ans)